MTH4101 Calculus II. Carl Murray School of Mathematical Sciences Queen Mary University of London Spring Lecture Notes

Transcription

1 MTH40 Calculus II Carl Murray School of Mathematical Sciences Queen Mary University of London Spring 20 Lecture Notes

2 Complex Numbers. Introduction We have already met several types of numbers. Natural numbers (or positive integers) (N):: If m and n are positive integers then N {0,, 2, 3,... } m + n p and mn q are also positive integers (i.e. the system is closed under addition and multiplication). This allows us to solve equations like 3 + n 7 but not equations like 7 + n 3. Integers (Z): Negative integers (and zero) allow us to solve equations like 7 + n 3. Z {..., 3, 2,, 0,, 2, 3,... } So we can solve equations like m + n p to find the third integer when we know the other two. But what about equations like mn q? Rationals (Q): Rational numbers can be expressed as the ratio of two integers. The set of rationals is denoted by Q. So if we know m and q we can find n q/m (provided m 0). Reals (R): Not all numbers can be expressed as the ratio of two integers. Consider the length of the hypotenuse of the triangle formed by two sides of a square and the diagonal (below). By Pythagoras theorem, h The solution is h 2 and this belongs to the set of real numbers, R but not Z or Q. 2

3 Figure : The diagonal of a square of unit radius Therefore, in Z, we can solve equations like where a Z. In Q, we can solve equations like where a, b Q (and a 0). x + a 0 () ax + b 0 (2) In R, we can solve equations like () and (2) as well as equations which are quadratic, such as ax 2 + bx + c 0 (3) provided a 0 and b 2 4ac 0. The solution of (3) is x b ± b 2 4ac 2a (4) But what do we do if b 2 4ac < 0? For this we need to invent a new set of numbers, C, the set of complex numbers. 3

4 .2 Complex Numbers The complex number, z, is defined in terms of an ordered pair of real numbers, (x, y), and the imaginary number such that i z x + iy where x is the real part and y is the imaginary part. We can write x Re(z) y Im(z). (Note that x and y are both real numbers.) Example: Solve the equation x 2 2x Using the formula for the solution of the quadratic with a, b 2 and c 2 we have x 2 ± ± 4 2 ± (4)( ) Hence the solution is x ± 4 ± i. 2 A consequence of the definition of i is that powers of i may be expressed in terms of ± and i. For example, i 2 i 3 i 2 i ( i)i i i 4 (i 2 ) 2 ( )( ) etc. 4

5 Figure 2: Representing a complex number in the Argand diagram Also i i i i i i i 2 i 2 i 2 i i i 3 i 3 i 2 i ( )i i i etc. Since a complex number is defined in terms of an ordered pair of real numbers (x, y), it follows that every complex number can be represented by one point in a plane. The converse is also true. In fact, there is a - correspondence between the set C and the points in a plane. This gives rise to the Argand diagram (above) where x and y lie along the real and imaginary axes, respectively. The Argand diagram repesents z x + iy both as a point P (x, y) and as a vector OP. From polar coordinates: x r cos θ y r sin θ and so z x + iy r(cos θ + i sin θ). 5

6 Note: The expression cos θ + i sin θ is sometimes written as cis θ. The quantity r (i.e. the length of the line joining O and P in the Argand diagram) is called the absolute value or the modulus of the complex number z. The modulus of z is written as z or mod z. Note that z is always positive and is zero only when z 0. The angle θ (i.e. the angle the line OP makes with the x or real axis) is called the argument of z. It is written as arg z. However, θ is not unique because the angles θ + 2kπ (where k is zero or any integer) are also arguments of the same complex number. We define the principal argument of a complex number to be that value of θ which satisfies π < θ π. From the Argand diagram: z r x 2 + y 2 and sin θ y x2 + y 2 cos θ x x2 + y 2 for all θ. Hence tan θ y x and θ tan y x Example: If z 2 + 3i find z and arg z. z r sin θ 3 3, cos θ 2 3 Hence tan θ sin θ cos θ 3 2 ; arg z θ tan (3/2)

7 y x+ iy y x x x iy Figure 3: Taking the complex conjugate corresponds to a reflection in the x axis in the Argand diagram If z x + iy is a complex number then the complex conjugate, z, is defined to be z x iy. Therefore finding the complex conjugate of a complex number is equivalent to changing the sign of the imaginary component. Since x iy x 2 + ( y) 2 x 2 + y 2, it follows that z z. Using the Argand diagram we can illustrate the effect of taking the complex conjugate of a complex number. It is clear that z is the mirror image of z in the real (i.e. x) axis..3 Operations with Complex Numbers Let z x + iy and z 2 x 2 + iy 2 be two complex numbers. The sum of z and z 2 is defined as z + z 2 (x + iy ) + (x 2 + iy 2 ) (x + x 2 ) + i(y + y 2 ). 7

8 In other words, we add the real parts and we add the imaginary parts. Hence, for any three complex numbers z, z 2 and z 3, we have z + z 2 z 2 + z (commutative) Note also that if z x + iy then z + (z 2 + z 3 ) (z + z 2 ) + z 3 (associative). z + z 2x and z z 2iy. We can use the Argand diagram to obtain a geometric interpretation of the addition of two complex numbers. Consider z 3 z + z 2 (x + x 2 ) + i(y + y 2 ). If we consider the corresponding Argand diagram we see that the resulting complex number, z 3 is obtained by completing the parallelogram with sides OP and OQ. This gives the point S(z 3 ) and the parallelogram OP SQ. There is a similar interpretation for the subtraction of two complex numbers. Let z x + iy and z 2 x 2 + iy 2 be two complex numbers. The product of z and z 2 is defined as z z 2 (x + iy )(x 2 + iy 2 ) (x x 2 y y 2 ) + i(x y 2 + x 2 y ) using i 2 Hence, for any three complex numbers z, z 2 and z 3, we have z z 2 z 2 z (commutative) z (z 2 z 3 ) (z z 2 )z 3 (associative). Also z z (x + iy)(x iy) x 2 + y 2 r 2 z 2 and since z is always positive, z z z Note: The fact that z z z 2, a positive, real quantity means that when a complex number is present in a denominator of a quotient, it can be removed by 8

9 y S( z ) 3 Q( z ) 2 O x x P( z) 2 3 x x Figure 4: Adding two complex numbers in the Argand diagram multiplying top and bottom by the complex conjugate of the denominator. We will make use of this property later on. Similarly, Likewise z n z n z n (z z) n/2 z n z z 2 2 z z 2 z z 2 z z z 2 z 2 z 2 z 2 2 and since all the absolute values (i.e. the moduli) have to be positive, this gives z z 2 z z 2. Note that z z 2 r r 2 (cos θ + i sin θ )(cos θ 2 + i sin θ 2 ) r r 2 (cos θ cos θ 2 sin θ sin θ 2 + i sin θ cos θ 2 + i cos θ sin θ 2 ) r r 2 {cos(θ + θ 2 ) + i sin(θ + θ 2 )} This has a geometrical interpretation in the context of the Argand diagram (see below). The result of multiplying z and z 2 is a complex number with modulus (absolute value) z z 2 r r 2 and an argument arg (z z 2 ) θ + θ 2. 9

10 Figure 5: The multiplication of two complex numbers There is a similar interpretation for the division of two complex numbers. It can easily be shown that z 2 r 2 {cos(θ 2 θ ) + i sin(θ 2 θ )}. z r Therefore the result of dividing z 2 by z is a complex number with modulus z 2 /z r 2 /r and an argument arg (z 2 /z ) θ 2 θ. Example: If z 3 + 5i and z 2 2 3i find the complex numbers (a) z + z 2, (b) z z 2, (c) z z 2 and (d) z /z 2. (a) (b) (c) (d) z + z i + 2 3i + 2i z z i (2 3i) 5 + 8i z z 2 ( 3 + 5i)(2 3i) ( ) + i( ) 9 + 9i z 3 + 5i ( 3 + 5i) (2 + 3i) z 2 2 3i (2 3i) (2 + 3i) 2 + i (using complex conjugate) 3 Note that z , z and z z Hence this confirms our previous result that z z 2 z z 2. 0

11 An important point to note about equations involving complex numbers is that there are really two equations: one where we can equate the real components and one where we can equate the imaginary components. Example: If z 5 + a + 3i and z 2 4 (2 + b)i where a, b R, find a and b when z z 2. If z z 2 then 5 + a + 3i 4 (2 + b)i. Equating the real parts gives 5 + a 4; hence a. Equating the imaginary parts gives 3 (2 + b); hence b 5. Note: If z z 2 then Re(z ) Re(z 2 ) and Im(z ) Im(z 2 )..4 Loci and Regions A locus (the plural is loci ) is a curve or other figure formed by all the points satisfying a particular equation. The use of modules and arguments can define loci or regions in the Argand diagram. Example: Describe the locus of the point defined by z 5. If we use z x + iy then z x 2 + y 2 r 5 since x 2 + y The curve described by the equation x 2 + y is a circle, radius 5 units, centred on the origin.

12 Example: Describe the locus of the point defined by arg z π/4. If we use z x + iy then arg z tan (y/x) π/4. Hence y x tan π and hence y x (x > 0). 4 This is the equation of a line from the origin making an angle π/4 with the x-axis. Example: Describe the locus of the point defined by z. If we use z x + iy then z x + iy. Hence z (x ) 2 + y 2 and hence (x ) 2 + y 2. This is the equation of a circle, radius, centred on x, y 0. Example: Describe the region defined by z. We have already shown (above) that the locus of z is a circle, radius, centred on x, y 0. Hence, the region defined by z has to be the interior of this circle. Example: Describe the locus of the point defined by z 2 z. If we use z x + iy then z 2 4 z 2. Hence x 2 + y 2 4((x ) 2 + y 2 ) 4(x 2 + y 2 + 2x). 2

13 Therefore 3x 2 + 3y 2 8x Dividing through by 3 and adding 6/9 to each side gives x x y which can be written as ( x 4 2 ( ) y 3) 2. 3 This is the equation of a circle, radius 2/3, centred on x 4/3, y 0. Example: Describe the locus of the point defined by arg (z a) π/4. If we use z x + iy then z a x a + iy and arg (z a) π ( ) y 4 tan. x a Hence y x a tan π and hence y x a. 4 This is the equation of a straight line, gradient, crossing the x axis at x a. Example: Describe the region defined by z z. If we use z x + iy then z x 2 + y 2 and z (x ) 2 + y 2. So, for z z, we have z 2 z 2. Hence x 2 + y 2 (x ) 2 + y 2 x 2 2x + + y 2. This gives 2x and so the region is given by x. This is the region to the 2 left of the vertical line through x /2. 3

14 z 5 arg z π/ 4 z z z 2 z z z Figure 6: Diagrams illustrating some of the solutions to the examples above 4

15 .5 Trigonometric Functions and Hyperbolic Functions Later in the course we will see that the power series a r x r a 0 + a x + a 2 x (5) r0 (where the a r are real constants and x is real) converges if x < R, where R is the radius of convergence defined by R lim a r. r If we replace x by a complex number z in (5) the power series will again converge provided z < R. The functions exp x, sin x, cos x can all be represented as power series in x; each has an equivalent series in z. These series in z converge for 0 z <. a r+ e z + z + z2 2! + z3 3! + z4 4! + (6) sin z z z3 3! + z5 5! z7 7! + (7) cos z z2 2! + z4 4! z6 6! +. (8) Now put z iθ in (6), where θ is any real number. We have: e iθ + iθ + i2 θ 2 2! + i3 θ 3 3! + i4 θ 4 4! + + iθ θ2 2! iθ3 + θ4 3! 4! + ) ) ( θ2 2! + θ4 4! + + i (θ θ3 3! + cos θ + i sin θ (using (7) and (8)). This is Euler s relation. But, as we showed above, we can also write z as z r(cos θ + i sin θ). Hence z r e iθ or, more generally, z r e i(θ+2kπ) 5

16 Figure 7: Argand diagrams for e iθ cos θ + i sin θ (a) as a vector and (b) as a point. where k is any integer. Replacing θ by θ in Euler s relation gives e iθ cos θ i sin θ. Hence and cos θ eiθ + e iθ 2 sin θ eiθ e iθ 2i (9) (0) Example: Write z (4 + 3i)e iπ/3 in the form u + iv where u and v are real numbers. From Euler s relation, e iπ/3 cos π 3 + i sin π i

