MATH2420 Multiple Integrals and Vector Calculus

Transcription

1 MATH242 Multiple Integrals and Vector Calculus Prof F.W. Nijhoff Semester 1, Full Course Notes (including examples) Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. In ordinary differential and integral calculus, you have already seen how derivatives and integrals interrelate. A derivative can be used as the opposite of an integration; it also occurs in changing variables in an integral. The same interrelation applies in multiple dimensions, but with more richness and variety. This module starts with a discussion of different coordinate systems in two and three dimensions. The use of Cartesian, plane polar, cylindrical polar and spherical polar coordinates will run through the whole module. The second section starts with a discussion of vector functions, which are the two- and threedimensional equivalents of the functions of ordinary calculus. These can be used to describe curves in space. Next we look at functions of several variables: that is, functions of a vector. With these two concepts we can introduce derivatives for fully three-dimensional functions (gradient, divergence and curl). This brings us to the halfway point of the module, and we will pause to review our new understanding before moving on to multiple-dimensional integrals. Here we extend the familiar idea of integration in one dimension to integration over an area or a volume. Finally, with the introduction of line and surface integrals we come to the famous integral theorems of Gauss and Stokes. These encompass beautiful relations between line, surface and volume integrals and the vector derivatives studied at the start of this module. Most real-life problems are not one-dimensional. The amount of heat stored in a piece of metal can be calculated by integrating its temperature in three dimensions; and the diffusion of dye in water is governed by differential equations based on three-dimensional derivatives. This is why a knowledge of vector calculus is essential for further study in many areas of applied mathematics.

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3 Chapter REVIEW.1 Calculus Differentiation The curve y f(x) has a slope at point x a given by the derivative of f with respect to x at a: f (a) df f(a + x) f(a) dx lim. (1) a x x A few particular derivatives are: f(x) ax n f(x) e λx f(x) u(x)v(x) f(x) u(x)/v(x) f (x) anx n 1 f (x) λe λx f (x) u (x)v(x) + u(x)v (x) f (x) [u (x)v(x) u(x)v (x)]/v 2 (x). Integration Integration is the opposite of differentiation: b a f (x)dx [f(x)] b a f(b) f(a) or f (x)dx f(x) (2) and it may also be seen as giving the area under a curve so the integral b f(x)dx gives the area under a the curve y f(x) between x a and x b. Some specific integrals are: x n dx x n+1 /(n + 1) for n 1 x 1 dx lnx e λx dx e λx /λ u(x)v (x)dx [u(x)v(x)] u (x)v(x)dx.2 Lines and Circles The vector equation of a straight line in three-dimensional space is x a + ub with u (, ) a real scalar. (3) 3

4 4 The equation of a circle of radius r centred about the origin is x 2 + y 2 r 2 on the point P(a, b) the the equation is (x a) 2 + (y b) 2 r 2. and if the circle is centred.3 Trigonometry Recall the functions sinx and cosx, with the identities: sin 2 x + cos 2 x 1 (4) tan x sinx/ cosx (5) and the derivatives (d/dx) sin x cos x (6) (d/dx)cosx sin x (7) (d/dx)tanx 1 + tan 2 x. (8) There are memorable values: x sinx cosx 1 π/2 1 π 1 3π/2 1 2π 1.4 Determinant The determinant of a 3x3 matrix A is a 11 a 12 a 13 A a 21 a 22 a 23 a 31 a 32 a 33 a 11(a 22 a 33 a 32 a 23 ) a 12 (a 21 a 33 a 31 a 23 ) + a 13 (a 21 a 32 a 31 a 22 ) (9).5 Vector Products We consider two vectors a (a 1, a 2, a 3 ) and b (b 1, b 2, b 3 ). Their scalar dot product is given by a b a 1 b 1 + a 2 b 2 + a 3 b 3 (1) and their vector cross product by i j k a b a 1 a 2 a 3 b 1 b 2 b 3 ([a 2b 3 b 2 a 3 ], [a 3 b 1 b 3 a 1 ], [a 1 b 2 a 2 b 1 ]). (11) Note that although for the scalar product a b b a, for the vector product we have a b b a. Recall that a b a b cosθ and a b a b sinθ (12) where θ is the angle between the vectors a and b.

5 Chapter 1 COORDINATE SYSTEMS 1.1 Cartesian Coordinates We are familiar with the rectangular Cartesian coordinates x, y in a plane, or x, y, z in space. If the unit vectors in the x, y and z directions are given by i, j and k then we can write the position vector r of a point as These two notations will be used interchangeably in this course. r (x, y, z) (1.1) xi + yj + zk. (1.2) 1.2 Plane Polar Coordinates In a fixed frame of reference the rectangular coordinates (x, y) are related to the polar coordinates (r, θ) through the relations x r cosθ, y r sin θ, r >, θ [, 2π) (1.3) ( y r (x 2 + y 2 ) 1/2, θ arctan. (1.4) x) r x θ y Figure 1.1: Polar coordinates 5

6 6 Express in polar coordinates the portion of the unit disc that lies in the first quadrant. The region may be expressed in polar coordinates as r 1 and θ π/ Cylindrical Coordinates In a fixed frame of reference the Cartesian coordinates (x, y, z) are related to the cylindrical coordinates (r, θ, z) through the relations: x r cosθ, y r sin θ, z z, r >, θ [, 2π) (1.5) ( y r (x 2 + y 2 ) 1/2, θ arctan, z z. x) (1.6) Express in cylindrical coordinates the solid bounded above by the cone z 2 (x 2 + y 2 ) 1/2 and bounded below by the disc : (x 1) 2 + y 2 1. The region may be expressed as (r cosθ 1) 2 + r 2 sin 2 θ 1 which is 2r cosθ + r 2 which becomes r 2 cosθ, and π/2 θ π/2. The z-limits are z 2 (x 2 + y 2 ) 1/2 which become z 2 r. 1.4 Spherical Coordinates In a fixed frame of reference the Cartesian coordinates (x, y, z) are related to the spherical coordinates (ρ, θ, φ) through the relations: x ρ sinθ cosφ, y ρ sin θ sin φ, z ρ cosθ, (1.7) ρ >, θ [, π), φ [, 2π) (1.8) ( y ρ (x 2 + y 2 + z 2 ) 1/2, φ arctan, θ arccos[z(x x) + y 2 + z 2 ) 1/2 ] (1.9) In the above equations, θ is the latitude or polar angle, and φ is the longitude. Remark: Spherical coordinates are commonly used in applications where there is a centre of symmetry. The centre of symmetry is then taken as the origin. Express in spherical polar coordinates the solid T that is bounded above by the cone z 2 x 2 +y 2, below by the xy-plane, and on the sides by the hemisphere z (4 x 2 y 2 ) 1/2. The solid is defined by the following inequalities: z 2 x 2 + y 2 z x 2 + y 2 + z 2 4

7 Course Notes 7 θ ρ x φ Figure 1.2: Spherical polar coordinates

8 8 Substituting the definitions of x, y and z in terms of ρ, θ and φ gives: so ρ 2 cos 2 θ ρ 2 sin 2 θ cos 2 φ + ρ 2 sin 2 θ sin 2 φ ρ cosθ ρ 2 sin 2 θ cos 2 φ + ρ 2 sin 2 θ sin 2 φ + ρ 2 cos 2 θ 4 ρ 2 cos 2 θ ρ 2 sin 2 θ ρ cosθ ρ 2 4 and we can use the fact that ρ and sinθ to deduce Given that θ π, this reduces to: ρ 2 cos θ 1 tan θ. ρ 2 π/4 θ π/2. In this case, where there is no information about φ contained in our limits, we use the whole permitted range: φ < 2π.

