Change of Variables in Double Integrals



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Change of Variables in Double Integrals Part : Area of the Image of a egion It is often advantageous to evaluate (x; y) da in a coordinate system other than the xy-coordinate system. In this section, we develop a method for converting double integrals into iterated integrals in other coordinate systems. Let s suppose that T (u; v) hf (u; v) ; g (u; v)i is a - map of a region in the uv-plane onto a region in the xy-plane, and let s suppose that is contained in a rectangle [a; b] [c; d] on which both f and g are di erentiable. A partition of [a; b] into n subintervals of width u i and [c; d] into m subintervals of width v j covers with a collection of rectangles. If we require u i < h and v j < h for some su ciently small h > ; then the image of the rectangles under T is a collection of regions covering which are approximately parallelograms. If A ij is the area of the image of the ij th rectangle, i ; : : : ; n; and j ; : : : ; m; then the area of is approximately Area of X X A ij However, in chapter 3, we learned that the area of the image of a rectangle is approximately A ij @ (u; v) u i v j (u i ;vj )

where u i ; v j is a point in the ij-th rectangle. Thus, the area of is approximately Area of X X @ (u; v) u i v j (u i ;vj ) and in the limit as h approaches, the double sum leads to a double integral. Thus, we have Area of @ (u; v) da uv where da uv dvdu if is type I and da uv dudv if is type II. EXAMPLE Find and describe the image of [; ] [; ] under the transformation T (u; v) hv cos (u) ; v sin (u)i and then nd the area of that image. olution: ince x v cos (u) and y v sin (u) ; we have x 4 + y v cos (u) + v sin (u) v which because u ranges from to is an ellipse centered at the origin. Moreover, these ellipses completely cover the region inside the ellipse corresponding to v ; which is the ellipse with semimajor axis and semi-minor axis. Moreover, the Jacobian determinant is @ @ v cos (u) v sin (u) @ (u; v) @u @v v sin (u) v cos (u) v @ @ v cos (u) v sin (u) @v @u

ince is type I, it follows that Area of Z Z Z Z @ (u; v) da uv v dvdu v du du Check your eading: What is the image of [; ] [; ] under T (u; v) hv cos (u) ; v sin (u)i Part : Change of Variable in Double Integrals Let (x; y) be continuous on a region that is the image under T (u; v) of a region in the uv-plane. Then the double integral over is of the form X X (x; y) da xy lim x i ; y j Aij h! where da xy da is the area di erential in the xy-plane. If we choose u i ; v j such that x i ; y j f u i ; vj ; g u i ; vj ; then similar to the discussion in part, we have X X (x; y) da xy lim x i ; y j Aij h! X X lim f u i ; v j ; g u i ; v j h! @ (u; v) u i v j (u i ;vj ) (f (u; v) ; g (u; v)) @ (u; v) da uv where da uv is the area di erential on the uv-plane. 3

That is, is is continuous on which is the image under T (u; v) hf (u; v) ; g (u; v)i of a region in the uv-plane, then (x; y) da xy (f (u; v) ; g (u; v)) @ (u; v) da uv () The formula () is called the change of variable formula for double integrals, and the region is called the pullback of under T: In order to make the change of variables formula more usuable, let us notice that implementing () requires 3 steps:. (a) i. Compute the pullback of ii. Find the Jacobian and substitute for da xy iii. eplace x and y by f (u; v) and g (u; v), respectively. Finally, evaluate the resulting double integral over ; using da uv dvdu if is type I and da uv dudv if is type II. EXAMPLE Evaluate (x + y) da where is the region with boundaries y x; y 3x; and x + y 4 Use the transformation T (u; v) hu v; u + vi. olution: To begin with, T (u; v) hu v; u + vi is equivalent to x u v; y u + v: Thus, the pullback of the boundaries is as follows: y x ) u + v u v ) v ) v y 3x ) u + v 3u 3v ) 4v u ) v u x + y 4 ) u v + u + v 4 ) u 4 ) u 4