17 Hence ( ) ( 3 z (4 + 3i) 2 + i 2 2 (4 + 3i)( + i 3) 4 3 ) ( 3 + i ) 3. 2 Note that e iπ/2 cos π 2 + i sin π 2 i e iπ cos π + i sin π e 3iπ/2 cos 3π 2 + i sin 3π 2 i e 2iπ cos 2π + i sin 2π In general, for any integer n, e nπi cos nπ + i sin nπ ( ) n e 2nπi cos 2nπ + i sin 2nπ Note: If we write z r e i(θ+2πk) then we can define the natural logarithm of z as ln z ln ( r e i(θ+2πk)) ln r + i(θ + 2πk) ln x 2 + y 2 + i (tan y ) x + 2πk. In (9) and (0) we defined cos and sin in terms of a complex exponential. Here we define two basic hyperbolic functions, cosh and sinh: cosh x ex + e x 2 sinh x ex e x 2 (pronounced cosh ) (pronounced shine ) (These should be compared with our previous expressions for cos θ 2 (eiθ + e iθ ) and sin θ 2i (eiθ e iθ ).) 7

18 We can also define several additional hyperbolic functions: tanh x sinh x cosh x ex e x e x + e x coth x tanh x ex + e x e x e x sech x cosh x 2 e x + e x cosech x sinh x 2 e x e x (pronounced than ) (pronounced coth ) (pronounced shec ) (pronounced coshec ) There are also several hyperbolic identities similar to the trigonometric ones. For example, cosh 2 x sinh 2 x sinh 2x 2 sinh x cosh x cosh 2x cosh 2 x + sinh 2 x..6 De Moivre s Theorem Consider the set of n complex numbers, z r (cos θ + i sin θ ) z 2 r 2 (cos θ 2 + i sin θ 2 ). z n r n (cos θ n + i sin θ n ) We have already shown that z z 2 r r 2 (cos(θ + θ 2 ) + i sin(θ + θ 2 )). Hence, z z 2 z 3 r r 2 r 3 (cos(θ + θ 2 + θ 3 ) + i sin(θ + θ 2 + θ 3 )). This can be generalised so that z z 2... z n r r 2... r n (cos(θ + θ θ n ) + i sin(θ + θ θ n )). Now set r r 2 r n and θ θ 2 θ n θ. This gives z n (cos θ + i sin θ) n cos nθ + i sin nθ 8

19 where n is any positive integer. This results is known as de Moivre s theorem. The result is also valid for n < 0 and n p/q (i.e. where n is a negative integer and where n is a rational number). For example, (cos θ + i sin θ) m (cos θ + i sin θ) m cos mθ + i sin mθ (cos mθ + i sin mθ) cos mθ i sin mθ which proves the relation. cos( mθ) + i sin( mθ) We now consider several applications of de Moivre s theorem..6. Expansion of cos n θ, sin n θ, cos nθ and sin nθ Let z cos θ + i sin θ. Then Hence z cos θ i sin θ. z + z z z 2 cos θ 2i sin θ. By de Moivre s theorem, and so z n (cos θ + i sin θ) n cos nθ + i sin nθ z n (cos θ + i sin θ) n cos nθ i sin nθ z n + z n 2 cos nθ z n z n 2i sin nθ. 9

20 Example: Find expressions for cos 6 θ and sin 5 θ in terms of cosines and sines of multiple angles, respectively. We have 2 cos θ z + z and so 2 6 cos 6 θ ( z + ) 6 z z 6 + 6z 5 z + 5z4 z z3 z + 3 5z2 z + 6z 4 z + 5 z 6 (z 6 + z ) + 6 (z 4 + z ) + 5 (z 2 + z ) cos 6θ + 2 cos 4θ + 30 cos 2θ Hence cos 6 θ (cos 6θ + 6 cos 4θ + 5 cos 2θ + 0). 32 Similarly, 2 5 i 5 sin 5 θ ( z ) 5 (z 5 z ) 5 (z 3 z ) ( + 0 z ) z 5 3 z and so giving 2 5 i sin 5 θ 2i sin 5θ 0i sin 3θ + 20i sin θ sin 5 θ (sin 5θ 5 sin 3θ + 0 sin θ). 6 The converse problem can also be solved using de Moivre s theorem. Example: Express cos 6θ and sin 6θ in terms of powers of cos θ and sin θ. 20

21 By de Moivre s theorem, cos 6θ + i sin θ (cos θ + i sin θ) 6 cos 6 θ + 6i cos 5 θ sin θ + 5i 2 cos 4 θ sin 2 θ + 20i 3 cos 3 θ sin 3 θ +5i 4 cos 2 θ sin 4 θ + 6i 5 cos θ sin 5 θ + i 6 sin 6 θ cos 6 θ + 6i cos 5 θ sin θ 5 cos 4 θ sin 2 θ 20i cos 3 θ sin 3 θ +5 cos 2 θ sin 4 θ + 6i cos θ sin 5 θ sin 6 θ. Remember that we essentially have two equations, one for the real parts and one for the imaginary parts. This gives Re(cos 6θ + i sin 6θ) cos 6θ cos 6 θ 5 cos 4 θ sin 2 θ + 5 cos 2 θ sin 4 θ sin 6 θ and Im(cos 6θ + i sin 6θ) sin 6θ 6 cos 5 θ sin θ 20 cos 3 θ sin 3 θ + 6 cos θ sin 5 θ. Note that we can simplify these expressions further using the fact that sin 2 θ + cos 2 θ..6.2 Roots of complex numbers and solution of equations We can use de Moivre s theorem to find the complex roots of numbers. Example: Find the n complex roots of the equation z n 0 where n is a positive integer. Remember that we can write unity (the number ) in a number of ways: e 2πki cos 2πk + i sin 2πk where k is any integer (or zero). de Moivre s theorem now gives z /n (cos 2πk + i sin 2πk) /n cos 2πk n 2 2πk + i sin n

22 which, letting k 0,, 2,..., (n ), has n distinct values z, z 2,..., z n. For example, when n 6 the six roots of z 6 0 are the six values of z cos 2πk 6 + i sin 2πk 6 obtained by putting k 0,, 2, 3, 4, 5. Taking each in turn gives (k 0) z cos 0 + i sin 0 (k ) z 2 cos π 3 + i sin π i 3 2 (k 2) z 3 cos 2π 3 + i sin 2π i 3 2 (k 3) z 4 cos π + i sin π (k 4) z 5 cos 4π 3 + i sin 4π 3 2 i 3 2 (k 5) z 6 cos 5π 3 + i sin 5π 3 2 i 3 2. Note that in the Argand diagram these roots are located on a circle of radius at angular separations of π/3. Also note that no new roots are obtained by giving k any other values. (E.g. Setting k 6 just reproduces the rot corresponding to k 0.) Figure 8 shows the location of the cube roots of the general complex number z r e iθ and the fourth roots of z 6. Example: Find the solutions of the equation z 4 i 3. z 4 + i ( ) i 2 { ( π ) ( π )} 2 cos 3 + 2πk + i sin 3 + 2πk 22

23 (a) (b) Figure 8: (a) The three cube roots of z r e iθ and (b) the four fourth roots of z 6. Hence, by de Moivre s theorem, ( π z 2 {cos /4 2 + πk ) 2 ( π + i sin 2 + πk )} 2 where k 0,, 2, 3. The four roots are: ( (k 0) z 2 /4 cos π 2 + i sin π ) 2 ( (k ) z 2 2 /4 cos 7π 7π + i sin 2 2 (k 2) z 3 2 /4 ( cos 3π 2 ) + i sin 3π 2 ( (k 3) z 4 2 /4 cos 9π 9π + i sin 2 2 ) z ) z 2 Note that in the Argand diagram these roots are located on a circle of radius 2 /4 at angular separations of π/2. Note: The principal root is the one closest to the positive x-axis. 23

24 Example: Find all the values of ( + i) i. Let z ( + i) i. Then ( ( ln z i ln( + i) i ln 2 e i(π/4+2kπ)) i ln ( π )) 2 + i 4 + 2kπ i ln ( π ) ( π ) kπ 4 + 2kπ + i ln 2. Hence z e (π/4+2kπ) e i ln 2 e {cos(ln (π/4+2kπ) 2) + i sin(ln } 2)..6.3 Evaluation of integrals Consider the integral e ax cos bx dx. We can write e ax cos bx Re ( e (a+ib)x) Hence ( ) ( ) e ax cos bx dx Re e (a+ib)x dx Re a + ib e(a+ib)x ( Re a + ib a ib ) a ib eax [cos bx + i sin bx] ( ) a ib Re a 2 + b 2 eax [cos bx + i sin bx] a 2 + b 2 eax Re (a cos bx + b sin bx ib cos bx + ia sin bx) e ax (a cos bx + b sin bx) a 2 + b2 24

25 An alternative way to do this question is to use integration by parts (twice!): I e ax cos bx dx eax sin bx sin bx a e ax dx b b eax sin bx a e ax sin bx dx b b eax sin bx a ( ) eax cos bx ( cos bx) a e ax dx b b b b eax sin bx + a b b eax cos bx a2 e ax cos bx dx b b 2 eax sin bx b + a b 2 eax cos bx a2 b 2 I. This gives us an equation in I which we can solve to find I. Gathering the terms in I gives ) I ( + a2 eax sin bx + a b 2 b b 2 eax cos bx and hence I as before. b2 a 2 + b 2 ( e ax sin bx b + a ) b 2 eax cos bx eax (a cos bx + b sin bx) a 2 + b2 2 Partial Derivatives 2. Functions of Several Variables Consider the following functions: V V (r, h) πr 2 h (volume of cylinder, radius r, height h) M M(r, ρ) 4 3 πr3 ρ (mass of sphere, radius r, density ρ) Each is a function of two variables. In the case of V, for example, the quantities r and h are the input variables and V is the output variable (independent variables dependent variable). 25

26 Figure 9: Example domains and ranges for function of two variables Figure 0: The domain of f(x, y) y x 2 consists of the shaded region and its bounding parabola y x 2 If f is a function of two independent variables, x and y, the domain of f is a region in the x-y plane. Example: Describe the domain of the function f(x, y) y x 2. Since f is defined only where y x 2 0, the domain is the closed unbounded region show in Fig. 0. The parabola y x 2 is the boundary of the domain. The points above the parabola make up the domain s interior. 26

27 Figure : Three of the parabolas that go to make up the surface defined by z x 2 + y 2, and the resulting paraboloid. There are two ways to visualise a function f(x, y):. Sketch the surface z f(x, y) in space. 2. Draw and label curves in the domain on which f has a constant value. As an example we will consider the function f(x, y) x 2 + y 2. To visualise the surface, consider the nature of f for a fixed value of y, y a say. In this case z x 2 + a 2 and z z(x). The equation z x 2 + a 2 defines a parabola in the plane y a, perpendicular to the y-axis. Each different value of a gives a different parabola. For example, for y a 0 we have z x 2. Therefore the required surface is made up of parabolas and forms a paraboloid as shown in Fig.. Examples of other surfaces are shown in Figure 2. Alternatively we can consider the curves where f has a constant value. The curves defined by f(x, y) x 2 + y 2 constant are a set of circles. The set of points in 27

28 Figure 2: The three dimensional surfaces defined by the functions (a) f(x, y) x 2 + y 2, (b) f(x, y) x 2 y 2, (c) f(x, y) x 2 + y and (d) f(x, y) y 2 x 2. 28

29 0 y x + y x + y x -5-0 Figure 3: Examples of two curves defined by f(x, y) x 2 + y 2 constant. the plane where a function f(x, y) has a constant value f(x, y) c is called a level curve of f. This is illustrated in Figure 3 for the function f(x, y) x 2 + y 2. Example: Graph the function f(x, y) 00 x 2 y 2 and plot the level curves f(x, y) 0, f(x, y) 5 and f(x, y) 75 in the domain of f in the plane. The domain is the entire x-y plane and the range is the set of real numbers 00. The graph is the paraboloid given by z 00 x 2 y 2 (see Fig. 4) When f(x, y) 0, we have 00 x 2 y 2 0 or x 2 + y This corresponds to a circle of radius 0. When f(x, y) 5, we have 00 x 2 y 2 5 or x 2 + y This corresponds to a circle of radius 7. When f(x, y) 75, we have 00 x 2 y 2 75 or x 2 + y This corresponds to a circle of radius 5. 29