9 Chapter 2 VECTOR CALCULUS 2.1 Vector Functions A vector-valued function f is a vector function whose components are single-valued functions (scalar valued functions). For example, given three single-valued functions f 1 (t), f 2 (t), f 3 (t) we can form the vector-valued function f(t) (f 1 (t), f 2 (t), f 3 (t)) f 1 (t)i + f 2 (t)j + f 3 (t)k (2.1) The magnitude of the vector-valued function f(t) is a scalar-valued function and is defined by s f(t) [ f 2 1 (t) + f2 2 (t) + f2 3 (t)] 1/2 (a) f 1 (t) cos(t), f 2 (t) sin(t), f 3 (t). (b) f 1 (t) t, f 2 (t) g(t), f 3 (t) for t [a, b]. In general, the graph of the vector function f(t) f 1 (t)i+f 2 (t)j +f 3 (t)k is a curve C, in the sense that, as t varies, the tip of the position vector f(t) traces out C. The equations (2.2) x f 1 (t), y f 2 (t), z f 3 (t) (2.3) corresponding to the components of f are the parametric equations of C. If one of the components is zero, e.g. f(t) f 1 (t)i + f 2 (t)j, then C is said to be a planar curve, otherwise C is a space curve Derivative of a vector function Given the vector function f(t) (f 1 (t), f 2 (t), f 3 (t)) the derivative of f is defined by Find f (t) for f (t sin(t), e t, t). f(t) (t sin(t), e t, t) f (t) (sin(t) + t cos(t), e t, 1) f (t) (2 cos(t) t sin(t), e t, ) f (t) ( 3 sin(t) t cos(t), e t, ). f (t) (f 1(t), f 2(t), f 3(t)). 9

10 1 Properties of the derivative (f + g) (t) f (t) + g (t) (αf) (t) αf (t) where α is a constant scalar (uf) (t) u(t)f (t) + u (t)f(t), where u is a scalar function (f g) (t) f(t) g (t) + f (t) g(t). (f g) (t) f(t) g (t) + f (t) g(t). (f(u)) (t) f (u(t))u (t) Integral of a vector function Given the vector function f(t) (f 1 (t), f 2 (t), f 3 (t)), the integral of f is defined by b a f(t)dt ( b a f 1(t)dt, b a f 2(t)dt, b a f 3(t)dt). Find f(t)dt for f(t) (t, (t + 1)1/2, e t ). f(t)dt ( t, (t + 1) 1/2, e t) dt ( t dt, ( [1 2 t2] 1 ( 1, [ ) (t + 1) 1/2 dt, e t dt 2 3 (t + 1)3/2] 1, [ e t] ) 1 ) 2, 2 3 (2)3/2 2 3 (1)3/2, e + 1 ( ) 1 2, 2 3 [23/2 1], 1 e Properties of the integral b a (f(t) + g(t))dt b a f(t)dt + b a g(t)dt b a (αf)(t)dt α b f(t)dt, where α is a constant scalar a b a (c f)(t)dt c(t) b a f(t)dt, where c is a constant vector. b a (c f)(t) c(t) b f(t)dt, where c is a constant vector. 2.2 Curves a 1. The equation of a straight line is parametrised by r(t) r + td, t (, ) (2.4)

11 Course Notes More generally, every vector function r(t) x(t)i + y(t)j + z(t)k (2.5) parametrises a curve in space (or a curve in plane if one of the components x(t), y(t) or z(t) is zero). It is also important to understand that a parametrised curve C is an oriented curve in the sense that as t increases on some interval of definition I, the tip of the position vector r(t) traces out C in a certain direction. For example, the unit circle parametrised by r(t) cos(t)i + sin(t)j, t [, 2π) is traversed in the anticlockwise direction, starting at the point (1,) Tangent vector, tangent line For a given curve C parametrised by r(t) (x(t), y(t), z(t)), the derivative vector r (t) (x (t), y (t), z (t)) is called the tangent vector to the curve C at the point P(x(t), y(t), z(t)), and r (t) points out in the direction of increasing t. In the case of a circle of radius a parametrised as r(t) a cos(t)i+a sin(t)j, show that the tangent vector r (t) is perpendicular to the radius vector r(t). The radius vector is r(t) (a cos(t), a sin(t)) which has r a. We differentiate to get which also has magnitude r a. Now r(t) r (t) r (t) a sin(t)i + a cos(t)j (a cos(t), a sin(t)) ( a sin(t), a cos(t)) a cos(t)( a sin(t)) + a sin(t)a cos(t) and since we know that if c and d but c d then c is perpendicular to d, so in this case r (t) is perpendicular to r(t). For a given curve C parametrised by r(t) (x(t), y(t), z(t)) the tangent line at a point t is the vector function R(λ) r(t) + λr (t), λ (, ) (2.6) Find a vector tangent to the twisted cubic r(t) ti + t 2 j + t 3 k at the point P(2, 4, 8), and then parametrise the tangent line.

12 12 The derivative along the curve is r (t) i + 2tj + 3t 2 k and at the point P(2, 4, 8) we have t 2 and so r i + 4j + 12k and the tangent line at the point P is R(λ) (2 + λ, 4 + 4λ, λ) Intersecting curves Two curves (C 1 ) : r 1 (t) x 1 (t)i + x 2 (t)j + x 3 (t)k, (C 2 ) : r 2 (u) y 1 (u)i + y 2 (u)j + y 3 (u)k, intersect iff there are numbers t and u for which r 1 (t) r 2 (u). The angle between two intersecting curves (which, by definition, is the angle between the corresponding tangent lines) can be obtained by examining the tangent vectors at the point of intersection. Show that the circles (C 1 ) : r 1 (t) cos(t)i + sin(t)j (C 2 ) : r 2 (u) cos(u)j + sin(u)k intersect at P(, 1, ) and Q(, 1, ). Find the angle between them at P. First we look for values of t and u at which the curves intersect. P(, 1, ): t π/2, u Q(, 1, ): t 3π/2, u π. Then we find the tangents to the two curves: (C 1 ) : r 1 (t) sin(t)i + cos(t)j (C 2 ) : r 2 (u) sin(u)j + cos(u)k and substitute in the values at the intersection point in question: r 1 P ( 1,, ) r 2 P (,, 1) These are both unit vectors so dotting them we obtain: r 1 P r 2 P r 1 P r 2 P cosθ cosθ so the vectors are perpendicular, the angle between them is π/2.

