Functions of Several Variables

Transcription

1 Chapter 1 Functions of Several Variables 1.1 Introduction A real valued function of n variables is a function f : R, where the domain is a subset of R n. So: for each ( 1,,..., n ) in, the value of f is a real number f( 1,,..., n ). For eample, the volume of a clinder: V πr h (i.e. V F (r, h)) is a function of two variables. If f is defined b a formula, we usuall take the domain to be as large as possible. For eample, if f is a function defined b f(, ) 9 cos () + sin( + ), we have a function of variables defined for all (, ) R. So R. However, if f is defined b f(,, z) z. Then f is a function of 3 variables, defined whenever + + z. This is all (,, z) R 3 ecept for (,, z) (,, ). Likewise, a multivariable function of m variables is a function f : R n, where the domain is a subset of R m. So: for each ( 1,,..., n ) in, the value of f is a vector f( 1,,..., n ) R n. For eample: 1. An object rotating around the origin in the -plane (sa at distance 5 from the origin) will have its position described b the function f(t) (5 cos (t), 5 sin (t)). This is a function from R to R.. An object spiralling around the ais (again at distance 5 from the ais) and travelling at constant speed might have its position given b f(t) (t, 5 cos (t), 5 sin (t)). This function is from R to R Surfaces The graph of a function f : R R, is the set {(,, z) R 3 : z f(, )}. This is a surface in R 3. The height z f(, ) over each point gives the value of f. The equation a + b + cz d represents a plane in R 3. This equation can be written less elegantl, epressing z as a function of and, as z a c b c + d, provided c. c When looking at functions of one variable f() it is possible to plot (, ) points to determine the shape of the graph. In the same wa, when looking at a function of two variables z f(, ), it is possible to plot the points (,, z) to build up the shape of a surface.

2 Functions of Several Variables Eample 1.1 raw the graph (or surface) of the function: z 9 (a circular paraboloid). For, z 9,, 1 z 8, 1, z 8, 1, 1 z 7, etc.... We can eventuall plot enough points to find the surface in Figure 1.1. z Figure 1.1: The paraboloid z 9. This method of drawing a surface is time consuming as we need to calculate man points before being able to plot the graph. Also some surfaces are quite comple and difficult to draw. We would like another wa of representing the graph so that the surface is easier to draw or visualize Contours and level curves Three dimensional surfaces can be depicted in two dimensions b means of level curves or contour maps. If f : R R is a function of two variables, the level curves of f are the subsets of : {(, ) : f(, ) c}, where cconstant. If f height, level curves are contours on a contour map. If f air pressure, level curves are the isobars on a weather map. The graph of f can be built up from the level sets: The slice at height z c, is the level set f(, ) c. Eample 1. For the elliptic paraboloid z +, for eample, the level curves will consist of concentric circles. For, if we seek the locus of all points on the paraboloid for which z 1, we solve the equation 1 + which is of course an equation of a circle. The locus of points units above the plane is just the origin, for if z, the equation becomes +, and this implies that. Eample 1.3 For the hperbolic paraboloid z the level curves are hperbolae ecept for which is the union of two lines.

3 1.1 Introduction z Figure 1.: Level curves of the elliptic paraboloid z Figure 1.3: Level curves of the hberbolic paraboloid z Partial derivatives These measure the rate of change of a function with respect to one of the variables, keeping all other variables fied. Let f : R R be a function of two variables, and (, ). efinition 1.1 The partial derivative of f with respect to is: f ( f( +, ) f(, ), ) lim (i.e. differentiate f with respect to, treating as a constant). The partial derivative of f with respect to is: f ( f(, + ) f(, ), ) lim (i.e. differentiate f with respect to, treating as a constant). Provided, of course, the limits eist. Referring to Figure 1.4, let us intersect the surface b a vertical plane constant, sa, to give a curve of intersection AP B. Then f (, ) gives the slope of the tangent P T to this curve at the point (,, z ).