30 Figure 4: The graph and selected level curves of the function f(x, y) 00 x 2 y 2. The curve in space in which the plane z c cuts a surface z f(x, y) is called the contour curve f(x, y) c. For example, Fig. 5 shows the contour curve produced where the plane z 75 intersects the surface z f(x, y) 00 x 2 y Limits and Continuity in Higher Dimensions For functions of one variable we say that f(x) approaches the limit L as x a whenever lim x a f(x) L. Therefore as x approaches a, so f(x) approaches L. The limits are the same as x a from both directions. For functions of one variable we say that f(x) is continuous at x a whenever lim x a f(x) exists, f(a) is defined and the limit L equals f(a). Therefore continuity of the function f(x) at x a lim x a f(x) f(a). 30

31 Figure 5: A plane z c parallel to the x-y plane intersecting a surface z f(x, y) produces a contour curve. 3

32 Figure 6: Definition of the limit of a function of two variables How do we extend the concepts of limits and continuity to functions of two variables? The definition of the limit of a function of two variables is given in Fig. 6. This definition leads to the following properties: If L, M and k are real numbers and then lim f(x, y) L and lim g(x, y) M (x,y) (x 0,y 0 ) (x,y) (x 0,y 0 ). lim (f(x, y) + g(x, y)) L + M (x,y) (x 0,y 0 ) 2. lim (f(x, y) g(x, y)) L M (x,y) (x 0,y 0 ) 3. lim (f(x, y) g(x, y)) L M (x,y) (x 0,y 0 ) 4. lim (kf(x, y)) kl (x,y) (x 0,y 0 ) 5. f(x, y) lim (x,y) (x 0,y 0 ) g(x, y) L M 6. If r and s are integers with no common factors, and s 0, then lim (f(x, (x,y) (x 0,y 0 ) y))r/s L r/s provided L r/s is a real number. For polynomials and rational functions the limit as (x, y) (x 0, y 0 ) can be cal- 32

33 culated by evaluating the function at (x 0, y 0 ) (provided the rational function is defined at (x 0, y 0 )). For example, lim (x,y) (0,) x xy + 3 x 2 y + 5xy y 0 (0)() (0) 2 () + 5(0)() () 3. 3 Example: Find lim (x,y) (0,0) x 2 xy x y. There is a problem with just setting x y 0 because x y 0 as (x, y) (0, 0). However, we can write lim (x,y) (0,0) x 2 xy x y lim (x,y) (0,0) x 2 xy ( x + y) x y x + y x(x y)( x + y) lim (x,y) (0,0) (x y) lim x( x + y) 0. (x,y) (0,0) Now we use limits to define continuity for a function of two variables (see Fig. 7). The Two-Path Test for Nonexistence of a Limit states that if a function f(x, y) has different limits along two different paths as (x, y) (x 0, y 0 ), then does not exist. lim f(x, y) (x,y) (x 0,y 0 ) Figure 8 illustrates this concept for paths approaching a point in radial and tangential directions. 33

34 Figure 7: Definition of continuity for a function of two variables (a) (b) Figure 8: To have continuity at a point we have to have the same limit as the point is approached from all directions, including (a) radial directions and (b) tangential directions. 34

35 Example: Show that the function has no limit as (x, y) (0, 0). f(x, y) 2x2 y x 4 + y 2 We cannot use substitution as it leads to 0/0. However, we can consider what happens as we approach (0, 0) along a family of different curves. Remember, the choice of curves is up to us as the Two-Path Test does not specify what the path should be. Consider the family of parabolas given by y kx 2 (x 0). Along the curves the function is f(x, y) ykx 2 2x2 y x 4 + y 2 ykx 2 2x2 (kx 2 ) x 4 + (kx 2 ) 2 2kx4 x 4 + k 2 x 4 Therefore, as we approach (0, 0) along any curve y kx 2, we have [ ] lim f(x, y) lim f(x, y) ykx 2 2k (x,y) (0,0) (x,y) (0,0) + k. 2 2k + k 2. Therefore the actual limit will depend on which path of approach we take (i.e. which parabola we are on which is determined by the value of k). Therefore, by the Two- Path Test there is no limit as (x, y) (0, 0). Sometimes it is more useful to use polar coordinates. Example: Determine the continuity of the function defined by { 2xy if (x, y) (0, 0) f(x, y) x 2 +y 2 0 if (x, y) (0, 0) In polar coordinates, i.e. x r cos θ, y r sin θ, the function can be written as f 2r2 cos θ sin θ r 2 (cos 2 θ + sin 2 θ) sin 2θ provided we are not at the origin (i.e. provided r 0). Therefore, as r 0, f 0 in all directions. For example, along θ π/4, f sin 2θ sin π/2 everywhere along the line. Therefore the function is not continuous. 35

36 2.3 Partial Derivatives For functions of one variable, y f(x), the derivative at a point is the gradient of the tangent to the curve at that point. For functions of two variables, z f(x, y), an infinite number of tangents exist at a point. However, if we fix y y 0 in f(x, y) and let x vary, then f(x, y 0 ) depends only on x. This leads to the following definition: The partial derivative of f(x, y) with respect to x at the point (x 0, y 0 ) is f f(x 0 + h, y 0 ) f(x 0, y 0 ) x lim f x h 0 (x0,y 0 ) h provided the limit exists. There is a similar definition for the partial derivative with respect to y: The partial derivative of f(x, y) with respect to y at the point (x 0, y 0 ) is f f(x 0, y 0 + h) f(x 0, y 0 + h) y lim f y h 0 (x0,y 0 ) h provided the limit exists. We can extend this to three (or more) dimensions, For example, if f(x, y) x 2 + y 2 then f x 2x, f y 2y. For example, if f(x, y, z) xy 2 z 3 then f x y 2 z 3, f y 2xyz 3, f z 3xy 2 z 2. Note how we treat the other variables as constants when we do partial differentiation. Example: Find f/ x and f/ y at the point (4, 5) for the function f(x, y) x 2 + 3xy + y. 36

37 At the point (4, 5) we have f x x (x2 + 3xy + y ) 2x + 3y f y y (x2 + 3xy + y ) 3x +. f x 7, f y 3. Example: Find z/ x if the equation yz ln z x + y defines z z(x, y). Hence (yz ln z) (x + y). x x y z x z z x + 0. This gives ( y ) z z x ; z x z yz. We can also obtain higher order derivatives. Example: If f(x, y) x cos y + y e x, find f xx 2 f x 2, f yx 2 f y x, f yy 2 f y 2 and f xy 2 f x y. The first step is to find the first partial derivatives: f x f y cos y + y ex x sin y + e x. 37

38 Now we take the partial derivatives of the first partial derivatives. This gives: 2 f x 2 y e x 2 f y x sin y + ex 2 f x y sin y + e x 2 f y 2 x cos y. The Mixed Derivatives Theorem states that if f(x, y) and its partial derivatives f x, f y, f xy and f yx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b) then f xy (a, b) f yx (a, b). The theorem can be extended to higher orders, provided the derivatives are continuous. Example: Find f yxyz if f(x, y, z) 2xy 2 z + x 2 y. Reversing the order we get f y 4xyz + x 2, f yx 4yz + 2x, f yxy 4z, f yxyz 4. The definition of a differential function is given in Figure 9. If f x and f y are continuous throughout an open region R, then f is differentiable at every point of R. If a function f(x, y) is differentiable at a point (x 0, y 0 ) then f is continuous at (x 0, y 0 ). 38

39 Figure 9: The definition of a differentiable function. 2.4 The Chain Rule If w f(x, y) has continuous partial derivatives f x and f y and if x x(t), y y(t) are differentiable functions of t then w f(x(t), y(t)) is a differentiable function of t and dw dt f dx x dt + f dy y dt. This is the Chain Rule. We can easily extend it to functions of three variables: where f f(x, y, z). dw dt f dx x dt + f dy y dt + f dz z dt We can use tree diagrams to illustrate the application of the Chain Rule (see Fig. 20). If w f(x, y) where x g(r, s) and y h(r, s) then w r w x x r + w y y r Also, if w f(x) and x g(r, s) then w r dw x dx r and and w s w x x s + w y y s. w s dw x dx s. 39

40 (a) (b) Figure 20: Tree diagrams are used to remember the Chain Rule. (a) To find dw/dt, start at w and read down each route to t, multiplying derivatives along the way; then add the products. (b) For functions of three variables there are three routes from w to t instead of two, but finding dw/dt is still the same: read down each route, multiplying derivatives along the way; then add. Example: Use the Chain Rule to find the derivative of w xy with respect to t along the path x cos t, y sin t. dw dt w dx x dt + w dy y dt y( sin t) + (cos t)x sin2 t + cos 2 t cos 2t. Note that we could have done this more directly by noting that w xy cos t sin t dw sin 2t ; 2 dt 2 cos 2t cos 2t. 2 Example: Express w/ r and w/ s in terms of r and s if w x + 2y + z 2, x r s, y r2 + ln s, z 2r. We have w r w x x r + w y y r + w z z r ( ) () + (2)(2r) + (2z)(2) s s + 2r 40

41 and w s w x x s + w y y s + w z z s ( ) ( ) r () + (2) + (2z)(0) 2 s 2 s s r s. 2 Suppose that F (x, y) is differentiable and that F (x, y) 0 defines y as a differentiable function of x. Then at any point where F y 0, dy dx F x F y. This is the Formula for Implicit Differentiation. Example: Find dy/dx if y 2 x 2 sin xy 0. F (x, y) y 2 x 2 sin xy dy dx F x ( 2x y cos xy) F y (2y x cos xy) 2x + y cos xy 2y x cos xy. 2.5 Directional Derivatives and Gradient Vectors The derivative of a function at a point in a particular direction is called the directional derivative and is defined in Fig. 2. The directional derivative is also denoted by (D u f) P0. This is the derivative of f at the point P 0 in the direction of the unit vector u. We can develop a more efficient formula for the directional derivative by considering the line x x 0 + su, y y 0 + su 2 through the point P 0 (x, y), parameterised with the arc length parameter s increas- 4

42 Figure 2: The definition of the directional derivative. Figure 22: The definition of the gradient vector. ing in the direction of the unit vector u u i + u 2 j. Then ( ) ( ) ( ) df f dx f ds u,p 0 x P 0 ds + dy (via the Chain Rule) y P 0 ds ( ) ( ) f f u + u 2 x P 0 y P [ 0 ( f ) ( ) ] f i + j [u i + u 2 j] x P 0 y P 0 which is the scalar product of the gradient of f at P 0 with the direction of the vector u. This leads to the definition of the gradient vector shown in Fig. 22. The expression f is called grad f, gradient of f or del f. 42

43 If we write i x + j y we can think of as a vector differential operator operating on a scalar function and returning a vector quantity. We can now write the directional derivative using the gradient: If f(x, y) is differentiable in an open region containing P 0 (x 0, y 0 ) then ( ) df ( f) ds P0 u. u,p 0 This is the scalar product of the grad f at P 0 and u. Example: Find the derivative of f(x, y) x e y + cos(xy) at the point (2, 0) in the direction of v 3i 4j. The unit vector is Now u v v v i 4 5 j. f x (2, 0) (e y y sin(xy)) (2,0) e 0 0 f x (2, 0) (xe y x sin(xy)) (2,0) 2e Hence and so f (2,0) f x (2, 0)i + f y (2, 0)j i + 2j (D u f) (2,0) f (2,0) u (i + 2j) ( 3 5 i 4 ) 5 j Note that: D u f f u f cos θ where θ is the angle between the vectors f and u. This implies the following: 43

44 f increases most rapidly when cos θ (i.e. u is parallel to f) f decreases most rapidly when cos θ (i.e. u is in opposite direction to f) f has zero change when cos θ 0 (i.e. u is orthogonal to f). The gradient has the following properties: (kf) k f for any number k (f + g) f + g (f g) f g (fg) f g + g f ( ) f g f f g g g 2 At every point (x 0, y 0 ) in the domain of a differentiable function f(x, y) the gradient of f is normal to the level curve through (x 0, y 0 ). This is illustrated in Fig. 23. A tangent line is always normal to the gradient. Therefore the equation of the tangent through a point P (x 0, y 0 ) is f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) 0. Example: at the point ( 2, ). Find the equation for the tangent to the ellipse x y2 2 The ellipse is a level curve of the function f(x, y) x2 4 + y2. The gradient of f at ( 2, ) is ( x ) f ( 2,) 2 i + 2yj i + 2j ( 2,) and hence the tangent to the line at this point is ( )(x + 2) + (2)(y ) 0 which simplifies to x 2y 4. This is illustrated in Fig

45 Figure 23: The gradient of a differentiable function of two variables is always normal to the function s level curve through that point. Figure 24: The tangent to the ellipse can be found by treating the ellipse as a level curve of the function f(x, y) (x 2 /4) + y 2. 45