13 Course Notes The unit tangent For a given curve C parametrised by r(t) (x(t), y(t), z(t)), the vector T : r (t) r (t) (2.7) is called the unit tangent vector to the curve C at the point P(x(t), y(t), z(t)). The unit tangent points in the direction of increasing t along the curve and is parallel to the curve. Questions How might we reverse the sense of a curve? What would happen to the unit tangent? Reversing the sense of a curve We make a distinction between the curve and the curve r r(t), t [a, b] (2.8) R(u) r(a + b u), u [a, b]. (2.9) Both vector functions trace out the same set of points, but the order has been reversed. Whereas the first curve starts at r(a) and ends at r(b), the second curve starts at r(b) and ends at r(a). Thus, for example, the vector function r(t) cos(t)i + sin(t)j, t [, 2π) gives the unit circle traversed anticlockwise while the reversed curve gives the unit circle traversed clockwise. Unit tangent R(u) cos(2π u)i + sin(2π u)j, u [, 2π) When we reverse the sense of a curve, the unit tangent T reverses direction (is multiplied by 1) because it always points in the direction of increasing t or u. 2.3 Arc Length The length of a continuously differentiable curve (C): r r(t), t [a, b] is given by L(C) b a r (t) dt. (2.1) Find the length of the curve r(t) 2t 3/2 i + 4tj from t to t 1. Find the length of the curve r(t) 2 cos(t)i + 2 sin(t)j + t 2 k from t to t π/2, and compare it to the straight-line distance between the endpoints of the curve.

14 14 The length of the curve r(t) 2t 3/2 i + 4tj from t to t 1 is given by and the derivative of r is which has magnitude and so L(C) L(C) The derivative vector along the curve is which has magnitude Thus the length of the curve is r (t) dt r (t) 3t 1/2 i + 4j r (t) (9t + 16) 1/2 [ (9t + 16) 1/2 2 dt 27 (9t + 16)3/2] [(25)3/2 (16) 3/2 ] [125 64] 27 r (t) 2 sin(t)i + 2 cos(t)j + 2tk r (t) (4 sin 2 t + 4 cos 2 t + 4t 2 ) 1/2 2(1 + t 2 ) 1/2. L(C) π/2 2(1 + t 2 ) 1/2 dt 2 π/2 (1 + t 2 ) 1/2 dt and we know from the example I showed in the first example class that b a so we have (1 + x 2 ) 1/2 dx 1 2 b(1 + b2 ) 1/2 1 2 a(1 + a2 ) 1/ arcsinhb 1 2 arcsinha L(C) 1 2 π(1 + π2 /4) 1/2 + arcsinhπ/2 The straight line distance between the ends of the curve is the distance between the points X (2,, ) and X π/2 (, 2, π 2 /4) which is D ( π 4 /16) 1/2.25(128 + π 4 ) 1/ Substituting in the value of π/2, we have L(C) 4.16, which is longer, as we would expect since a straight line is the shortest possible distance between two points.

15 Chapter 3 FUNCTIONS OF SEVERAL VARIABLES 3.1 Introduction Since we live in a three-dimensional world, in applied mathematics we are interested in functions which can vary with any of the three space variables x, y, z and also with time t. For instance, if the function f represents the temperature in this room, then f depends on the location (x, y, z) at which it is measured and also on the time t when it is measured, so f is a function of the independent variables x, y, z and t, i.e. f(x, y, z, t). 3.2 Geometric Interpretation For a function of two variables, f(x, y), consider (x, y) as defining a point P in the xy-plane. Let the value of f(x, y) be taken as the length PP drawn parallel to the z-axis (or the height of point P above the plane). Then as P moves in the xy-plane, P maps out a surface in space whose equation is z f(x, y). : f(x, y) 6 2x 3y The surface z 6 2x 3y, i.e. 2x + 3y + z 6, is a plane which intersects the x-axis where y z, i.e. x 3; which intersects the y-axis where x z, i.e. y 2; which intersects the z-axis where x y, i.e. z 6. : f(x, y) y 2 x 2 In the plane x, there is a minimum at y ; in the plane y, there is a maximum at x. The whole surface is shaped like a horse s saddle; and the picture shows a structure for which (, ) is called a saddle point Plane polar coordinates Since the variables x and y respresent a point in the plane, we can express that point in plane polar coordinates simply by substituting the definitions: x r cosθ y r sin θ. 15

16 x y 3 Figure 3.1: The surface z x 2 y x y 3 Figure 3.2: The surface z x 2 + y 2

17 Course Notes 17 : f(x, y) x 2 + y 2 The surface z x 2 +y 2 may be drawn most easily by first converting into plane polar coordinates. Substituting x r cosθ and y r sin θ gives z r 2. The surface is symmetric about the z-axis and its cross-section is a parabola. [Check with the original function and the plane y.] Thus the whole surface is a paraboloid (a bowl). Another way to picture the same surface is to do as map-makers or weather forecasters do and draw contour lines (or level curves) produces by taking a section, using a plane z const. and projecting it onto the xy-plane. For z x 2 + y 2 as above, the contour lines are concentric circles. 3.3 Partial Differentiation For a function f(x) that depends on a single variable, x, we can form the ordinary derivative df/dx. For example, if f(x) 3x 4 +sin(x) then df/dx 12x 3 + cos(x). Similarly, for a function f that depends on several variables x, y,...we can differentiate with respect to each of these variables. This process is called partial differentiation (partial derivatives). If f(x, y) yx 4 + sin(x) then we treat y as a constant as we did 3 for the function of x, and have f/ x 4yx 3 + cos(x). To find a partial derivative we hold all but one of the independent variables constant and differentiate with respect to that one variable using the ordinary rules for one-variable calculus. Notation: If f depended on x, y,... but if only x is allowed to vary, the partial derivative of f with respect to x is denoted by f/ x or by f x. (Distinguish carefully between df/dx the ordinary derivative with straight d and f/ x with curly.) Similarly, we can denote the partial derivative with respect to y by f/ y or by f y, etc. Calculate the partial derivatives of the functions: (a) f(x, y) x 2 + 2xy 2 + y 3 ; (b) f(x, y, z) xz + e yz + sin(xy). (a) Holding y constant gives f/ x 2x + 2y 2 +. Holding x constant gives f/ y + 4xy + 3y 2. (b) Holding both y and z constant gives f x z + + y cos(xy). Holding both x and z constant gives f y + ze yz + xcos(xy). Holding both x and y constant gives f z x + ye yz +. If z is a function of two independent variables x and y, and z satisfies xz + lnz 2x + 3y, find z/ x in terms of x, y and z. Differentiating each term in the equation with respect to x, holding y constant, and treating z as a function of x, we obtain x z x + z + 1 z z x 2 so that z z(2 z) x (1 + xz).