4 4 Functions of Several Variables z tangent line B surface: zf(,) p A (, ) plane Figure 1.4: Tangent to surface in the direction. z surface: zf(,) q B C tangent line (, ) plane Figure 1.5: Tangent to surface in the direction.

5 1.1 Introduction 5 Similarl, f (, ) gives the slope of the tangent Q T to the curve of intersection BQ C of the surface with a vertical plane (see figure 1.5). Therefore f f and give the slope of a surface at a point in the directions of the and aes respectivel. Similarl, for a function f of n variables 1,... n we can define partial derivatives, f f 1, f f,..., f f n. 1 n Eactl the same rules of differentiation appl as for a function of one variable. If we have a function of two variables f(, ) we treat as a constant when calculating f, and treat as a constant when calculating f Higher partial derivatives are themselves functions of two variables, so the can also be partiall differenti- Notice that f ated. and f For a function of two variables f : R there are four possiblilties: f ( ) f f, f ( ) f f, f ( ) f f, f ( ) f f. Higher order partial derivatives are defined similarl: Usuall, but certainl not alwas, 3 f ( f f ( )) f f. For the functions we will be encountering the mied partial derivatives will generall be equal. In fact, this is true whenever f and f are continuous. Eample 1.4 Find all the second partial derivatives of Solution: f(, ) 3 e + cos () f 3 e sin (), f 3 e 3 cos (),

6 6 Functions of Several Variables ( ) f ( 3 e sin () ) 6e cos (), ( ) f ( 3 e 3 cos () ) 4 3 e cos (), ( ) f ( 3 e 3 cos () ) 6 e + 3 sin (), ( ) f ( 3 e sin () ) 6 e + 3 sin (). f f f f ifferentiable functions For a function of one variable f() is differentiable at means: 1. The graph of f has a tangent line at (see Figure 1.6). The equation of this tangent line is f() tangent line Δf errorf( +Δ)-f()-f () Δ Δ aaaaaaaaaaa +Δ Figure 1.6: A differentiable function has a tangent line. f ( )( ).. The change in f near is well approimated b a linear function: f f ( ) + error, where the error is small compared with or more precisel: error as. These ideas can be generalized to a function of two variables. A function f is differentiable at (, ) means that f has a well defined tangent plane at (, ). All tangents to the surface at (, ) lie in the one plane (see Figure 1.7). Or more formall efinition 1. A function f(, ) if differentiable at (, ) if the change in f near (, ) is well approimated b a linear function: where f f (, ) + f (, ) + error, error as (, ) (, ). Here ( ) + ( ) f f( +, + ) f(, ). The equation of the tangent plane to z f(, ) at (, ) is z z f (, ) + f (, )

7 1.1 Introduction 7 normal line tangent plane p M Figure 1.7: The tangent plane and normal line to the surface z f(, ). or z f(, ) + f (, )( ) + f (, )( ). Eample 1.5 Find the cartesian equation of the tangent plane to the surface z 1 at the point (1,, 4). Solution: We first find the partial derivatives At (1,, 4) Therefore b using the formula the tangent plane is z, z, z, z 4. z 4 + ( 1)( ) + ( )( 4), , z 6 4. If we rearrange this equation into the usual form for a plane we have f (, ) + f (, ) z f(, ) + f (, ) + f (, ) so the normal vector for the tangent plane is n (f (, ), f (, ), 1). From this we can find the equation to the normal line to the tangent plane at (, ) as or f (, ) f (, ) z z 1 (,, z) (,, z ) + t (f (, ), f (, ), 1).