46 Figure 25: The definitions of the tangent plane and the normal line. 2.6 Tangent Planes and Differentials The definitions of the tangent plane and the normal line are shown in Fig. 25. Therefore the equation of the tangent plane is f x (P 0 )(x x 0 ) + f y (P 0 )(y y 0 ) + f z (P 0 )(z z 0 ) 0 and the equation of the normal line is x x 0 + f x (P 0 )t, y y 0 + f y (P 0 )t, z z 0 + f z (P 0 )t. Example: Find the tangent plane and normal line of the surface f(x, y, z) x 2 + y 2 + z 9 0 (a circular paraboloid) at the point P 0 (, 2, 4) (see Fig. 26). f P0 (2x i + 2y j + k) (,2,4) 2 i + 4 j + k where at the point P 0 we have f x (P 0 ) 2, f y (P 0 ) 4 and f z (P 0 ). Therefore the equation of the tangent plane is 2(x ) + 4(y 2) + (z 4) 0 which simplifies to 2x + 4y + z 4. 46

47 Figure 26: The tangent plane and normal line to the surface x 2 + y 2 + z 9 0 at P 0 (, 2, 4). 47

48 The normal line to the surface at P 0 is x + 2t, y 2 + 4t, z 4 + t. Example: The surfaces f(x, y, z) x 2 + y (cylinder) and g(x, y, z) x + z 4 0 (plane) meet in an ellipse E. Find parametric equations for the line tangent to E at the point P 0 (,, 3) (see Fig. 27). The tangent line is orthogonal to both f and g at P 0 and hence is parallel to the vector v f g. Hence Hence f (,,3) (2xi + 2yj) (,,3) 2i + 2j g (,,3) (i + k) (,,3) i + k. v (2i + 2j) (i + k) Therefore the equation of the tangent line is i j k i 2j 2k. x + 2t, y 2t, z 3 2t. The estimated change, df, in the value of a function f when we move a small distance ds from a point P 0 in a direction u is df ( f P0 u ) ds. Example: Estimate how much the value of f(x, y, z) y sin x + 2yz will change if the point P (x, y, z) moves 0. units from P 0 (0,, 0) towards P (2, 2, 2). 48

49 Figure 27: The cylinder f(x, y, z) x 2 + y and the p;lane g(x, y, z) x + z 4 0 intersect in an ellipse. 49

50 Figure 28: The definitions of the linearisation and standard linear approximation of a function f(x, y). First we need to find a vector describing the displacement. We have P 0 P P P 0 (2, 2, 2) (0,, 0) 2i + j 2k. Hence the unit vector in this direction is u P 0 P P 0 P P 0 P i + 3 j 2 3 k. Also Therefore Hence f (0,,0) ((y cos x)i + (sin x + 2z)j + 2yk) (0,,0) i + 2k. f P0 u (i + 2k) df ( f P0 u ) ds ( 2 3 i + 3 j 2 ) 3 k ( 2 ) (0.) units. 3 Figure 28 gives the definitions of the linearisation and standard linear approximation of a function f(x, y). Linearisation is illustrated in Fig

51 Figure 29: If f is differentiable at (x 0, y 0 ), then the value of f at any point (x, y) nearby is approximately f(x 0, y 0 ) + f x (x 0, y 0 ) x + f y (x 0, y 0 ) y. Example: Find the linearisation of f(x, y) x 2 xy + 2 y2 + 3 at the point (3, 2). We first evaluate f, f x and f y at the point (x 0, y 0 ) (3, 2): ( f(3, 2) x 2 xy + ) 2 y giving (3,2) f x (3, 2) ( x 2 xy + ) x 2 y2 + 3 f y (3, 2) ( x 2 xy + ) y 2 y2 + 3 (3,2) (3,2) (2x y) (3,2) 4 ( x + y) (3,2) L(x, y) f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) 8 + (4)(x 3) + ( )(y 2) 4x y 2. Hence the linearisation of f at (3, 2) is L(x, y) 4x y 2. 5

52 If f has continuous first and second partial derivatives throughout an open set containing a rectangle R centred at (x 0, y 0 ) and if M is any upper bound for the values of f xx, f yy and f xy on R, then the error E(x, y) incurred in replacing f(x, y) on R by its linearisation satisfies the inequality E(x, y) 2 M ( x x o + y y 0 ) 2. Example: Find an upper bound for the error in the linear approximation of f(x, y) x 2 xy + 2 y2 + 3 over the rectangle R defined by x 3 0., y 2 0. near (3, 2). This is the same function investigated in the previous example. We have f xx 2 2, f xy, f yy. The largest absolute value of the second partial derivatives is 2 and so we take M 2 in the expression for the error. This gives E(x, y) 2 2 ( x 3 + y 2 )2 ( x 3 + y 2 ) 2. Hence E(x, y) ( ) Figure 30 contains the definition of the total differential of a function f. Example: The volume V πr 2 h of a cylinder is to be calculated from measured values of r (the radius) and h (the height). Suppose that r is measured with an error of no more than 2% and h with an error of no more than 0.5%. Estimate the resulting possible percentage error in the calculation of V. 52

53 Figure 30: The definition of the total differential. and so Hence dv V dv V r dr + V h dh 2πrh dr + πr 2 dh dv V 2πrh dr + πr2 dh πr 2 h 2 dr r + dh h. 2 dr r + dh h 2 dr r + dh h 2(0.02) Therefore the error is no more than 4.5%. 2.7 Extreme Values and Saddle Points When we investigated extreme values for functions of one variable we looked for points where the graph had a horizontal tangent line. For functions of two variables we look for points where the surface defined by z f(x, y) has a horizontal tangent plane. This leads to the definition of local maxima, local minima, critical points and saddle points (see Fig. 3). Local maxima correspond to mountain peaks on the surface z f(x, y) and local minima correspond to valley bottoms (see Fig. 32). From the definition critical points are locations where both f x and f y are zero or where one or both of f x and f y do not exist. Therefore local maxima and minima are critical points but critical points can also include saddle points (see the definition). An example of a saddle point is the origin in the surface shown in Fig. 33. (See also Fig. 2d.). 53

54 Figure 3: The definitions of local maximum, local minimum, critical point and saddle point for a function f(x, y). 54

55 Figure 32: A local maximum corresponds to a mountain peak and a local minimum to a valley low. Therefore, finding critical points of a function is not sufficient to identify the type of critical point (local maximum, local minimum or saddle point). To do this we need to make use of second partial derivatives. Theorem: The Second Derivative Test for Local Extreme Values Suppose that f(x, y) and its first and second partial derivatives are continuous throughout a disk centred at (a, b) and that f x (a, b) f y (a, b) 0. Then:. f has a local maximum at (a, b) if f xx < 0 and f xx f yy f 2 xy > 0 at (a, b) 2. f has a local minimum at (a, b) if f xx > 0 and f xx f yy f 2 xy > 0 at (a, b) 3. f has a saddle point at (a, b) if f xx f yy f 2 xy < 0 at (a, b) 4. The test is inconclusive at (a, b) if f xx f yy f 2 xy 0 at (a, b). In this case we must find some other way to determine the behaviour of f at (a, b). 55

56 Figure 33: The origin is a saddle point of the function f(x, y) y 2 x 2. There are no local extreme values. The quantity f xx f yy fxy 2 f xx f xy f xy f yy is called the discriminant or Hessian of the function f. Example: Find the local extreme values of f(x, y) xy x 2 y 2 2x 2y + 4 and determine the nature of each. At extreme values f x and f y are zero. This gives the two simultaneous equations f x y 2x 2 0 ; f y x 2y 2 0. The solution of these equations is x y 2. Hence ( 2, 2) is the only point where f may take an extreme value. Now take the second derivatives: At the point ( 2, 2), f xx 2 < 0, f yy 2, f xy. f xx f yy f 2 xy ( 2)( 2) 2 3 > 0. 56

57 So f xx < 0 and f xx f yy f 2 xy > 0. Therefore f has a local maximum at ( 2, 2). The value of f at this point is f( 2, 2) 8. Example: Find the local extreme values of f(x, y) xy. f x y 0 and f y x 0 at extreme values. Therefore the origin (0, 0) is a possible extreme value. Taking second derivatives, f xx 0, f yy 0 and f xy. Hence f xx f yy f 2 xy < 0. Therefore f(x, y) has a saddle point at (0, 0). Hence f(x, y) xy has no local extreme values. 2.8 Lagrange Multipliers Suppose that f(x, y, z) and g(x, y, z) are differentiable and g 0 when g(x, y, z) 0. To find the local maximum and minimum values of f subject to the constraint g(x, y, z) 0, we need to find the values of x, y, z and λ that simultaneously satisfy the equations f λ g and g(x, y, z) 0. This is the Method of Lagrange Multipliers. For functions of two variables the condition is similar but without the variable z. We will see how the method works by considering two examples. Example: Find the greatest and smallest values that the function f(x, y) xy takes on the ellipse x y2 2. (see Fig. 34). 57

58 Figure 34: The ellipse defined by x 2 /8 + y 2 /2. We need to find the extreme values of f(x, y) xy subject to the constraint g(x, y) x2 8 + y First, find the values of x, y and λ for which f λ g and g(x, y) 0. f f x i + f y j yi + xj g g x i + g y j x 4 i + yj. Hence yi + xj λ xi + λyj. 4 Comparing components gives Therefore y λ 4 x, x λy. y λ λ2 (λy) 4 4 y. Hence y 0 or λ ±2 and there are two cases to consider. 58

59 Figure 35: When subjected to the constraint g(x, y) x 2 /8 + y 2 /2 0 the function f(x, y) xy takes on extreme values at the four points (±2, ±). These are the points on the ellipse when f (red) is a scalar multiple of g (blue).. If y 0, then x y 0. But (0, 0) does not lie on the ellipse. hence y If y 0, then λ ±2 and x ±2y. Substituting in g(x, y) 0 gives (±2y) y2 2 4y2 + 4y 2 8 y ±. Therefore f(x, y) has its extreme values on the ellipse at the four points (±2, ), (±2, ). The extreme values are xy 2 and xy 2. Example: Find the maximum and minimum values of the function f(x, y) 3x + 4y on the circle x 2 + y 2. f(x, y) 3x + 4y, g(x, y) x 2 + y 2. 59

60 The Lagrange multiplier condition states that f λ g and hence 3i + 4j 2λxi + 2λj x 3 2λ, Therefore x and y have the same sign. The condition g(x, y) 0 gives x 2 + y 2 0 y 2 λ (λ 0). and this gives ( ) λ ( ) λ This gives 9 4λ λ λ2 λ ± 5 2. Hence x 3 2λ ±3 5, y 2 λ ±4 5. Therefore the function f(x, y) 3x+4y has extreme values at (x, y) ±(3/5, 4/5). 3 Multiple Integrals 3. Double Integrals Consider a function f(x, y) defined on a rectangular region R : a x b, c y d (see Fig. 37). The area of a small rectangle with sides x k and y k is A k x k y k. At each point the function has a particular value, e.g. f(x k, y k ). As in Section 2 on Partial Derivatives, we can consider this as defining the height z at the point (i.e. z f(x, y)). 60

61 Figure 36: The function f(x, y) 3x + 4y takes on its largest value on the unit circle g(x, y) x 2 + y 2 0 at the point (3/5, 4/5) and its smallest value at the point ( 3/5, 4/5). At each of these points, f is a scalar multiple of g. The figure shows the gradient at the first point. 6

62 Figure 37: Rectangular grid partitioning the region R into small rectangles of area A k x k y k. Figure 38: Approximating solids with rectangular boxes leads us to define the volumes of more general solids as double integrals. The volume of the solid shown here is the double integral of f(x, y) over the base region R. 62

63 Figure 39: As n increases, the Riemann sum approximations approach the total volume of the solid shown in Fig. 38. The product f(x k, y k ) A k is the volume of a solid with base area A k and height f(x k, y k ) (see Fig. 38). The Riemann sum, S n, of these solids over R is S n n f(x k, y k ) A k. k Now consider what happens as A k 0 (as n ). When the limit of these sums exists the function f is said to be integrable and the limit is called the double integral of f over R written as f(x, y) da or f(x, y) dx dy R Also, the volume of the portion of the solid directly above the base A k f(x k, y k ) A k. Hence the total volume above the region R is Volume lim S n f(x, y) da n where A k 0 as n. Figure 39 shows how the Riemann sum approximations of the volume become more accurate as the number n of boxes increases. Consider the calculation of the volume under the plane z 4 x y over the rectangular region R : 0 x 2 and 0 y in the x-y plane. R R is 63