18 Second-order partial derivatives For f(x, y) we can form f/ x and f/ y. Each of these can then be differentiated again with respect to x or y to form the second-order derivatives / x( f/ x) denoted by 2 f/ x 2 or f xx ; / y ( f/ x) denoted by 2 f/ y x or f xy ; / y ( f/ y) denoted by 2 f/ y 2 or f yy ; / x( f/ y) denoted by 2 f/ x y or f yx. f xy and f yx are called mixed derivatives. The concept can be extended, e.g. if f depends on x, y and z we can form / z ( 2 f/ x 2), written as 3 f/ z x 2 or f xxz, etc. If f(x, y) x 4 y 2 x 2 y 6 then f x 4x3 y 2 2xy 6 f y 2x 4 y 6x 2 y 5 2 f x 2 12x 2 y 2 2y 6 2 f y x 8x3 y 12xy 5 2 f y 2 2x 4 3x 2 y 4 2 f x y 8x 3 y 12xy 5 Note that in this example f xy f yx and this is true in general. The Mixed Derivatives Theorem states that if f xy and f yx are continuous then f xy f yx. Thus to calculate a mixed derivative we can calculate in either order. [Think about calculating / x( f/ y) if f(x, y) xy + 1/(sin(y 2 ) + e y ).] For third-order derivatives the mixed derivatives theorem gives f xxy f xyx f yxx (check for yourself in the last example). For a function f(x), the ordinary derivative df/dx gives the slope of the tangent to the curve at any point P. For a function f(x, y), the partial derivative f x is evaluated holding y constant, and so gives the slope of the tangent to the surface at P in a plane y const Summary We form partial derivatives with respect to one variable by holding all the other variables constant and differentiating. The order of mixed derivatives is not important. If f(x, y) represents a surface above the xy plane then f/ x is the slope of a section taken in the x direction at a point P and f/ y the slope of a section in the y direction. Between them these two tangents define a plane which (we will see later) is the tangent plane to the surface at P.

19 Chapter 4 GRADIENTS In this chapter we introduce derivatives in multiple dimensions. Although we limit ourselves to differentiating known functions, the real use of these derivatives is in differential equations. When we learnt calculus, there was a big gap between starting to study differentiation and solving our first differential equation. In this chapter we meet three-dimensional derivatives. Because this is a major concept we won t get to any physical applications; but almost no physical problem can be solved without using these tools. 4.1 Gradient and Directional Derivative For a function of three variables f(x, y, z), the partial derivatives f x, f f y and z measure the rates of change of f along x, y and z directions, respectively. We now ask how we can calculate the rate of change of f in any direction in space. The answer lies in the vector ( f f x, f y, f ) f z x i + f y j + f z k (4.1) call the gradient of f. From its definition, the component of f along i is ( f i) f x rate of change along i and similarly for the y and z directions. But for any direction in space we are free to temporarily call it the i-direction and carry over the above analysis. Thus in general: For any direction in space defined by a unit vector u the rate of change of f along u is given by ( f u) and is called the directional derivative of f along u. If f(x, y, z) x 2 + xy + z, find f. What is the rate of change of f along the direction i+2j + 2k at the point P(1, 1, 1)? f ( f x, f y, f ) (2x + y, x, 1). z Now at the point P(1, 1, 1), we have f (3, 1, 1). To find the rate of change of f along a vector v, we need to calculate the dot product of f with the unit vector of v. Here, v (1, 2, 2) which has modulus ( ) 9 3 so the unit vector is ˆv (1/3, 2/3, 2/3). f ˆv (3, 1, 1) (1/3, 2/3, 2/3) 1 + 2/3 + 2/3 7/3. 19

20 2 If f ( 6xy + y cos(xy), 3x 2 + xcos(xy), 3z 2), determine f(x, y, z) up to an arbitrary constant. In effect we are solving three simultaneous equations: f/ x 6xy + y cos(xy); f/ y 3x 2 + xcos(xy); f/ z 3z 2. Integrating the first of these with respect to x gives f 3x 2 y + sin(xy) + A(y, z) which can be differentiated w.r.t. y and z to give: f/ y 3x 2 + xcos (xy) + A(y, z)/ y; f/ z A(y, z)/ z. These combine with the last two of the simultaneous equations to give A(y, z)/ y ; A(y, z)/ z 3z 2. Thus A is a function of z only (from the first), and A z 3 + C (from the second). This gives f 3x 2 y + sin (xy) + z 3 + C and we can check the individual terms of the gradient by differentiating this function Two further properties of the gradient We look at cases where u is parallel or perpendicular to f. The change df in f due to a change in the position P by dr dru is given by change in f (rate of change with distance) (distance), i.e. df ( f u)dr f (udr) f dr f dr cos(θ) (4.2) where θ is the angle between the vectors dr and f. From this equation it can be seen that the direction dr for which df is a maximum is obviously that for which cos(θ) 1, or θ, i.e. the direction of f. Thus Property 1. At any point, f points in the direction in which f is increasing most rapidly and its magnitude f gives this maximum rate of change. Again from eq. (4.2), df corresponds to θ π/2, when f and dr are perpendicular. But df means that f has not changed - so the displacement dr is along the surface f const.. Thus Property 2. At any point, f is perpendicular to the surface f const. through that point. In the previous example f 3i+j +k at P(1, 1, 1). This is moving away from P in the direction in which f is increasing most rapidly and the maximum rate of change is f (11) 1/2. Find a unit vector perpendicular to the surface z x 2 + y 2 at the point A(1, 2, 5). Deduce the equation of the tangent plane to the surface at A. The normal to the surface f const. at any point is in the direction of f at that point. In this case the formulation f const. is given as f(x, y, z) x 2 + y 2 z. f 2xi + 2yj k so at the point A we have f 2i + 4j k. The unit normal is n 1 (21) 1/2 (2i + 4j k). The tangent plane at A is perpendicular to the normal at A. The equation of the plane is r n const. where r xi + yj + zk, i.e. in this case (x, y, z) 1 (21) 1/2 (2, 4, 1) const.. We can multiply by (21) 1/2 (which changes the value of the constant) to have 2x + 4y z const.. This plane passes through the point A(1, 2, 5) so the constant has the value , hence the equation of the tangent plane is 2x + 4y z 5.

21 Course Notes 21 Summary For a function f(x, y, z) at any point P (a) The gradient f ( f x rate of change of f; )i + ( f y f )j + ( z )k gives the magnitude and the direction of the maximum (b) The direction derivative f u gives the rate of change of f along the direction of the unit vector u; (c) The gradient f is perpendicular to the surface f const. through P. 4.2 Linear Approximations (Tangents) Motivation: Many functions arising in applications are difficult to deal with. We thus need ways of approximating such functions by others which are easier to handle. The most useful approximations are polynomials. We consider first the single variable case, f(x), which will guide us into the treatment of the two-variable case One-variable case (tangent line) The tangent to the curve y f(x) at A where x a has the slope f (a) and therefore has the equation y f (a)x + const.. It passes through A, i.e. for x a, y f(a), so const. f(a) f (a)a. Thus the equation of the tangent line (the line parallel to the curve) is y f(a) + (x a)f (a). (4.3) For points close to A the tangent gives a close approximation to the curve. The approximation is called the linear approximation to f(x) near x a. Find the linear approximation to f(x) 1 + x 2 near x 2. f(x) f(a) + (x a)f (a) (4.4) If f(x) 1 + x 2 then f (x) 2x. At the point x 2 we have f 5 and f 4. Therefore the linear approximation is f(x) 5 + 4(x 2) 4x Two-variables case (tangent plane) Above we saw that, for a function of one variable, approximating its curve by a tangent line gave a linear approximation. For a function of two variables, f(x, y), the corresponding linear approximation arises when we approximate a surface by its tangent plane. Suppose the surface is z f(x, y). The tangent plane at A is perpendicular to the normal at A, i.e. is perpendicular to the gradient vector at A. Now the surface is f(x, y) z, or F(x, y, z) const., whose gradient is F f x i + f yj k. Thus the tangent plane r n const. at A is x ( ) f x A + y ( ) f z const. (4.5) y A