8 8 Functions of Several Variables Errors and approimations If the function z f(, ) is differentiable at (, ), then for (, ) near (, ) f(, ) f(, ) + f (, ) + f (, ). where,. The right hand side of this equation is called the linear approimation to f near (, ). This equation ma also be written as which is sometimes written as which is called the differential of f. f f + f df f d + f d Some applications of this linear approimation to f are 1. Estimating values of functions.. Estimating errors in measurements. Eample 1.6 Estimate the value of f(, ) 1 + when.1,.. Solution: Use the linear approimation to f at (, ): f f 1 (1 + ) 1/ ( 1) f f 1 (1 + ) 1/ () At (, ) : f 1, f 1. The linear approimation for (, ) near (, ) is Taking.1,. gives f(, ) f(, ) + f (, )( ) + f (, )( ) f(.1,.) 1 1 (.1) + (.) 1.15 The size of the error depends on the second order derivatives of f. 1. Space Curves 1..1 Vector valued functions A curve in R n is a vector valued function c : [a, b] R n where c(t) (c 1 (t), c (t),..., c n (t)). You can think of c(t) as the position at time t. For eample the vector function c(t) (cos (t), sin (t)) describes the unit circle in R. This is equivalent to the parametric equations cos (t), sin (t), t R.

9 1. Space Curves 9 The derivative of c is dc dt c (t) (c 1(t), c (t),..., c n(t)) that is we differentiate each component of c. The derivative can also be defined as the limit c c(t + t) c(t) (t) lim. t t The derivative gives the tangent vector to the curve at the point c(t). If c(t) represents the position at time t, then c (t) dc dt and c (t) d c dt at time t. velocit vector, acceleration vector, For eample if c(t) (cos (t), sin (t)) represents the unit circle in R then c (t) ( sin (t), cos (t)) velocit, c (t) ( cos (t), sin (t)) acceleration. + 1 c (t) c (t) c(t) t aaa Figure 1.8: The velocit and acceleration of c(t) (cos (t), sin (t)). 1.. The chain rule If is a differentiable function of i.e. f() and is a differentiable function of t i.e. g(t), then is a function of t d f(g(t)) and dt d d d dt If z is a function of and i.e. z f(, ) and and are both functions of the same variable t, i.e. f f(, ) and g(t), h(t)

10 1 Functions of Several Variables then z is a function of t: For eample, if z, z f ( g(t), h(t) ). sin (t) and cos (t) then z sin (t) cos (t). We would like to be able to find the rate of change of f with respect to t. This can be found from the chain rule for two variables df dt f d dt + f d dt. Eample 1.7 If z, sin (t), cos (t), find dz dt where t π 3. Solution: We can do this b two methods the second can be used to check our first answer. Method 1. Using the chain rule At t π 3, Therefore dz dt z, d d cos (t), dt sin(t) z, sin (t), dt 3 and cos(t) 1. z d dt + z d dt ()(cos (t)) + ( )( sin (t)) cos (t) + sin (t) Method. Using substitution to find z z(t) Therefore dz dt z sin (t) cos (t). sin (t) cos (t) cos (t) ( sin (t)) 4 sin (t) cos (t) 4 which is the same answer we obtained b method 1. The chain rule can be used for functions of more than two variables: 3 1 3, Given a function f( 1,,..., n ) defined at points of R n, consider the values of f along a curve 1 1 (t), (t),..., n n (t). Here t R is a parameter along the curve (e.g. time or arc length). Let w f( 1,,..., n ) function of t. If f, 1,,..., n are differentiable, then where each w i is evaluated at ( 1,,..., n ). dw dt w d 1 dt +... w d n n dt