64 Figure 40: To obtain the cross-sectional area A(x), we hold x fixed and integrate with respect to y. First consider a slice perpendicular to the x-axis (see Fig. 40). The volume under the plane is x2 x0 A(x) dx () where A(x) is the cross-sectional area at x. For each value of x we may calculate A(x) as the integral A(x) y y0 (4 x y) dy (2) which is the area under the curve z 4 x y in the plane of the corss-section at x. In calculating A(x), x is held fixed and the integration takes place with respect to y. 64

65 Combining () and (2) we have Volume x2 x0 x2 ( y x0 x2 x0 [ 7 2 x x2 2 A(x) dx ) (4 x y) dy dx y0 ] y x2 [4y xy y2 dx 2 y0 x0 ] 2 ( ) (0 0) ( ) 7 2 x dx We can write Volume (4 x y) dy dx. This is an iterated or repeated integral. The expression states that we can get the volume under the plane by (i) integrating 4 x y with respect to y from y 0 to y, holding x fixed, and then (ii) integrating the resulting expression in x from x 0 to x 2. In other words, first do the dy integral and then do the dx integral. Now consider the plane perpendicular to the y-axis (see Fig. 4). We have A(y) The volume is then as before. Volume x2 x0 y y0 (4 x y) dx A(y) dy y y0 ] x2 [4x x2 2 xy 6 2y. x0 (6 2y) dy [ 6y y 2] 0 5 This leads to the Fubini s Theorem (First Form) (see Fig. 42). Example: Calculate f(x, y) da for R f(x, y) 6x 2 y where R : 0 x 2, y. 65

66 Figure 4: To obtain the cross-sectional area A(y), we hold y fixed and integrate with respect to x. Figure 42: The statement of the First Form of Fubini s Theorem. 66

67 R f(x, y) da 2 0 ( 6x 2 y) dx dy (2 6y) dy [ 2y 8y 2] [ x 2x 3 y ] x2 x0 dy ( 6) ( 0) 4. Alternatively, we could reverse the order: 2 f(x, y) da ( 6x 2 y) dy dx R as before [ y 3x 2 y 2] y y dx [ ( 3x 2 ) ( 3x 2 ) ] 2 dx 2 dx [2x] Example: Calculate the volume V under z f(x, y) x 2 y with base R where R is the rectangle x 2, 3 y 4. V R x2 x x 2 y da [ x 2 y 2 2 x2 ( y4 x ] y4 y 3 Changing the order gives: V x 2 y da as before. R y4 y 3 [ x 3 y 3 ] x2 x dx x 2 y dy y 3 x2 x y4 ( x2 y 3 dy x y4 y 3 ) dx [ 7x 2 7x 3 2 dx 6 ) x 2 y dx dy 7y 3 dy [ 7y 2 6 ] x2 ] y4 x y In this last example we could have separated the integrand into its x and y parts: x2 ( y4 ) ( x2 ) ( y4 ) V x 2 y dy dx x 2 dx y dy 7 x y 3 x y

68 Figure 43: The statement of the Stronger Form of Fubini s Theorem. More generally, if f(x, y) g(x) h(y), (i.e. the function is separable) and the region is rectangular then xb ( yd ) g(x) h(x) da g(x) h(y) dy dx R xa ( xb xa yc ) ( yd g(x) dx yc ) h(y) dy. Now consider the case where the region R is not rectangular. We can make use of a stronger form of Fubini s Theorem (see Fig. 43). Example: Find the value of (3 x y) da where R is the region defined by R 0 x, x and y x. The region of integration in the x-y plane and the volume defined by z 3 x y is shown in Fig. 44(a). In order to do the double integral we will first consider the approach where we fix the value of x and do the y integral (see Fig. 44(b)). We 68

69 Figure 44: (a) Prism with a triangular base in the xy-plane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may (b) integrate first with respect to y and then with respect to x, or (c) integrate first with respect to x and then with respect to y. 69

70 have (3 x y) da R x yx x0 0 y0 (3x 3x2 2 (3 x y) dy dx ) [ 3x 2 dx 2 x3 2 0 ] 0 ] yx [3y xy y2 2. We can also change the order of the integration where we fix the value of y and do the x integral (see Fig. 44(c)). We have R (3 x y) da y x y0 0 0 (3 x y) dx dy [3x x2 xy 0 2 xy ((3 2 ) )) y (3y y2 2 y2 dy ( 5 2 4y + 3 ) [ ] y 5 2 y2 dy 2 y 2y2 + y3 2 y0 y0 ] x. xy dx dy In some cases the order of integration can be crucial to solving the problem. Example: Calculate (sin x)/x da where R is the triangle in the x-y plane R bounded by the x-axis, the line y x and the line x. The region of integration is shown in Fig. 45. Taking vertical strips (i.e. keeping x fixed and allowing y to vary) gives 0 ( x 0 sin x x ) dy dx 0 [ y sin x ] yx dx x y0 0 sin x dx [ cos x] 0 cos + cos 0 cos. However, if we reverse the order of integration we get 0 y sin x x 70 dx dy

71 Figure 45: The region of integration defined by the x-axis, the line y x and the line x. and (sin x)/x dx cannot be expressed in terms of elementary functions making the integral difficult to do. A key part of the process of double (and multiple) integration over a region is to find the limits of the integration. We can illustrate the procedure by considering the double integral of a function over the region R given by the intersection of the line x + y with the circle x 2 + y 2 (see Fig. 46).. The first step (Fig. 46(a)) is to sketch the region of integration and label its boundary curves. 2. The second step (Fig. 46(b)) is to find the y-limits of integration. Imagine a vertical line through the region, R, and mark the points where it enters and leaves R. In this case such a line would enter at y x and leave at y x The third step (Fig. 46(c)) is to find the x-limits of integration. Choose the x-limits that include all vertical lines through R. In this case lower limit is x 0 and the upper limit is x. 7

72 Figure 46: Steps in finding the limits of integration in double integrals. (a) Sketch. (b) Find the y-limits of integration. (c) Find the x-limits of integration. (d) Reverse the order of integration if necessary. 4. The fourthstep (Fig. 46(d)), which may not be necessary, is to consider reversing the order of integration. If we were to reverse the order of integration, the x-limits would be from x y to x y 2 and then the y-limits would be from y 0 to y. Example: Sketch the region of integration for the integral 2 2x 0 x 2 (4x + 2) dy dx 72

73 Figure 47: An example of how to changing the order of integration changes the limits on the integrals. and write an equivalent integral with the order of integration reversed. Evaluate the integral. As written, the order of integration would imply that we do the y-integral first, from y x 2 to y + 2x, followed by the x-integral from x 0 to x 2. However, we are told to reverse the order of integration. This means we do the x-integration first, from x y/2 to x y, followed by the y-integral from y 0 to y 4 (see Fig. 47). In other words, 2 2x 0 x 2 (4x + 2) dy dx 4 y 0 y/2 (4x + 2) dx dy We can evaluate the integral using either ordering so let us revert to the original: 2 2x 0 x 2 (4x + 2) dy dx [4xy + 2y] 2x x 2 dx 2 0 ( 8x 2 + 4x 4x 3 2x 2) dx ( 4x 3 + 6x 2 + 4x ) dx [ x 4 + 2x 3 + 2x 2]

74 y (a) y (b) y 2 y 2x x y/2 x y 0 y 0 x x y 0 y 0 x x Figure 48: The triangular region bounded by y 0, x and y 2x, with (a) vertical strips for fixed x and varying y, and (b) horizontal strips for fixed y and varying x. Double integrals have the following properties: c f(x, y) da c f(x, y) da for any number c R R (f(x, y) ± g(x, y)) da f(x, y) da ± g(x, y) da R R R f(x, y) da 0 if f(x, y) 0 on R R f(x, y) da g(x, y) da if f(x, y) g(x, y) on R R R f(x, y) da f(x, y) da + f(x, y) da R R if R R R 2 R 2 Example: Find f(x, y) da for the triangular region bounded by y 0, x R and y 2x for f(x, y) x 2 y 2. The region and the two possibilities for the order of integration are shown in Fig

75 (a) y a y + a 2 -x 2 (b) x - a 2 -y 2 y a x + a 2 -y 2 -a 0 a fixed x x -a 0 a x Figure 49: The semi-circular region bounded by x 2 + y 2 a 2 and y 0, with (a) vertical strips for fixed x and varying y, and (b) horizontal strips for fixed y and varying x. First consider the order where we fix x and vary y giving us vertical strips for the first integration (Fig. 48(a)). We have x ( y2x ) x [ ] x f(x, y) da x 2 y 2 2 y 3 y2x dy dx dx R x0 y0 x0 3 y0 x [ ] 8x 5 8x 6 x 3 dx x0 Alternatively, if we fix y and vary x we have horizontal strips for the first integration (Fig. 48(b)). This gives y2 ( x ) y2 [ ] x f(x, y) da x 2 y 2 3 y 2 x dx dy dy R y0 xy/2 y0 3 xy/2 y2 ( ) [ ] y 2 3 y5 y 3 y2 dy 24 9 y y0 x0 y0 Note that this example is not separable because it is a non-rectangular region (i.e. the limits on the x and y integrals now depend on the region of integration). Now consider the region R inside the semi-circle defined by x 2 +y 2 a 2 and y 0. The region and the two possible strips that could be taken are shown in Fig. 49. If we fix x then the vertical strips have y-limits given by 0 y + a 2 x 2 75

76 as in Fig. 49(a). We can then allow x to vary within the limits a x a so that the whole region is covered. Hence ( x+a ) y+ a 2 x 2 f(x, y) da f(x, y) dy dx. R x a y0 Alternatively, if we fix y then the horizontal strips have x-limits given by a 2 y 2 x + a 2 y 2 as in Fig. 49(b). We can then allow y to vary within the limits 0 y a so that the whole region is covered. Hence ( ya ) x+ a 2 y 2 f(x, y) da f(x, y) dx dy. x a 2 y 2 R y0 Note that we will obtain the same answer in each case. In practice we need to use the way that provides the simpler integration. Example: Find the region R for the double integration x2a y 6a 2 x 2 x0 and hence evaluate the integral. y ax 2xy dy dx In the second integral the limits of y are ax y 6a2 x 2. Therefore the lower limit defines a parabola, y 2 ax, while the upper limit defines a circle, x 2 + y 2 6a 2 (the circle s radius is 6a). These two curves intersect at 6a 2 x 2 ax x 2 + ax 6a 2 0 x 2a. We are now in a position to draw the two curves. These are shown in Fig

77 y y ax y 6a2-x2 x 0 x 2a x Figure 50: The region bounded by the curve y x and the curve y 6a 2 x 2, with vertical strips for fixed x and varying y. Therefore the integral can be written as x2a 6a 2 x 2 x0 y ax 2xy dy dx x2a x0 x2a x0 [ xy 2 ] y 6a 2 x 2 y ax dx ( 6a 2 x ax 2 x 3) dx [3a 2 x 2 ax3 3 x4 4 ] x2a x0 6 3 a4. In this case it is actually possible to do the integration using horizontal strips keeping y fixed for the first integration. However, the integration would have to be done in two parts. In the first part the limit for x are 0 x y 2 /a with y going from 0 to 2a. In the second part the limits are 0 x 6a 2 y 2 with y going from 2a to 6a. There are always two ways to do a double integral; choose the simpler because the other may be impossible! Double integrals can also be calculated over unbounded regions. 77

78 Example: Evaluate the integral x e (x+2y) dx dy. 0 0 We have 0 0 x e (x+2y) dx dy e 2y x e x dx dy (integrate by parts with u x, dv e x dx) { [ x e ] 2y e x ( ) } 0 e x dx dy e ( 2y (0 0) + [ e x] ) 0 dy 0 [ ] 2 e 2y ( ) Area The area, A, of a closed, bounded plane region R is given by A da. This is equivalent to calculating f(x, y) da with f(x, y). R R Example: Find the area of the region R bounded by y x and y x 2 in the first quadrant. The diagram of the region of integration is shown in Fig. 5. Taking vertical strips, the area A is given by A x 0 0 dy dx [y] x x 2 x 2 0 [ x (x x 2 2 ) dx dx 2 x3 3 ]