22 22 This plane passes through A(a, b) so the constant in eq. (4.5) has the value a and thus (4.5) becomes z f(a, b) + (x a) ( ) ( ) f x + b f y f(a, b) A A ( ) ( ) f f + (y b). (4.6) x A y A Thus the linear approximation to the surface close to A(a, b) is given by the tangent plane (4.6), so f(x, y) f(a, b) + (x a) and we note that we can rewrite this using the gradient f as ( ) ( ) f f + (y b) x A y A (4.7) f(x, y) f(a, b) + ((x a), (y b)) ( f) A. (4.8) Find the linear approximation to f(x, y) x 2 + y 2 near the point (1, 2). If f(x, y) x 2 + y 2 then f/ x 2x and f/ y 2y. At the point (1, 2) we have f 5, f/ x 2 and f/ y 4. Thus f(x, y) 5 + (x 1)2 + (y 2)4 2x + 4y The Chain Rule For a function of one variable, f(x), if x also depends on another variable t, then f depends on t and the chain rule gives df dt df dx (4.9) dx dt We now generalise this result for a function of several variables. For f(x, y), suppose x and y depend on t. Let t increase by t and so x and y increase by x and y. Then from our earlier work on linear approximations we obtain f(t + t) f(t) + x ( ) f + y x ( ) f y (4.1) so if we rearrange and let t we obtain the chain rule for a function of two variables, which is, ( ) ( ) df f dx f dy dt x dt + y dt. (4.11) Note that f depends on x and y [so partial derivatives f/ x, f/ y] whilst x and y depend on the single variable t [so ordinary derivatives dx/dt, dy/dt]. Thus f depends on t and has the derivative df/dt given by the chain rule (4.11). If f(x, y) x 2 + y 2, where x sin(t), y t 3, then df dt ( ) f dx x dt + ( ) f dy y dt 2xcos(t) + 2y3t2 2 sin(t)cos(t) + 6t 5.

23 Course Notes 23 Of course in this simple example we can check the result by substituting for x and y before differentiation to give f(t) (sin(t)) 2 + (t 3 ) 2, so df dt 2 sin(t)cos(t) + 6t5 as before. The chain rule extends directly to functions of three or more variables. If f(x, y, z) ln (2x 3y + 4z), where x e t, y lnt, z cosht, then df dt ( f x ) dx dt + ( f y 2e t 2x 3y + 4z ) dy dt + 2et 3/t + 4 sinht 2e t 3 lnt + 4 cosht. ( f z 3(1/t) 2x 3y + 4z + ) dz dt 4 sinht 2x 3y + 4z Extended chain rule For f(x, y) suppose that x and y depend on two variables s and t. Then changing either s or t changes x and y, so changes f, i.e. producing f f s and t according to the extended chain rule f(x, y) x 2 y 3, where x s t 2, y s + 2t. Then f s f x x s + f y y s, f t f x x t + f y y t. (4.12) f x 2xy3 and f y 3x2 y 2 and f s f x x s + f y y s 2xy3 + 3x 2 y 2 (s t 2 )(s + 2t) 2 (5s + 4t 3t 2 ) f t f x x t + f y y t 2xy3 ( 2t) + 3x 2 y 2 (2) 2(s t 2 )(s + 2t) 2 (3s 2st 7t 2 ). (i) If f is a function of x and y, where x e s cost, y e s sin t, prove that sint f f s +cost t es f y. (ii) If f is a function of z/x and x/y, prove that x f x + y f y + z f z. (iii) For a function f(x, y), if the independent variables x and ( y) are changed to polar coordinates r and θ, so x r cosθ and y r sin θ, show that 1 r r r f r f r 2 θ 2 f 2 x + 2 f 2 y 2 (i) If x e s cost and y e s sin t then x s es cost x t es sin t y s es sin t y t es cost.

24 24 It follows that f s f t Combining these two equations we have f x es cost + f y es sint f x es sint + f y es cost sint f s + cost f t f y es. (ii) Let u z/x and v x/y. Then u/ x z/x 2, u/ z 1/x, v/ x 1/y and v/ y x/y 2. f x f y f z f z u x 2 + f 1 v y f x v y 2 f 1 u x. Substituting the latter two of these into the first we obtain as required. f x f z z x y f x y (iii) We calculate the derivatives in terms of the polar coordinates: f r f θ and the second derivatives we need: ( r f ) ( x f r r x and simplify: ( r f ) r r and add: 1 r 2 f θ 2 r 2 f θ 2 x f f cosθ + x y sin θ f f r sin θ + x y r cosθ x + y f y ( y f x + x f y ( f x + f x 2 x 2 + y 2 f x y ( y 2 f x 2 + f y + x 2 f x y r ) cosθ + ( x f y ) ( r sin θ) + y ) ( cosθ + x 2 f ) sin θ + r x + y f y ) sin θ ( y f x + x f y ) (r cosθ) ) sinθ x y + f y + f y 2 y 2 ( f x y 2 f x y + f x 2 y 2 ( r f ) f r r 2 θ 2 1 [( 2 ) ( f r x f 2 ) ] f y 2 xcosθ + x f y 2 y sin θ 2 f x f y 2 ) cosθ

25 Course Notes The vector differential operator (grad) The vector differential operator is defined formally by setting ( x, y, ). (4.13) z By formally we mean that this is not an ordinary vector but its components are differentiation symbols. As the term operator suggests, is thought of as something that operates on things. What sort of things? Scalar functions and vector functions. Suppose that f(x, y, z) is a scalar function. Then operates on f as follows: f This is the gradient of f, which we have just discussed. Question How does operate on vector functions? ( x, y, ) ( f f z x, f y, f ). (4.14) z Answer Using either of the vector products we already know dot or cross Divergence If q(x, y, z) (q 1 (x, y, z), q 2 (x, y, z), q 3 (x, y, z)) is a vector function, then by definition div(q) q q 1 x + q 2 y + q 3 z. (4.15) This product, q, defined in imitation of the ordinary dot product, is a scalar called the divergence of q. Calculate q for the vector function q(x, y, z) (x y, x + y, z). q (x y) x + (x + y) y + z z Physical Meaning Consider a fluid whose velocity everywhere is given by the vector function u(x, y, z) (u 1, u 2, u 3 ). Then imagine we have a cubic volume [a, a + x] [b, b + y] [c, c + z] and we want to know how much fluid is flowing out of it per unit time. z y c + z c x a a + x

26 26 The fluid flowing through one face is the velocity normal to the face multiplied by the area of that face. So for the face on which x a, u 1 gives the velocity pointing in and the face has area y z, so the flow out is y zu 1 (x a). Looking at the face on which x a + x, the area of the face is still y z but the velocity out is u 1 (since the inside of the cube is on the other side of this face). The contribution from this face is Similarly, the other 4 contributions are y zu 1 (x a + x). x zu 2 (y b) x yu 3 (z c) x zu 2 (y b + y) x yu 3 (z c + z). We add together the six contributions to find Rate of flow out y z(u 1 (x a + x) u 1 (x a)) + x z(u 2 (y b + y) u 2 (y b)) + x y(u 3 (z c + z) u 3 (z c)) x y z[(u 1 (x a + x) u 1 (x a))/ x + (u 2 (y b + y) u 2 (y b))/ y + (u 3 (z c + z) u 3 (z c))/ z] and if we let the x etc. terms tend to zero we have Rate of flow out volume ( u 1 / x + u 2 / y + u 3 / z) volume div(u). Thus the divergence represents the flow out of the volume, per unit time, per unit volume. This is where the name comes from it is the rate at which the fluid is diverging Curl For the vector q q 1 i + q 2 j + q 3 k, we also have i j k curl(q) q det x y q 1 q 2 q 3 z ( q3 y q 2 z, q 1 z q 3 x, q 2 x q 1 y ). (4.16) This second product, q, defined in imitation of the ordinary cross product, is a vector called the curl of q. Calculate q for (from above) q(x, y, z) (x y, x + y, z). q i j k / x / y / z x y x + y z i() + j() + k(1 + 1) 2k.