11 1.3 Gradient Vectors and irectional erivatives Gradient Vectors and irectional erivatives Gradient and Gradient vector Consider the function of two variables f(, ). Its graph represents a surface in three dimensions. If and are themselves a function of another variable t then ((t), (t)) is a curve C c(t) in the plane. The function f((t), (t)) will then represent a curve on the surface of f(, ) directl above curve C c(t) in the plane. z curve on surface zf(,) c(t), curve in - plane Figure 1.9: The curve on the surface of z f(, ) directl above the curve C c(t) in the plane. The chain rule epresses df along the curve C as the dot product of the two vectors dt then v ( d dt, d ) dt and f ( f, f ). v f f d dt + f d dt df dt i.e. the chain rule. The two vectors v and f have onl two components therefore the lie in the plane. v is tangent to the curve ((t), (t)) and is called the tangent vector. f is called the gradient vector of f. These ideas can also be etended for functions of more than two variables. efinition 1.3 If f( 1,,..., n ) is a function of n variables then the gradient of f is the vector valued function ( f f( 1,,..., n ),..., f ). 1 n Notice that if f : R n R, then f : R n R n. This is an eample of a vector field on R n.

12 1 Functions of Several Variables 1.3. irectional derivatives The partial derivatives f and f give the rate of change of f in directions parallel to the and ais. What about other directions? Let P (, ) be a fied point in the plane. Let l be a line in the plane that passes through P. The point P (, ) moves along the line l. irectl above it, the point Q moves along the surface z f(, ), tracing out a curve C. What is the rate at which the z coordinate of Q changes with the distance s between P and P? z curve on surface Q zf(,) Q s P line in - plane P Let u be the unit vector Figure 1.1: Curve on the surface of z f(, ) above the line l. u u 1 i + u j with initial point P (, ) and pointing in the direction of motion of P (, ). Then P P su ( )i + ( )j su 1 i + su j therefore Hence + su 1 and + su. z f(, ) f( + su 1, + su ). B the chain rule, dz ds z dw ds + z d ds f u 1 + f u f u.

13 1.3 Gradient Vectors and irectional erivatives 13 Thus, the rate of change of f(, ) at (, ) in the direction of the unit vector u is given b the directional derivative u f(, ) f(, ) u This can be etended to functions of n variables. Let f( 1,..., n ) be a differentiable function of n variables, u be a unit vector in R n and P be a point in R n. efinition 1.4 The directional derivative of f in the direction of u at P is: Taking (t) P + tu, d dt u f(p ) lim t f(p + tu) f(p ) t d dt f(p + tu) t rate of change of f in the direction of u. u in the chain rule u f d d f((t)) f f u. dt dt Thus u f(p ) f(p ) u. Eample 1.8 Find the rate of change of at the point (1, ) in the direction of the vectors: (i) a: a unit vector 45 to the ais, (ii) b (, 1). f From the previous eample the gradient vector at (1, ) is f ( 1, ) (i) Unit vector: u 1 a a 1 (1, 1). Therefore the directional derivative is u f 1 (1, 1) ( 1, ) 1 (ii) Unit vector: u b (, 1). Therefore the directional derivative is i.e. there is no change in f in this direction. u f (, 1) ( 1, )

14 14 Functions of Several Variables irections of fastest increase and decrease Given f( 1,..., n ) a differentiable function of n variables, and a point P in R n, we can consider u f as a function of the unit vector u. If f, then the directional derivative atp has a maimum value of f in the direction of f. That is the maimum rate of increase in f is f in the direction of u f f. The directional derivative at P has a minimum value of f in the direction of f. That is the maimum rate of decrease of f at P is given b f and is in the dirction of f or u f f Level curves and gradient Let (t), (t) be a level curve of f(, ) so that Then b the chain rule So f((t), (t)) const. df dt f d dt + f d dt. f v hence f(, ) is normal to the level curve at (, ) Figure 1.11: The gradient vector f is perpendicular to the level curves.