79 Figure 5: The region bounded by the curve y x and the curve y x 2. Example: Find the area of the region R enclosed by the parabola y x 2 and the line y x + 2. The region of integration is shown in Fig. 52. Determining the points of intersection is essential to determining the limits on the integrations. We can find the points by setting x 2 x + 2 which gives x 2 x 2 (x + )(x 2) 0, giving x and x 2. The corresponding values of y are y and y 4. So the points of intersection are (, ) and (2, 4). If we use vertical strips (i.e. fix x and vary y) for the first integral we will not have to split up the region of integration. From the diagram we see that the lower and upper limits for the first integration are therefore y x 2 and y x + 2. This gives A 2 x [y] x+2 x dx 2 dy dx x 2 ( ) [ x + 2 x 2 x 2 dx 2 ] 2 x3 + 2x

80 Figure 52: Calculating the area of the region bounded by the line y x+2 and the curve y x 2 takes (a) two double integrals if the first integration is with respect to x, but (b) only one if the first integration is with respect to y. Double integrals can also be used to find the average value of functions over different regions. The average value of the function f(x, y) over the region R is defined to be f f(x, y) da. area of R R Example: Find the average value of f(x, y) x cos xy over the rectangle R : 0 x π, 0 y. The area of the region R is just π, the product of the length of the two sides of the rectangle. We just need to find f(x, y) da and then divide by π. R π 0 0 x cos xy dy dx Hence f 2/π. π 0 π 0 [sin xy] y y0 dx (sin x 0) dx [ cos x] π

81 3.3 Change of Variables in Double Integrals For functions of one variable it is often useful to integrate by a change of variable, e.g. x x(u). The rule is to replace x by x(u) and dx by (dx/du)du and then alter the x-limits to the u-limits. This is integration by substitution. This gives I xb xa f(x) dx uu2 uu f(x(u)) dx du du where the limits u and u 2 correspond to the limits a and b such that a x(u ) and b x(u 2 ). This procedure is fine if x(u) increases with u. If x(u) is a decreasing function of u the u-limits are then reversed and therefore we have a change of sign: I xb xa f(x) dx uu2 uu f(x(u)) dx du du. But dx/du < 0 in this case, so we can combine both cases in one formula: xb uu2 f(x) dx f(x(u)) dx du du. xa uu Note that on the right-hand side of this equation the function f(x) is expressed as f(x(u)). Also, the right-hand side of the equation includes a magnification factor dx/du, multiplying the du; this comes from transforming from dx to du. For functions of two variables one would similarly expect that the change in variables x x(u, v), y y(u, v) (for example, for polar coordinates u r and v θ) would result in a change in the area by a magnification factor M such that dx dy M du dv. As an example consider a linear change of coordinates: x x(u, v) au + bv, y y(u, v) cu + dv or ( ) x y ( ) ( ) a b u c d v 8 (3)

82 v (u0,v) e 2 (a) (u,v) y (b) (b, d) (a+b, c+d) e 2 P e (u,v0) u e (a, c) x Figure 53: (a) The unit square in the (u, v) coordinate system. (b) The transformed unit square in the (x, y) coordinate system. where a, b, c and d are constants. Now write M for the transformation matrix composed of a, b, c and d and recall that a unit square in (u, v) variables has sides ( ) ( ) ( ) ( ) u u 0 e v 0, e c 2 as shown in Fig. 53(a). Now, to see what happens to this unit square under the transformation M, just apply M. This gives ( ) ( ) ( ) a b a M e e c d 0 c ( ) ( ) ( ) a b 0 b M e 2 e 2 c d d where (a, c) and (b, d) represent the coordinates of the new corners in the (x, y) plane (see Fig. 53(b)). Therefore, under the transformation M we find unit square in (u, v) based on e, e 2 parallelogram in (x, y) based on e, e 2. Note from the matrix and the diagram that the point (, ) in (u, v) transforms to the point (a + b, c + d) in (x, y). 82

83 y b a c R c T 2 d T b a P T 2 c b R b T d c x Figure 54: Individual areas in the transformed unit square. Now consider the area of the parallelogram P (see Fig. 54). We have Area P [Total area of rectangle] [Area of 2 pairs of equal triangles T and T 2 ] [Area of 2 rectangles R]. Therefore, Area P (a + b)(c + d) 2 2 ac 2 bd 2bc ( ) 2 a b ad bc det det M c d Since the unit square (area) gets multiplied by a factor of det M, a small rectangle of sides δu and δv, with area δu δv also gets multiplied by the same factor det M. Hence, replacing u and v in Eq. (3) by δu and δv gives the corresponding changes δx and δy as ( ) ( ) ( ) δx a b δu δy c d δv or, δx a δu + b δv, δy c δu + d δv,. 83

84 Therefore, for a linear change of variables: (Rectangular area du dv in (u, v) plane) ( Parallelogram area, i.e. (det M)δu δv in (x, y) plane). Now let us consider a nonlinear change of coordinates. We take the transformation to have the following form: x x(u, v), y y(u, v) where, neglecting small errors, the increments in x and y are given by δx x x δu + u v δv δy y y δu + u v δv or, in matrix form, ( ) δx δy ( x/ u x/ v y/ u y/ v ) ( δu δv ). The Jacobian matrix is defined to be ( ) ( ) x, y x/ u x/ v M u, v y/ u y/ v and the Jacobian determinant is defined to be ( ) (x, y) x, y (u, v) det M u, v. So, for a nonlinear change of variables: (Rectangular area du dv in (u, v) plane ) (Parallelogram area, i.e. (det M)δu δv in (x, y) plane). This is illustrated in Fig

85 (a) v v+δv v R (b) y R' v+δv v u+δu u u u+δu u x Figure 55: The transformation of the rectangular area R in the (u, v) plane to the area R in the (x, y) plane. Therefore, the required formula for double integrals under a change of variables is: f(x, y) dx dy f(x(u, v), y(u, v)) (x, y) R (u, v) du dv R where (x, y) (u, v) det M can be thought of as the magnification factor. Note that the factor (x, y) (u, v) is det M, the absolute value of the determinant of the matrix M. Note that we take the modulus as in the one variable case. Note that is the analogue of dx dy (x, y) (u, v) du dv dx dx du du. This should help when trying to remember the formula, (x, y) (u, v) x/ u x/ v y/ u y/ v. 85

86 In matrix notation, vertical lines on either side of a matrix denote the determinant. However, vertical lines on either side of an expression also denote the absolute value. For example, if we let ( ) a b A c d then and det A a c b d ad bc det A ad bc. Example: Evaluate the integral I (x 2 + y 2 ) dx dy where R is a circle x 2 + y 2 a 2, by changing to polar coordinates. In polar coordinates we have x r cos θ, y r sin θ. Therefore, taking u r and v θ, we can write the Jacobian matrix as ( ) ( ) x/ r x/ θ cos θ r sin θ M y/ r y/ θ sin θ r cos θ R and the Jacobian determinant is (x, y) det M (r, θ) cos θ sin θ r sin θ r cos θ r ( cos 2 θ + sin 2 θ ) r which is always positive and so we do not need to take the absolute value. The original area R and the transformed area R are shown in Fig. 56. Note that the circle in the (x, y) plane transforms into a rectangle in the (r, θ) plane. Note that here R is the region given by x 2 + y 2 a 2 and R is the region given by 0 r a, 0 θ 2π. Therefore ( I (x 2 + y 2 ) dx dy ) r 2 (r) dr dθ R R 86

87 (a) y θ 2π (b) a R x 0 R' a r Figure 56: The transformation of the circular area R in the (x, y) plane to the rectangular area R in the (r, θ) plane. where the r 2 on the right-hand integral comes from the transformed x 2 +y 2 and the r dr dθ is from the transformed dx dy with r coming from the Jacobian determinant det M. Hence ra θ2π ( ra ) ( θ2π ) I r 3 dθ dr r 3 dr dθ πa4 θ0 2 r0 θ0 r0 where we note that the integral is separable. Example: Find the area enclosed by the lemniscate r 2 4 cos 2θ. The curve is shown in Fig. 57. The symmetry of the curve about the x- and y-axes means that we only need to calculate the shaded area and multiply it by 4. Note that r 0 when θ π/4, 3π/4, 5π/4, 7π/4 and that r 2 when θ 0, π. The total area is A 4 4 π/4 4 cos 2θ 0 0 π/4 0 [ r 2 2 r dr dθ ] r 4 cos 2θ r0 [4 sin 2θ] π/ dθ 4 since dx dy r dr dθ) π/4 0 2 cos 2θ dθ

88 Figure 57: To integrate over the shaded region, we run r from 0 to 4 cos 2θ and θ from 0 to π/4. Example: Evaluate the double integral 4 xy/2+ 0 xy/2 2x y 2 dx dy by applying the transformation u (2x y)/2, v y/2 and integrating over an appropriate region of the x-y plane. We can write x and y in terms of u and v as x u + v and y 2v. The Jacobian of the transformation is J(u, v) (x, y) (u, v) x/ u x/ v y/ u y/ v (u + v)/ u (u + v)/ v (2v)/ u (2v)/ v Note that: (x, y) (u, v) ( ) (u, v). (4) (x, y) 88

89 Figure 58: The equations x u + v and y 2v transform G into R. Reversing the transformation by the equations u (2x y)/2 and v y/2 transforms R into G. This can be useful in solving some problems. Example: Evaluate the integral I R dx dy (i.e. the area of the region R) where R is enclosed by y 2 x, y 2 2x, xy and xy 2. The region R bounded by the curves is shown in Fig. 59. To solve the integral consider the change of variables defined by u y 2 /x, v xy. Then we can write the four bounding curves as y 2 x u, y 2 2x u 2, xy v, xy 2 v 2. So the region becomes a square (the region R in Fig. 59(b)). Now, for the Jacobian determinant it is easier to use Eq. (4). So, to calculate (x, y)/ (u, v) we first calculate (u, v)/ (x, y) and then take the inverse. Using 89

90 (a) y 2 y 2 2x v v2 (b) R y 2 x R' xy2 v xy x 0 0 u u u2 Figure 59: The transformation of (a) the area R in the (x, y) plane bounded by the curves y 2 x, y 2 2x, xy 2 and xy to (b) the square R in the (u, v) plane bounded by u, u 2, v and v 2. u y 2 /x and v xy we have (u, v) (x, y) u/ x u/ y v/ x v/ y y 2 /x 2 y 2y/x x 3y2 x 3u. Therefore, using Eq. (4), (x, y) (u, v) ( ) (u, v) (x, y) 3u. Hence I 3 3 R dx dy (x, y) R (u, v) du dv 3u du dv u2 v2 dv du 3 u v u [ v v2 du u] R u2 u u2 u v u du [ln u]u2 u 3 ln

91 Example: Evaluate the integral e x2 /2 dx. If we call this integral I, we can write ( ) ( ) I 2 e x2 /2 dx e y2 /2 dy e (x2 +y 2 )/2 dx dy. Now transform to polar coordinates with the limits 0 r < and π θ π. This gives π I 2 e r2 /2 (x, y) π π 0 (r, θ) dr dθ r e r2 /2 dr dθ π 0 π [ ] π π e r2 /2 dθ ((0) ( )) dθ dθ 2π. Hence I 2π. π 0 π π Note that the probability density function for a normal (or Gaussian) distribution is ϕ(x) σ 2π e (x µ)2 /(2σ2 ) for mean µ and standard deviation σ. If we write t (x µ)/σ (i.e. express the displacement from the mean in terms of the standard deviation) then the total probability is P σ e (x µ)2 /(2σ 2) dx 2π σ e t2 /2 σ dt 2π e t2 /2 σ dt. (by our previous result) 2π In fact, more generally, the error function, erf z, is defined as erf z 2 π z 0 e t2 dt. 9

92 3.4 Triple Integrals Triple integrals are integrations where the region of integration is a volume. The basic concepts are similar to those we introduced for two-dimensional (double) integrals, but now we have lim δx,δy,δz 0 ( f(x, y, z) δx δy δz ) where δv δx δy δz are now small volumes (see Fig. 60(a)). The limit as the size of the volume element δv 0 is written as f(x, y, z) dx dy dz f(x, y, z) dv V where V is the three dimensional region being integrated over. The integrals are, as in the two-dimensional case, evaluated by repeated integration where we integrate over one variable at a time. For example, we could start by integrating over z first (see Fig. 60(b)). The procedure is as follows. V Fix (x, y) and integrate over the allowed values of z in the region V. The z-integral limits are the small, filled circles at the bottom and the top of the dashed line with, say, z z (x, y) at the bottom and z z 2 (x, y) at the top as shown in Fig. 60(b). Therefore we are summing vertically over the boxes shown in Fig. 60(b). 2 This result depends on the choice of (x, y) and is defined in the region R of the (x, y) plane which is the projection of V on to this plane as shown in Fig. 60(c). This now defines the region over which we must do the x and y integrations. Now we can take the double integral of the result of the z-integration over the region R in the (x, y) plane (see Fig. 60(d)). Therefore V f(x, y, z) dv xb yy2 (x) zz2 (x,y) xa yy (x) zz (x,y) f(x, y, z) dz dy dx. 92

93 Figure 60: (a) The volume of integration, V, and the volume element δv δx δy δz. (b) The lower (z z ) and upper (z z 2 ) limits on z for the first integration over the volume of integration, V ; this leads to a stack of volume elements from z z to z z 2. (c) The projection of the volume V on to the (x, y) plane defines the region R over which we must do the x and y integrations. (d) The region R in the (x, y) plane over which we must do the x and y integrations. 93

94 z z (a) (b) y y x x Figure 6: The volume formed by the intersection of the planes x 0, y 0, z 0 and x + y + z. (a) The lower and upper bounds on z (at each end of the dashed line) for fixed x and y. (b) Vertical strips from the (x, y) plane to the tetrahedron plane defined by z x y as well as strips in the (x, y) plane (z 0). Consider the evaluation of the integral T f(x, y, z) dv over the tetrahedron T bounded by the planes x 0, y 0, z 0 and x+y+z. Note that the plane x + y + z passes through x (putting y z 0) and similarly through y and z. The relevant planes are shown in Fig. 6. Now evidently for fixed (x, y) the z-limits are the heavy dots corresponding to z 0 at the bottom and z x y at the top. This gives our z-limits. The projection R of T on to the (x, y) plane is the triangle on which the tetrahedron rests, i.e. the triangle given by x 0, y 0 and x + y (obtained by setting z 0). So I x y x z x y x0 y0 z0 For example, if f(x, y, z) then I dv T 94 T f(x, y, z) dz dy dx. dv volume of T.