27 Course Notes 27 Physical Meaning Consider a solid body which is rotating about the z-axis at rate ω, so that each point (x, y, z) inside it has velocity u ( ωy, ωx, ). We can express this as u r, where r (x, y, z) and (,, ω) is called the rotation vector. Then the curl of the velocity field (the vector field, i.e. three-dimensional vector function u, defining the velocity) is u i j k / x / y / z ωy, ωx k(ω + ω) 2ωk (,, 2ω). We have shown that for a rotation, the curl is related to the rotation vector. This is true in general so if we had a fluid velocity which was partly rotating and partly moving in some other way, the curl would extract the rotation part (the circulation) Grad and div in polar coordinates The forms of grad and div in polar coordinates are not examinable in this module, but you may find them useful in later years. The forms for curl are too unpleasant to be quoted here; do not be fooled into carrying out a simple cross product with the gradient vector. Plane polar coordinates The unit vector in the r direction is e r ; the unit vector in the θ direction is e θ. The values of these two vectors vary according to location. We can represent any vector v in terms of these two unit vectors: Then we have Cylindrical polar coordinates v v r e r + v θ e θ. f f r e r + 1 r (rv r ) r v 1 r In a similar fashion to plane polar coordinates, we obtain Spherical polar coordinates f f r e r + 1 r (rv r ) r v 1 r f θ e θ + 1 r v θ θ f θ e θ + f + 1 r z e z v θ θ + v z z Here we denote the unit vector in the ρ direction as e ρ and similar for θ and φ, and a vector v can be written v v ρ e ρ + v θ e θ + v φ e φ.

28 28 Grad and div are given by: Basic Identities f f ρ e ρ + 1 f ρ θ e θ + 1 f ρ sin θ φ e φ v 1 ρ 2 (ρ 2 v ρ ) ρ + 1 (sin θv θ ) + 1 v φ ρ sin θ θ ρ sinθ φ For vectors we have that a a. Is it true that? Define ( )f by ( )f ( f). Theorem (The curl of a gradient is zero) If f(x, y, z) is a scalar function then ( f). Proof i j k ( f) / x / y / z f x f y f z i(f yz f zy ) + j(f zx f xz ) + k(f xy f yx ) and by the mixed derivatives theorem, ( f). For vectors we have that a (a c). The analogous operator formula ( q) is also valid. Theorem (The divergence of a curl is zero) If q(x, y, z) is a vector function then ( q). Proof i j k ( q) / x / y / z q 1 q 2 q 3 ( q2 z q 3 y, q 3 x q 1 z, q 1 y q ) 2 x 2 q 2 x z 2 q 3 x y + 2 q 3 y x 2 q 1 y z + 2 q 1 z y 2 q 2 z x. The following identities are left for you as an exercise. Theorem If f(x, y, z) is a scalar function and q(x, y, z) is a vector function then show that: (i) (fq) ( f) q + f( q) (ii) (fq) ( f) q + f( q).

29 Course Notes The Laplacian From the operator we can construct other operators. The most important of these is the Laplacian (or the Laplace operator) 2 which operates on scalar functions f(x, y, z) according to the rule: Calculate the Laplacian for f(x, y, z) x 2 + y 2 + z 2. 2 f ( f) 2 f x f y f z 2. (4.17) 2 f 2 f x f y f z 2 x 2x + y 2y + 2z z The forms for the Laplacian in different coordinate systems are more complex (cf. the final example of 4.3.1). They are given by: 2 f 1 r 2 f 1 r ( r f r r r 2 f 1 ρ 2 ρ ( r f r ) ( ρ 2 f ρ f r 2 in plane polar coordinates θ2 ) + 1 r 2 2 f ) + θ f in cylindrical polar coordinates z2 ( 1 ρ 2 sinθ f ) 1 + sinθ θ θ ρ 2 sin 2 θ 2 f in spherical polar coordinates φ2 4.5 Jacobian The Jacobian of the transformation x x(s, t), y y(s, t) is the determinant (x, y) (s, t) x y s t x y t s (4.18) of the Jacobian matrix ( x/ s y/ s x/ t y/ t ). (4.19) Find the Jacobian of the coordinate transformation x s 2 /t, y t 2 /s. The Jacobian matrix is ( x/ s y/ s x/ t y/ t ) ( 2s/t t 2 /s 2 s 2 /t 2 2t/s ) which has determinant so the Jacobian is 3. (x, y) (s, t) 2s/t t 2 /s 2 s 2 /t 2 2t/s Question: If x x(s, t) and y y(s, t), can we solve for s and t in terms of x and y?

30 3 Answer: Yes, provided that the Jacobian (x,y) (s,t). This is the implicit function theorem. Using x s 2 /t and y t 2 /s as above, solve for s and t in terms of x and y. To find s, we eliminate t. x 2 s 4 /t 2 so x 2 y s 3. Similarly, y 2 t 4 /s 2 so xy 2 t 3, and s (x 2 y) 1/3 t (xy 2 ) 1/3. Exercise Prove that the Jacobian of the reverse transformation is Jacobian in terms of the extended chain rule We recall from section that the extended chain rule is: f s f x x s + f y y s, f t f x x t + f y y t where f depends on x and y which depend on s and t. We can write this in matrix notation as ( ) ( )( ) f/ s x/ s y/ s f/ x f/ t x/ t y/ t f/ y (4.2) (4.21) so we can see that the Jacobian matrix can be used to convert one set of partial derivatives into another Jacobians of the standard coordinate transformations The following Jacobians may be quoted as standard results: (x, y) (r, θ) (x, y, z) (r, θ, z) (x, y, z) (ρ, θ, φ) r for plane polar coordinates r for cylindrical polar coordinates ρ 2 sin θ for spherical polar coordinates.