15 1.3 Gradient Vectors and irectional erivatives 15 Eample 1.9 Find the level curve of f corresponding to f 1 and sketch the gradient vector at the points (1, ), ( 1, ). Solution: The level curve is 1 or ± 1 The gradient vector is sketched in Figure 1.1. f (, ), at (1, ) f (, ) i, ( 1, ) f (, ) i. 4 f f Figure 1.1: The level curves and gradient vector f of f. In general, if f( 1,..., n ) is differentiable at a point P and f(p ) c, then f(p ) is perpendicular to the level set f() c containing P Tangent planes to surfaces Theorem 1.5 If F (,, z) is differentiable, F (P ), and F (P ) c, then the level set F (,, z) c has well defined tangent plane at P which is orthogonal to F (P ). If F or F is not defined at P, there generall won t be a nice tangent plane. For eample the point at the tip of the cone in Figure aaaa Figure 1.13: A point where there is no nice tangent plane. Therefore as F is orthogonal to the tangent plane, the equation for the tangent plane at P is given b F (P ) ( P )

16 16 Functions of Several Variables and the equation for the normal line S {(,, z) : F (,, z) c} at P is (t) P + t F (P ). These equations also work in the n dimensional case where S hpersurface : F ( 1,..., n ) c (dimension n 1). In the three dimensional case for F (,, z) constant these equations can also be written: The tangent plane at P (,, z ) is f (P )( ) + f (P )( ) + f z (P )(z z ). The normal line at P is (t) + t f (P ) (t) + t f (P ) z(t) z + t f (P ). z 1.4 Etreme Values and Saddle Point Maima and minima: Critical points Let f( 1,..., n ) f() be a function of n variables. Let ( 1,..., n ) R n. efinition 1.6 f has a local maimum at if f() f( ) for all near. f has a local minimum at if f() f( ) for all near. Theorem 1.7 If f( 1,..., n ) has a local maimum or local minimum at P, then either: (i) f f... f at P (i.e. f(p ) ), or 1 n (ii) One or more of the partial derivatives f 1,..., f n does not eist at P. efinition 1.8 We sa P is a critical point of f if f(p ). Note: Not all critical points are local maima and minima. e.g. (i) For the function f(, ) + the critical points satisf: so f and f and. Therefore and is the onl critical point of f and this is a global minimum. (ii) For the functions such as f(, ) the critical points satisf: f, f so,. Therefore, is the onl critical point of f and this is not a local maimum or local minimum but a saddle p

17 1.4 Etreme Values and Saddle Point 17 z Figure 1.14: Paraboloid f(, ) +. aaaa Figure 1.15: Tpical saddle point. Eample 1.1 Find the critical point(s) of the function: g(, ) Solution: We need to solve g [ ] and [ g ]. Therefore and 5 5 1, Therefore there is a critical point at (, 1).

18 18 Functions of Several Variables 1.4. Classification of critical points Let (, ) be a critical point of f(, ) so that f (, ) f (, ) and assume f has continuous second order derivatives near (, ). [ ] f f H, evaluated at ( f f, ) X, and is called the Hessian of f at (, ). In order to determine the tpe of critical points we need to look more closel at H. Second erivative Test: Let (, ) be a critical point of f(, ) so that f (, ) f (, ) and assume f has continuous second order derivatives near (, ). Then there are four cases depending upon det(h) and f (, ): 1. If det (H) f f > at (, ) and f (, ) >, then a minimum occurs at (, ). f f. If det (H) > at (, ) and f (, ) <, then a maimum occurs at (, ). 3. If det (H) < then a saddle point occurs at (, ). 4. If det (H) then the test is inconclusive and the nature of the critical point (, ) can t be determined from the second derivatives. In order to determine the tpe of critical point it is necessar to stud higher order derivatives. The functions f(, ) 4 + 4, f(, ) 4 4, f(, ) 4 4 all have a critical points at (, ) with det H. But these points are a local minimum, a local maimum and a saddle point respectivel. Eample 1.11 Find and classif the critical points of Solution: First finding the critical points f(, ) f, f, ( 1), and, or 1 and. Therefore for, and for 1, 1 ±1. Thus we have three critical points (, ), (1, 1) and ( 1, 1). Checking for tpe of critical points f, f, f. [ H ] At (, ), det H ( )() therefore the test is inconclusive. At (1, 1), det H ()( ) 4 therefore (1,1) is a saddle point. At ( 1, 1), det H ()( ) ( ) 4 therefore (-1,1) is a saddle point Eistence of maima and minima In general, global maima and global minima need not eist.