95 Therefore, in this case I x y x z x y x0 y0 x y x x0 x x0 y0 z0 dz dy dx ( x y) dy dx ( x) 2 dx 2 6 and this is the volume of the tetrahedron. x y x [z] z x y x0 y0 x ] y x x0 [y xy y2 2 z0 dy dx We can picture f dv as a four-dimensional volume, but it is easier to V think of f(x, y, z) as a mass density (i.e. mass per unit volume), so that f dv total mass of V. V y0 dx Example: Find the volume V of the region D enclosed by the surfaces z x 2 + 3y 2 and z 8 x 2 y 2. Where the surfaces intersect x 2 + 3y 2 8 x 2 y 2 and so at the intersection we have x 2 + 2y 2 4, z > 0. This is illustrated in Fig. 62. The boundary of the region R (the projection of D on to the x-y plane) is an ellipse with equation x 2 + 2y 2 4. We now have all the information necessary to 95

96 Figure 62: The region enclosed by the two paraboloids and its projection on to the x-y plane. 96

97 do the integral. We have V D dz dy dx 2 2 (4 x 2 )/2 (4 x 2 )/2 8 x 2 y 2 x 2 +3y 2 dz dy dx (4 x 2 )/2 ( (8 x 2 y 2 ) (x 2 + 3y 2 ) ) dy dx (4 x 2 )/2 (4 x 2 )/2 ( 8 2x 2 4y 2) dy dx (4 x 2 )/2 [(8 2x 2 )y 43 y3 ] (4 x 2 )/2 (4 x 2 )/2 {( (4 (8 2x 2 x2 ) ) 4 ( ) ) 4 x 2 3/ ( (4 (8 2x 2 x2 ) ) 4 ( ) )} 4 x 2 3/2 dx ( (4 2(8 2x 2 x2 ) ) 8 ( ) ) 4 x 2 3/2 dx ( ( ) 4 x 2 3/2 8 8 ( ) ) 4 x 2 3/2 dx π/2 dx ( 4 x 2 ) 3/2 dx [since (8 8/3)/(2 3/2 ) 4 2/3] π/2 π/2 4 3/2 ( cos 2 θ ) 3/2 2 cos θ dθ [using subst. x 2 sin θ] 6 cos 4 θ dθ 4 2 π/2 3 2 [3θ + 2 sin 2θ + 4 ] π/2 sin 4θ ( π π ) 8 2 π. 2 π/2 6 π/2 π/2 (3 + 4 cos 2θ + cos 4θ) dθ 8 Triple integrals can also be used to find the average value of a function over a volume. For example, the function F (x, y, z) defined over a volume D has an 97

98 Figure 63: The cube defined by the planes x 2, y 2 and z 2. average value F (x, y, z) volume of D D F (x, y, z) dv Example: Find the average value of F (x, y, z) xyz over the cube bounded by the planes x 2, y 2 and z 2 in the first octant. The solid of integration is shown in Fig. 63. The volume of the cube is The integral is xyz dx dy dz [ x 2 2 yz ] 2 dy dz 0 2yz dy dz 2 4z dz [ 2z 2] [ y 2 z ] 2 0 dz 98

99 Therefore the average value of F (x, y, z) xyz over the cube is F (x, y, z) xyz dv volume of cube 8 8. cube 3.5 Change of Variables in Triple Integrals Changing variables in triple integrals is similar to the procedure used for double integrals. Suppose x x(u, v, w), y y(u, v, w), z z(u, v, w). We define the Jacobian matrix for change of variables from (x, y, z) to (u, v, w) to be ( ) x/ u x/ v x/ w x, y, z M y/ u y/ v y/ w. u, v, w z/ u z/ v z/ w We can also define the Jacobian determinant (x, y, z) (u, v, w) det M such that the transformation for volume is dx dy dz (x, y, z) (u, v, w) du dv dw. As before, for reversible transformations (x, y, z)/ (u, v, w) 0 and we have (x, y, z) (u, v, w) ( ) (u, v, w). (x, y, z) The integral under the change of variables becomes f(x, y, z) dx dy dz V f(x(u, v, w), y(u, v, w), z(u, v, w)) (x, y, z) V (u, v, w) du dv dw where V is the transformed volume in (u, v, w) coordinates. 99

100 z P θ r O y φ P' x Figure 64: The spherical polar coordinate system (r, θ, φ) for a point P in relation to the cartesian coordinate system (x, y, z). Example: Find an expression for dx dy dz for the coordinate transformation from cartesian coordinates to spherical polar coordinates given by x x(r, θ, φ) r sin θ cos φ y y(r, θ, φ) r sin θ sin φ z z(r, θ, φ) r cos θ. The relationship between the coordinate systems is shown in Fig. 64. Note that OP r sin θ. Let u r, v θ and w φ. Then (x, y, z) (u, v, w) (x, y, z) (r, θ, φ) sin θ cos φ r cos θ cos φ r sin θ sin φ sin θ sin φ r cos θ sin φ r sin θ cos φ cos θ r sin θ 0 r 2 sin θ. 00

101 Therefore dx dy dz r 2 sin θ dr dθ dφ. Example: A volume V in the first octant is bounded by the six surfaces xy, xy 2, yz, yz 2, xz and xz 2. Using the change of variables, r xy, s yz, t xz evaluate the integral V xyz dx dy dz. The new limits are r to r 2, s to s 2 and t to t 2. The Jacobian determinant is (r, s, t) r/ x r/ y r/ z (x, y, z) s/ x s/ y s/ z t/ x t/ y t/ z y x 0 0 z y z 0 x y z y 0 x x 0 y z x y(xz) + x(yz) 2xyz. But and so V xyz dx dy dz ( ) (x, y, z) (r, s, t) (r, s, t) (x, y, z) 2xyz V xyz r2 2xyz dx dy dz 2 [r]2 [s]2 [t] r s2 t2 s t dt ds dr 2 0

102 Figure 65: The definition of an infinite sequence. 4 Infinite Sequences and Series 4. Sequences A sequence is a list of numbers in a given order: a, a 2, a 3,..., a n,.... Each of the a, a 2, etc. represents a number; these are therms of the sequence. For example 2, 4, 6, 8,..., 2n,... has first term a 2, second term a 2 4 and nth term a n 2n. The integer n is called the index of a n and denotes where a n occurs on the list. We can consider the sequence a, a 2, a 3,..., a n,... as a function that sends to a, 2 to a 2, etc. and in general sends the positive integer n to the nth term a n. The definition of an infinite sequence is shown in Fig. 65. Sequences can be described by rules or by listing terms. For example, { } a n n {an }, 2, 3,..., n,... b n ( ) n+ (/n) {b n } {, 2, 3, 4,..., n } ( )n+,... { c n (n )/n {c n } 0, 2, 2 3, 3 4,..., n } n,... d n ( ) n+ {d n } {,,,,..., ( ) n+,... } 02

103 Figure 66: Sequences can be represented as points on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis a n is its value. Sequences can be illustrated graphically either as points on a real axis or as the graph of a function defining the sequence. This is illustrated in Fig. 66. Consider the following sequences: {, 2, 3, 4,..., n,... } {0, 2, 23, 34,..., n,... } {, 2, 3, 4,..., n,... } terms approach 0 as n gets large terms approach as n gets large terms get larger than any number as n increases {,,,,..., ( ) n+,... } terms oscillate between and, never converging to a single value This leads to the definition of convergence, divergence and a limit shown in Fig. 67. The concept of a limit is illustrated in Fig. 68. Here a n L if y L is a horizontal asymptote of the sequence of points {(n, a n )}. 03

104 Figure 67: The definition of convergence, divergence and a limit. Figure 68: a n L if y L is a horizontal asymptote of the sequence of points {(n, a n )}. In this figure, all the a n s after a N lie within ɛ of L. 04

105 We will now consider two examples of the application of the definitions. () We want to prove that: lim n n 0. Let ɛ > 0 be given. We need an integer N such that for all n, n > N, n 0 < ɛ. This condition will be satisfied provided /n < ɛ, which means n > /ɛ. Therefore if N is any integer greater than /ɛ, the implication will hold for all n > N. Hence lim n (/n) 0. For example, suppose we take ɛ 0.0 then the condition is just n > 00. (2) We want to prove that the sequence {,,,,..., ( ) n+,... } diverges. Assume that the sequence converges to some number L. Choose ɛ in the 2 definition of the limit and so all terms a n of the sequence with n larger than some N must lie within ɛ of L. Since is in every term in the sequence, must lie 2 within ɛ of L. Hence L < or 2 2 < L < 3 2. Then is also in every other term and so we must have L ( ) < 2 or 3 2 < L < 2. However, both conditions cannot be satisfied simultaneously. Therefore no such limit exists and so the sequence diverges. The definition of a sequence that diverges to infinity, or diverges to negative infinity is given in Fig. 69. The sum, difference, product, constant multiple and quotient rules for sequences are given in Fig. 70. We can use these rules to help us to calculate limits of sequences. Example: lim n Find lim n (n )/n. ( ) ( n lim n n n ) 05 lim lim n n n 0.