31 Chapter 5 DOUBLE AND TRIPLE INTEGRALS 5.1 Multiple-Integral Notation Previously ordinary integrals of the form f(x)dx b J a f(x) dx (5.1) where J [a, b] is an interval on the real line, have been studied. Here we study double integrals f(x, y)dxdy (5.2) where is some region in the xy-plane, and a little later we will study triple integrals f(x, y, z)dxdy dz (5.3) where T is a solid (volume) in the xyz-space. T y f(x) a b Figure 5.1: Integration of y f(x) 31

32 32 z f(x, y) 5.2 Double Integrals Properties (1) Area property Figure 5.2: Integration of z f(x, y) dxdy Area of. In particular if is the rectangle [a, b] [c, d] then dxdy (b a)(d c). (2) Linearity [αf(x, y) + βg(x, y)] dxdy α where α and β are constants. f(x, y)dxdy + β g(x, y)dxdy (5.4) (3) Additivity If is broken up into a finite number of nonoverlapping basic regions 1,..., n, then f(x, y)dxdy f(x, y)dxdy f(x, y)dxdy. (5.5) 1 n Geometric Interpretation The double integral over gives the volume of the solid T whose upper boundary is the surface z f(x, y) and whose lower boundary is the region in the xy-plane: f(x, y) dx dy volume of T. (5.6) The Evaluation of Double Integrals by Repeated Integrals If an ordinary integral b f(x)dx proves difficult to evaluate, it is not because of the interval [a, b] a but because of the integrand f. Difficulty in evaluating a double integral f(x, y)dxdy can come from two sources: from the integrand f or from the domain. Even such a simple looking integral as 1 dxdy is difficult to evaluate if is complicated. In this section we introduce a technique for evaluating double integrals over domains that have special shapes. The key idea is that double integrals over such special domains can be reduced to a pair of ordinary integrals.

33 Course Notes 33 Horizontally simple domain a b The projection of the domain onto the x-axis is a closed interval [a, b] and consists of all points (x, y) with Then a x b, and φ 1 (x) y φ 2 (x). (5.7) f(x, y)dxdy b a ( ) φ2(x) f(x, y)dy dx. (5.8) Here we first calculate φ 2(x) φ 1(x) f(x, y)dy by integrating f(x, y) with respect to y from y φ 1(x) to y φ 2 (x). The resulting expression is a function of x alone, which we then integrate with respect to x from x a to x b. Evaluate (x4 2y)dxdy, where the domain consists of all points (x, y) with 1 x 1 and x 2 y x 2. φ 1(x) (x 4 2y)dxdy x1 yx 2 x 1 x1 x 1 y x 2 (x 4 2y)dy dx x1 2x 6 dx [2x 7 /7] x1 x 1 4/7 x 1 [x 4 y y 2 ] yx2 y x 2 dx Vertically simple domain d c The projection of the domain onto the y-axis is a closed interval [c, d] and consists of all points (x, y) with c y d, and ψ 1 (y) x ψ 2 (y). (5.9) Then f(x, y)dxdy d c ( ) ψ2(y) f(x, y)dx dy. (5.1) Here we first calculate ψ 2(y) ψ 1(y) f(x, y)dx by integrating f(x, y) with respect to x from x ψ 1(y) to x ψ 2 (y). The resulting expression is a function of y alone, which we then integrate with respect to y from y c to y d. The integrals in the right-hand sides of formulae (5.8) and (5.1) are called repeated integrals. Remark 1 Sometimes a domain can be expressed both as a horizontally simple domain: a x b, φ 1 (x) y φ 2 (x), and as a vertically simple domain: c y d, ψ 1 (y) x ψ 2 (y). Then ( b ) φ2(x) ( d ) ψ2(y) f(x, y)dxdy f(x, y)dy dx f(x, y)dx dy. (5.11) a φ 1(x) c ψ 1(y) ψ 1(y)

34 34 Therefore we can, at least in theory, perform the integration in either order. However, there are situations where one order is preferable over the other. Evaluate (x1/2 y 2 )dxdy, where the domain consists of all points (x, y) with x 1 and x 2 y x 1/4, which is the same as the set of all points (x, y) with y 1 and y 4 x y 1/2. Carrying out the x-integral first, (x 1/2 y 2 )dxdy y1 xy 1/2 y y1 y xy 4 (x 1/2 y 2 )dxdy y1 y [ 2 3 x3/2 xy 2 ] xy1/2 xy y3/4 y 5/ y6 dy [ 8 21 y7/4 2 7 y7/ y7 ] y1 y whereas carrying out the y integral first, (x 1/2 y 2 )dxdy x1 yx 1/4 x x1 x which is clearly the same as above. yx 2 (x 1/2 y 2 )dy dx [x 1/2 y y 3 /3] yx1/4 yx 2 dx x1 x dy 2 3 x3/4 x 5/2 1 3 x6 /3 dx Remark 2 Finally, if, the domain of integration, is neither horizontally nor vertically simple, then it is usually possible to break it up into a finite number of domains, say 1,..., n, each of which is either horizontally or vertically simple. Then we can use the additivity property given above Evaluating Double Integrals Using Polar Coordinates Let be a domain formed with all points (x, y) that have polar coordinates (r, θ) in the set Γ : α θ β, ρ 1 (θ) r ρ 2 (θ) (5.12) where β α + 2π. Then f(x, y)dxdy f(r cos(θ), r sin(θ))r dr dθ Γ β ρ2(θ) α ρ 1(θ) f(r cos(θ), r sin(θ))r dr dθ. (5.13) Use polar coordinates to evaluate xy dxdy, where is the portion of the unit disc that lies in the first quadrant.

35 Course Notes 35 The region may be expressed in polar coordinates as r 1 and θ π/2. Thus we have π/2 π/2 xy dxdy r cosθr sin θr dr dθ cosθ sinθ[r 4 /4] 1 dθ 1 4 π/2 cosθ sin θ dθ 1 4 sin π/2 in which we used the substitution u sin θ with du cos θ dθ. u du 1 4 [u2 /2] Calculate the volume of the solid bounded above by the cone z 2 (x 2 + y 2 ) 1/2 and bounded below by the disc : (x 1) 2 + y 2 1. The region may be expressed as r 2 cosθ and π/2 θ π/2 [see 1.3]. The integral may be expressed as 2 (x 2 + y 2 ) 1/2 dxdy which then becomes 2 (x 2 + y 2 ) 1/2 dxdy π/2 2 cos θ π/2 π/2 π/2 (2 r)r dr dθ [4 cos 2 θ 8 cos 3 θ/3] dθ π/2 π/2 We use the fact that cos 2 θ 1 2 (1 + cos2θ) and substitute u sin θ, du cosθ dθ: π/2 π/2 2 (x 2 + y 2 ) 1/2 dxdy 4 cos 2 θ dθ 8 3 cos 3 θ dθ 5.3 Triple Integrals Properties (1) Volume property 2 π/2 π/2 π/2 1 + cos2θ dθ 8 3 π/2 1 [r 2 r 3 /3] 2cos θ dθ (1 u 2 )du 2[θ sin2θ]π/2 π/2 8 3 [u 1 3 u3 ] 1 1 2π T dxdy dz Volume of T. In particular if T is the box T [a, b] [c, d] [e, f] then T (2) Linearity T [αf(x, y, z) + βg(x, y, z)] dxdy dz α f(x, y, z)dxdy dz + β where α and β are constants. T dxdy dz (b a)(d c)(f e). T g(x, y, z)dxdy dz (5.14)