19 1.4 Etreme Values and Saddle Point 19 For functions of one variable: A continuous function f : [a, b] R defined on a closed, bounded interval has a global maimum and global minimum in [a, b]. If f is differentiable, the maimum or minimum ma occur at a critical point or boundar point a or b. Similarl for functions of n variables: If f : R n R is continuous, and K is a closed, bounded subset of R n, then f has a global maimum and minimum on K. If f is differentiable then the maimum and minimum ma occur at a critical point or boundar point of K. (The proof of this can be found in real analsis tetbooks.) efinition K R n is bounded if it lies within a finite distance from the origin. (i.e. There eists c, such that < c for all K.). R n is a boundar point of K if there are points arbitraril close to which are in K, and points arbitraril close to which are not in K (see Figure 1.16). 3. K is closed if it contains all its boundar points. interior points aaaa boundar points Figure 1.16: Boundar points in a set. Eamples: 1. Closed disc + r is closed and bounded (see Figure 1.17). boundar points circle +r Figure 1.17: Closed disc + r.. Open disc + < r is bounded but not closed (see Figure 1.18). Figure 1.18: Open disc + < r. 3. Half plane {(, ) : } is closed but not bounded (see Figure 1.19).

20 Functions of Several Variables 1.5 Multiple Integrals One Variable Recall from that Figure 1.19: Half plane {(, ) : }. b a f() d gives the area between the graph f() and the -ais for a b (if f() ). We approimate the area under the graph b rectangular strips. f() f( i ) a i b curaa Δ i Figure 1.: ividing a region into rectangular strips. efinition 1.1 The sum is called a Riemann sum. Then b a f() d f( i ) i i lim ma i ( ) f( i ) i. i Question How can we calculate the volume of the region in R 3 ling above a set in the plane (z ) and below the graph z f(, )? One approach is to divide the region into small rectangular boes and add up the volumes of all the boes. Then take a limit. The result will be called integral of f over, and written f da.

21 1.5 Multiple Integrals 1 z zf(,) ( i, i, f( i, i )) volume area height ΔA i f( i, i ) curaa ΔA i Δ i Δ i Figure 1.1: ividing a region into rectangular boes rectangle R i point ( i, i ) curaa Figure 1.: ividing into rectangles efinition of double integral Let be a bounded region in R, f : R a function of two variables defined on. efine the integral of f over using Riemann sums in the following wa. 1. Cover b a rectangular grid.. Let A 1, A,..., A k be the rectangles inside. Then 3. Now look at Riemann sums A i area of A i i i ( i, i ) point inside A i. k f( i, i ) A i i1 k f( i, i ) i i i1 Let A be the maimum length of edges in the rectangles A i. Theorem 1.11 If f is continuous on, and is bounded b curves of finite total length, then all such Riemann sums approach the same limit provided A c.

22 Functions of Several Variables So we can define f da lim A i1 k f( i, i ) A i. This is called Riemann integral of f over the region, also written f d d Properties of double integrals (f + g) da cf da c f da f da + g da. f da, where c is a constant. g da, if f g at all points of. f da f da + 1 f da, if is the union of two non overlapping regions 1 and. 1 e.g Interpretations of double integrals 1. If f : R is continuous, f(, ) then in the plane and below the graph of f.. If f(, ) 1 for all, then we obtain the area of. Area() 1 d d. 3. Integral of densit is the total mass Integral of charge densit is the total charge. f da is the volume of solid ling above region ouble Integrals over Rectangular Regions The partial derivatives of a function f(, ) are calculated b holding one of the variables fied and differentiating with respect to the other variable. Consider the reverse of differentiation, partial integration. The smbol b a f(, ) d is a partial definite integral with respect to. It is evaluated b holding fied and integrating with respect to.