106 Figure 69: The definition of a sequence that diverges to infinity. Figure 70: The sum, difference, product, constant multiple and quotient rules for sequences. 06

107 Figure 7: The Sandwich Theorem for Sequences. Example: Find lim n 5/n 2. lim n 5 n 5 lim 2 n n lim n n The Sandwich Theorem for Sequences (see Fig. 7) provides another method for finding the limits of sequences. Note that if b n c n and c n 0 as n, then b n 0 also, because c n b n c n. Example: Find lim n (sin n)/n. By the properties of the sine function we must have sin n for all n. Therefore n sin n n sin n lim n n n 0 because lim n ( /n) lim n (/n) 0 and the use of the Sandwich Theorem. Example: Find lim n /2 n. /2 n must always lie between 0 and /n (e.g. <, <, <, <,... ) Therefore 0 2 lim n n n 2 0. n 07

108 Figure 72: The Continuous Function Theorem for Sequences. Figure 73: A theorem making the formal connection between lim n a n and lim x f(x). The Continuous Function Theorem for Sequences (see Fig. 72) can be used to determine the limits of sequences. Example: Determine the limit of the sequence { 2 /n} as n. We already know that the sequence { n} converges to 0 as n. Let an /n, f(x) 2 x and L 0 in the continuous function theorem for sequences. This gives 2 /n f(/n) f(l) 2 0 as n. Hence the sequence { 2 /n} converges to. We can also make use of l Hôpital s Rule to find the limits of sequences. To do so we need to make use of the theorem shown in Fig

109 Example: Show that lim n ((ln n)/ n) 0. ln n /n lim lim n n n (/2)n /2 lim n 2 n/2 n (using l Hôpital s Rule) 2 lim n n 0. /2 Example: Does the sequence whose nth term is a n ((n+)/(n )) n converge? If so, find lim n a n. If we just take the straightforward limit we get. Typically with questions of this type we take the logarithm. This gives: ( ) n ( ) n + n + ln a n ln n ln. n n Hence ( ) n + lim ln a n lim n ln lim n n n n lim (n ) (n+) n+ (n ) 2 n /n 2 lim n 2/(n 2 ) /n 2 n ln ( n+ ) n /n (using l Hôpital s Rule) 2n 2 lim n n 2 2. This implies lim n ln a n 2 and since f(x) e x is continuous we have a n e ln an e 2 as n. Therefore the sequence {a n } converges to e 2. Figure 74 gives some common results for the limits of sequences. The first result can be proved using l Hôpital s rule. The second and third results can be proved using logarithms and applying the theorem given in Fig. 73. Proofs of the remaining results are given in Appendix 3 of Thomas. 09

110 Figure 74: Some limits that frequently arise. Example: Show that lim n n n 2. lim n n n2 lim n n 2/n lim n ( n /n ) 2 () 2. Example: Show that lim n n sin(/n). ( ) lim n sin sin(/n) ( /n 2 ) cos(/n) lim lim n n n (/n) n ( /n 2 ) ( ) lim cos. n n (using l Hôpital s Rule) Figure 75 contains the definition of a nondecreasing sequence; this is a sequence where each term is at least as large as its predecessor. Examples of nondecreasing sequences are the sequence of natural numbers, {, 2, 3,..., n,... } and constant sequences such as {3, 3, 3,... }. Figure 76 has the definitions of sequences that are bounded from above, an upper bound, and the least upper bound. 0

111 Figure 75: The definition of a nondecreasing sequence. Figure 76: The definition of bounded from above, upper bound and least upper bound. From these definitions we can state that the sequence of natural numbers given by, {, 2, 3,..., n,... }, has no upper bound and that the sequence {, 2, 3,..., n,... } n+ is bounded from above by M. This leads on to the Nondecreasing Sequence Theorem given in Fig. 77. This implies that a nondecreasing sequence converges when it is bounded from above and it diverges to infinity if it is not bounded from above. Figure 77: The Nondecreasing Sequence Theorem.

112 Figure 78: The definitions of an infinite series, the nth term in a series, partial sums, converges, sum and diverges. 4.2 Infinite series An infinite series is the sum of an infinite sequence of numbers a + a 2 + a a n +. Figure 78 gives the definitions of an infinite series, the nth term in a series, partial sums, converges, sum and diverges. A geometric series has the form a + ar + ar ar n + 2 ar n n ar n n0

113 where a and r are fixed real numbers and a 0. The quantity r is called the ratio of the geometric series and can be positive or negative. In the special case where r the nth partial sum is s n a + a + a a n na and the series diverges because lim n s n ± depending on the sign of a. If r the series diverges because either s n 0 or s n 0 depending on the value of n. Now consider the case of a geometric series with r. We have s n a + ar + ar ar n rs n ar + ar ar n + ar n s n rs n a ar n or s n ( r) a( r n ) s n a( rn ) (r ). r Therefore, if r < then r n 0 as n and hence s n a/( r). If r > then r n and the series diverges. So we have ar n n a r for r < and the geometric series converges. For example, n 9 ( ) n (/9) 3 (/3) 6 (a /9, r /3) and n0 ( ) n 5 4 n 5 + (/4) 4 (a 5, r /4). Example: Find the sum of the series n /((n(n + )). 3

114 Note that we can use partial fractions to write Hence the sum of the first k terms is k n(n + ) n n(n + ) n n +. k n ( n ) n + and so the kth partial sum is ( s k ) ( ) ( ) ( k ) k + ( ) ( ) ( k + ) k k + Hence s k as k and so the series converges giving n(n + ). n Suppose the series n a n converges to a sum S and the nth partial sum of the series is s n a + a a n. When n is large, both s n and s n are close to S and therefore their difference, a n, is close to zero. Using the Difference Rule for sequences we have a n s n s n S S 0 as n. Hence, if n a n converges, then a n 0. This is stated in Fig. 79. Note that the converse of this is false: If a n 0 this does not imply that the series n a n converges. This, in turn, leads to the nth-term Test for Divergence shown in Fig. 80. For example, we can apply the nth Term Test in the following cases: n 2 diverges because n 2 n n n n n + n diverges because n + n ( ) n+ diverges because lim n ( ) n+ n 2n + 5 diverges because 4 lim n does not exist n 2n

115 Figure 79: Property of the general term in a convergent series. Figure 80: Statement of the nth Term Test for Divergence. Consider the unusual case of a series where a n 0 but the series itself diverges. The series is n + 2 n n + where there are two terms of /2, four terms of /4,..., 2 n terms of /2 n, etc. In this case each grouping of terms adds up to so the partial sums must increase without bound and so the series diverges, even though the terms of the series form a sequence that converges to 0. If we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series. This is stated formally in Fig. 8. Example: n Find n (3n )/6 n. 3 n ( 6 n 2 ) n 6 n n (/2) (/6) n 2 n 6 n n (two geometric series)

116 Figure 8: The Sum, Difference and Constant Multiple Rules for convergent series. We can add a finite number of terms or delete a finite number of terms without altering the convergence or divergence of a series but if the series is convergent this will usually alter the sum. Consider the series, a n a + a a k + n a n. If n a n converges, then nk a n converges for any k >. Conversely, if nk a n converges for any k >, then n a n converges. Note that re-indexing a series (e.g. changing the starting value of the index) does not alter its convergence, provided the order of the terms is preserved. nk 4.3 The Integral Test For a given series, a n we want to know: () Does it converge? (2) If it converges, what is its sum? A corollary of the Nondecreasing Sequence Theorem (see Fig. 77) is that the series n a n of non-negative terms converges if and only if its partial sums are bounded from above. Consider the harmonic series: n n +. n 6

117 Figure 82: The sum of the areas of the rectangles under the graph of f(x) /x 2 is less than the area under the graph. This series is actually divergent even though the nth term, /n 0 as n. However, the series has no upper bound for its partial sums. We can see this by writing the series as + ( ) ( ) ( ) +. 6 Now + > 2, > 4, > 8 and so on. Therefore the sum of the 2 n terms ending with /2 n+ is > 2 n /2 n+ /2. Therefore the sequence of partial sums is not bounded from above, and so the harmonic series diverges. Now consider the series, n n n 2 + Does it converge or diverge? To answer this question we will consider a new approach involving the use of integration. What we need to do is compare series n /n2 with the integral x /x2 dx (see Fig. 82). From Fig. 82, 7

118 Figure 83: The Integral Test. s n n 2 f() + f(2) + f(3) + f(n) n < f() + x dx 2 < + x 2 dx Therefore [ s n < + x dx + ] 2. 2 x Thus s n < 2 for all n and the partial sums are bounded from above (by 2) and therefore the series converges. (In fact, the sum of the series is π 2 /6.) Note that the series and the integral need not have the same value in the convergent case. The approach we have just taken leads us to the Integral Test (see Fig. 83). We will consider the proof for the case N and we start with the asumption that f is a decreasing function with f(n) a n for every n. In Fig. 84(a) the areas of the rectangles a, a 2,..., a n enclose more area than that under the curve y f(x) between x and x n +. Therefore we can write n+ f(x) dx a + a a n. Now consider the rectangles as shown in Fig. 84(b). If we ignore the first rectangle we can write a 2 + a a n or, adding the area a to each side gives n a + a 2 + a a n a + 8 f(x) dx n f(x) dx

119 Figure 84: Subject to the conditions of the Integral Test, the series n a n and the integral f(x) dx both converge or both diverge. 9

120 Combining the two inequalities gives n+ f(x) dx a + a a n a + These inequalities will hold as n. n f(x) dx. Therefore, if n f(x) dx is finite, the right-hand part of the inequality shows that an is also finite. Similarly, if n f(x) dx is infinite, then a n is infinite by the left-hand part of the inequality. The Integral Test can be used to show that the p-series, n /np converges if p >, and diverges if p (see Thomas for proof). Example: Show that the series n /(n2 + ) converges by the integral test. The function f(x) /(x 2 + ) is positive, continuous and decreasing for x. Also [ dx lim tan x ] b x 2 + lim [ b tan b tan ] b π 2 π 4 π 4 and so the series converges (but we do not know its sum). 4.4 The Ratio Test Let a n be a series with positive terms and suppose that Then: a n+ lim ρ. n a n (a) The series converges if ρ <. 20

121 (b) The series diverges if ρ > or ρ is infinite. (c) The test is inconclusive if ρ. This is the Ratio Test. Proofs of the above results are given in Thomas. The two series we looked at in the last section are good examples of cases where ρ and the test is inconclusive. n : a n+ a n a n+ : n 2 a n /(n + ) /n /(n + )2 /n 2 n n + ( ) 2 n 2. n + In each case ρ (i.e. the test is inconclusive) and yet we know that /n diverges whereas /n 2 converges. Example: series: (a) Use the Ratio Test to investigate the convergence of the following (a) n 2 n n, (b) n (2n)! (n!), (c) n! 2 n. n n a n 2n n ; a n+ 2n n+ ; ( ) n a n+ (2n+ + 5)/3 n+ a n (2 n + 5)/3 n 3 2n n n a n+ a n < as n and the series converges. 3 (b) a n (2n)! (n!) 2 ; a n+ a n+ a n 4n + 2 n /n + /n (2(n + )!) ((n + )!) 2 ; (2n + 2)! n! n! (n + )!(n + )! (2n)! (2n + 2)(2n + ) (n + )(n + ) 4 > and the series diverges. 2

122 Figure 85: The definition of power series about x 0 and x a. (c) a n n! n n ; a n+ (n + )! (n + ) n+ ; a n+ (n + )! nn a n (n + ) n+ n! (n + )n n (n + ) n (n + ) ( ) n n n ( ) n n (n + ) n n + + /n and we cannot determine the convergence or the divergence. 4.5 Power Series A power series is an infinite series in powers of some variable, usually x. Such series can be added, subtracted, multiplied, differentiated and integrated to give new power series. Power series about x 0 and x a are defined in Fig. 85. Consider the case where the coefficients in () in Fig. 85 are all unity.: c n x n n0 x n + x + x x n +. n0 This is just a geometric series with first term and ratio x. We know from the 22

123 Figure 86: The graphs of f(x) /( x) and four of its polynomial approximations. properties of geometric series that it converges to /( x) for x <. Hence x + x + x2 + + x n +, < x <. We can think of the right-hand side of this equation a sequence of partial sums which are polynomials P n (x) that approximate the function on the left. f(x) x ; P 0(x) (horizontal line) P (x) + x (straight line, slope ) P 2 (x) + x + x 2 (quadratic curve [parabloa]). etc. The successive polynomials are illustrated in Fig. 86. Now consider the power series 2 (x 2) + 4 (x 2)2 + + ( 2) n (x 2) n +. This matches the form of (2) in Fig. 85 with a 2, c n ( /2) n. This is a geometric series with the first term and ratio r (x 2)/2. This series 23

124 Figure 87: The graphs of f(x) 2/x and its first three polynomial approximations. converges for (x 2)/2 < or 0 < x < 4. The sum is Hence 2 x (x 2) 2 + r + (x 2)/2 2 x. (x 2)2 4 + ( 2) 2 (x 2) n +, 0 < x < 4. Again we can consider the series as a sequence of partial sums which are polynomials P n (x) that approximate 2/x: f(x) 2 x ; P 0(x) P (x) 2 (x 2) 2 x 2 P 2 (x) 2 (x 2) + 4 (x 2)2 3 3x 2 + x2 4. etc. The successive polynomials are illustrated in Fig. 87. A series a n converges absolutely if the corresponding series of absolute values, an, converges. The Convergence Theorem for Power Series is shown in Fig. 88 and a corollary to it in Fig

125 Figure 88: The Convergence Theorem for Power Series. Figure 89: A corollary to the Convergence Theorem for Power Series. A series that converges but does not converge absolutely, converges conditionally. Here R is called the radius of convergence and the interval of radius R centred at x a is called the interval of convergence. When studying the convergence of power series such as these, alternating series frequently arise. Here we can make use of an additional test. The Alternating Series Test states that the series ( ) n+ u n u u 2 + u 3 +u 4 u 5 + n converges if all three of the following conditions hold: (i) The u n are all positive, (ii) u n u n+ for all n N, for some integer N and (iii) u n 0 as n. We can test a power series for convergence using several methods: 25