36 36 (3) Additivity If T is broken up into a finite number of nonoverlapping basic regions T 1,...,T n, then T f(x, y, z)dxdy dz f(x, y, z)dxdy dz T 1 f(x, y, z)dxdy dz. T n The Evaluation of Triple Integrals by Repeated Integrals (5.15) Let T be a solid whose projection onto the xy-plane is labelled xy. Then the solid T is the set of all points (x, y, z) satisfying (x, y) xy, χ 1 (x, y) z χ 2 (x, y). (5.16) The triple integral over T can be evaluated by setting ( ) χ2(x,y) f(x, y, z)dxdy dz f(x, y, z)dz dx dy. (5.17) T xy χ 1(x,y) In eq. (5.17) we can evaluate the integration with respect to z first and then evaluate the double integral over the domain xy as studied for double integrals. In particular if xy is horizontally simple, say then the solid T itself is the set of all points (x, y, z) such that a x b, φ 1 (x) y φ 2 (x). (5.18) a x b, φ 1 (x) y φ 2 (x), χ 1 (x, y) z χ 2 (x, y) (5.19) and the triple integral over T can be expressed by three ordinary integrals as: [ b ( φ2(x) ) ] χ2(x,y) f(x, y, z)dxdy dz f(x, y, z)dz dy dx. (5.2) T a φ 1(x) χ 1(x,y) Here we first integrate with z [from z χ 1 (x, y) to z χ 2 (x, y)], then with respect to y [from y φ 1 (x) to y φ 2 (x)], and finally with respect to x [from x a to x b]. There is nothing special about this order of integration. Other orders of integration are possible and in some cases more convenient. Suppose for example that the projection of T onto the xz-plane is a domain xz of the form z 1 z z 2, φ 1 (z) x φ 2 (z). (5.21) If T is the set of all (x, y, z) with then T z 1 z z 2, φ 1 (z) x φ 2 (z), ψ 1 (x, z) y ψ 2 (x, z) (5.22) f(x, y, z)dxdy dz z2 z 1 [ ( φ2(z) ) ] ψ2(x,z) f(x, y, z)dy dx dz. (5.23) φ 1(z) ψ 1(x,z) In this case we integrate first with respect to y, then with respect to x, and finally with respect to z. Still four other orders of integration are possible. Evaluate the triple integral 2 x 4 x 2 xyz dz dy dx.

37 Course Notes 37 2 x 4 x 2 xyz dz dy dx 2 x x 4 x 2 y z dz dy dx x x(4 x 2 ) 2 y dy dx 1 2 x 3 (4 x 2 ) 2 dx x 2 x y[ 1 2 z2 ] 4 x2 dy dx x(4 x 2 ) 2 [ 1 2 y2 ] x dx 16x 3 8x 5 + x 7 dx 1 4 [4x4 4x 6 /3 + x 8 /8] [64 256/3 + 32] 8/3 Use triple integration to find the volume of the tetrahedron T shown in the figure. z (,, 1) (, 1, ) y x (1,, ) The region of integration is bounded above by the plane x + y + z 1 so it may be expressed as Volume of T 1 2 z z z z y y x dxdy dz [(1 z)y 1 2 y2 ] 1 z y dz z z z z y 1 2 (1 z)2 dz (z 2 2z + 1)dz 1 2 [1 3 z3 z 2 + z] 1 z 1 6. (1 z y)dy dz Evaluating Triple Integrals Using Cylindrical Coordinates Let T be a solid whose projection onto the xy-plane is labelled xy. Then the solid T is the set of all points (x, y, z) satisfying (x, y) xy, χ 1 (x, y) z χ 2 (x, y). (5.24) The domain xy has polar coordinates in some set rθ and then the solid T in cylindrical coordinates is some solid S satisfying (r, θ) rθ, χ 1 (r cos(θ), r sin(θ)) z χ 2 (r cos(θ), r sin(θ)). (5.25)

38 38 Then T f(x, y, z)dxdy dz rθ xy ( ) χ2(x,y) f(x, y, z)dz dxdy χ 1(x,y) ( ) χ2(r cos(θ),r sin(θ)) f(r cos(θ), r sin(θ), z)dz r dr dθ χ 1(r cos(θ),r sin(θ)) Evaluate using cylindrical coordinates the triple integral 2 2 The region is the same as the region Substituting in the polar coordinates we have < z < 4 x 2 y 2 (4 x 2 ) 1/2 < y < (4 x 2 ) 1/2 2 < x < 2 S < z < 4 r 2 r 2 < 4. f(r cos(θ), r sin(θ), z)r dr dθ dz. (5.26) (4 x 2 ) 1/2 (4 x 2 ) 1/2 4 x 2 y 2 (x 2 +y 2 )dz dy dx. (x 2 + y 2 )dz dy dx 2π 2 4 r 2 2π 2 2π r 3 dz dr dθ r 3 (4 r 2 )dr dθ 2π 2 2π [r 3 z] 4 r2 z [r 4 r 6 /6] 2 dθ [16 32/3] dθ 2π[16 32/3] 32π/3 dr dθ Use cylindrical coordinates to find the volume of the solid T bounded above by the plane z y and below by the paraboloid z x 2 + y 2. The region x 2 + y 2 < z < y only exists where x 2 + y 2 < y, i.e. r 2 < r sinθ, which means r < sin θ which can only happen in the quadrants where sin θ is positive, θ π. Thus the integral may be written as V T π θ sin θ r sin θ r zr 2 r dz dr dθ which is easy to solve by repeated integration, giving the final answer V T π/32.

39 Course Notes Evaluating Triple Integrals Using Spherical Coordinates Let T be a solid in xyz-space with spherical coordinates in the solid S of ρθφ-space. Then f(x, y, z)dxdy dz f(ρ sinθ cosφ, ρ sin θ sin φ, ρ cosθ)ρ 2 sin θdρ dθ dφ. (5.27) T Evaluate T (x2 + y 2 + z 2 )dxdy dz, where T is the sphere of radius 1. We convert to spherical coordinates to have 2π (x 2 + y 2 + z 2 )dxdy dz T S 1 5 π φ θ ρ 2π π φ θ ρ 4 sin θ dρ dθ dφ sin θ dθ dφ 1 5 2π φ 2π π φ θ [ cosθ] π dφ 2 5 [ρ 5 /5] 1 sinθ dθ dφ 2π dφ 4 5 π. Find the volume of the solid T that is bounded above by the cone z 2 x 2 + y 2, below by the xy-plane, and on the sides by the hemisphere z (4 x 2 y 2 ) 1/2. The volume is the same as that discussed in 1.4, and can be expressed as T ρ 2 sin θ dρ dθ dφ ρ 2, π/4 < θ < π/2 and φ < 2π. 2π π/2 sin θ π/4 π/2 π/4 2 2π ρ 2 dφdρ dθ 2π π/2 π/4 π/2 sin θ[ρ 3 /3] 2 dθ 16 3 π sin θ dθ π/ π[ cosθ]π/2 π/ π(2 1/2 ) 8 3 π 2 sinθ 2 ρ 2 dρ dθ 5.4 Jacobians and changing variables in multiple integration During the course of the last few sections you have met several formulae for changing variables in multiple integration: to polar coordinates, to cylindrical coordinates, to spherical coordinates. The purpose of this section is to bring some unity to that material and provide a general description for other changes of variable Change of variables for double integrals Consider the change of variables x x(u, v) and y y(u, v), which maps the points (u, v) of some domain Γ into the points (x, y) of some other domain. Then The area of (x, y) (u, v) du dv. (5.28) Suppose now that we want to integrate some function f(x, y) over. If this proves difficult to do directly, then we can change variables (x, y) to (u, v) and try to integrate over Γ instead. Then f(x, y)dxdy f(x(u, v), y(u, v)) (x, y) (u, v) du dv. (5.29) Γ Γ