23 1.5 Multiple Integrals 3 This being the case, we can consider the following tpes of calculations: [ d ] b f(, ) d d which are respectivel written c b a a [ ] d f(, ) d d c d b c a b d a c f(, ) d d f(, ) d d These are eamples of iterated (in this case, double) integrals. Eample 1.1 Use a double integral to find the volume of the solid that is bounded above b the plane z 4 and below b the rectangle R {(, ) : 1, } Solution: The volume we want is shown in Figure 1.3. The volume is given b 4 z z4 1 Figure 1.3: Volume under the plane z 4. V 1 (4 ) d d [ 4 1 ( ) 7 d d [ 7 1 ] 1 5 ] 1 d Alternativel, V 1 (4 ) d d 5.

24 4 Functions of Several Variables ouble Integrals over Non-rectangular Regions The regions involved in double integrals can be divided into groups according to their boundaries. Tpe 1 Region: g () g 1 () aaaaaaaaaaa a We see that is the region defined b g 1 () g () where a b. b Here i.e. f da b g() a g 1() f(, ) d d 1. Integrate f(, ) with respect to, keeping fied.. Integrate the result with respect to from a to b. Tpe Region: d h () h 1 () c aaaaaaaaaaa N ow is the region defined b h 1 () h () where c d. Here i.e. f da d h() c h 1() f(, ) d d 1. Integrate f(, ) with respect to, keeping fied.. Integrate the result with respect to from c to d. Eample 1.13 Evaluate ( 3 + 4) da where is the region enclosed b the graphs and.

25 1.5 Multiple Integrals 5 Solution: First, find the intersection points of these curves and sketch. ( ) or, 4 respectivel This can be evaluated in two was: 1. As a tpe region where the region is described as, So that the integral becomes ( 3 + 4) da ( ) ( 3 + 4) d d [ 3 + ] d ( ) d ( 8 5) d [ ] 3 3. As a tpe 1 integral 1, 4. The integral is then (using opposite order of integration) ( 4 ) ( 3 + 4) d d Changing the Order of Integration As we have seen double integrals can be evaluated b integrating with respect to and then or with respect to and then. It ma be that one of these methods gives a simpler epression to integrate and we ma wish to change the order of integration to take advantage of this. Eample 1.14 Evaluate the integral 1 1 e d d Solution: We can t evaluate the inner integral 1 e d eplicitl. So we need to convert this to a double integral and change the order of integration. First find the region of integration.

26 6 Functions of Several Variables We have 1, 1. So is the triangle bounded b below, b 1 on the right, and b as the hpotenuse on the left. We can rewrite the region as Therefore the integral is, 1. 1 e d d This is a much simpler integral since can be considered a constant in the inner integral. Therefore we have 1 e d d 1 1 [e ] d ( e ) d This can now be done using the substitution u and du d. Therefore e d 1 eu du e u e [ ] 1 1 e 1 (e 1) Triple Integrals Just as a double integral can be evaluated b two single integrations, a triple integral can be evaluated b three single integrations. If a region in R 3 is defined b the inequalities and f is continuous on the region, then a b c d k z l b d l a c k d b l c a k b l d a k c l b d k a c l d b k c a d l b c k a f(,, z) dz d d f(,, z) dz d d f(,, z) d dz d f(,, z) d d dz f(,, z) d d dz f(,, z) d dz d Eample 1.15 Evaluate 1 z 3 dv over the region G defined b the inequalities G 1 3 z

27 1.5 Multiple Integrals 7 Solution: There are 6 possible forms of the integral to use. We choose z 3 dz d d [ 3 z 4] d d 48 d d 43 d [ 16 ] The other 5 possible forms give the same result. We can also evaluate triple integrals over more general regions. In fact if f(,, z) 1 then the triple integral f(,, z) dv will give the volume of the region G. G

28 8 Functions of Several Variables