That is, is the region bounded by v ; v u; u : The Jacobian of the transformation is @ (y; v) @x @y @u @v @x @y @v @u ( ) Thus, da dvdu and () implies that ZZ ZZ (x + y) da xy (u v + u + v) da uv Z Z u 63 4u dvdu In example, we used the notation da to state the problem and da xy in working the problem. This re ects the convention that if working solely within the xycoordinate system, it is understood that da da xy : Check your eading: Is in example a type I or type II region? Part 3: Converting Type I and Type II integrals We can convert a type I or type II integral into di erent coordinates by rst converting into a double integral and then using (). EXAMPLE 3 Use the transformation T (u; v) hu; u + vi to evaluate the iterated integral Z Z x+3 x+ dydx p xy x 5

olution: To do so, we notice that as a double integral we have Z Z x+3 dydx p p da x+ xy x xy x where is the region bounded by x ; x ; y x + and y x + 3: ince T (u; v) hu; u + vi implies that x u; y u + v; we have y x + ) u + v u + ) v y x + 3 ) u + v u + 3 ) v 3 x ) u x ) u Thus, the pullback of is u ; u ; v ; v 3: To compute the Jacobian, we notice that x u; y u + v implies that @ (u; v) @x @y @u @v @x @y @v @u Thus, da dudv, so that we transform the double integral into p da p da uv xy x u (u + v) u As a result, we have Z Z x+3 x+ dydx p xy x Z Z 3 Z Z 3 Z Z 3 p u + uv p uv dudv u dudv u v dudv 4 p p 3 8 4 p 3 + 4 p EXAMPLE 4 Use T (u; v) u ; v to evaluate Z Z p x ye px dydx olution: The region of integration is bounded the curves x ; x ; y ; and y p x: ince x u and y v; we have x ) u x ) u y ) v y p x ) v u 6

Moreover, x u and y v implies that @ @ @ (u; v) @u u @v v Thus, da ududv; so that Z Z p x ye px dydx @ @v u Z Z u Z @ @u v u ye px da xy ve u u da uv u 3 e u du e + 6 ve u u dvdu Check your eading: What method was used in the nal step of example 4? Part 4: Change of Variable for Numerical Purposes Many algorithms for approximating double integals numerically require the use of special types of regions and sophisticated methods for partitioning those regions. In such instances, the change of variables formula is often used as a tool for transforming a double integral into a form that is more amenable to numerical approximation, such as converting to integration over a rectangular region since a rectangle can be easily partitioned. Most importantly, however, a change of variable might lead to the reduction of a double integral to a single integral, in which case only a single integral need be approximated numerically. In general, numerical methods for single integrals are preferable to numerical methods for multiple integrals. EXAMPLE 5 Transform the following using x v cosh (u) ; y v sinh (u) : Z p 3 Z p y +9 sin x y x y dydx y Evaluate one of the integrals, and then approximate the remaining integral numberically. 7

olution: To do so, we notice that as a double integral we have Z p 3 Z p y +9 sin x y x y dxdy y sin x y x y da xy where is the region bounded by y, x y; and x y + 9: picture The pullback of to is accomplished by letting x v cosh (u) and y v sinh (u): y ) v sinh (u) ) v or sinh (u) ) v or u x y ) v cosh (u) v sinh (u) ) tanh (u) :5 ) u tanh (:5) x y + 9 ) v cosh (u) v sinh (u) 9 ) v 9 Thus, the pullback of is u ; u tanh () ; v ; v 3: We next compute the Jacobian determinant: @ (u; v) @x @y @x @y @u @v @v @u v sinh (u) sinh (u) cosh (u) v cosh (u) v cosh (u) sinh (u) v Thus, da xy j vj da uv vda uv since v is inside of : ince x y v ; the double integral becomes sin x y x y da xy sin v v Z 3 Z tanh (:5) Z 3 sin v u v tanh (:5) Z 3 v da uv sin v dudv v tanh (:5) dv sin v dv v Finally, the remaining integral can be estimated numerically so that Z p 3 Z p y +9 sin x y x y dydx tanh (:5) y Z 3 sin v dv :4573 v 8

Exercises Find and describe the image of the given region under the given transformation. Then compute the area of the image.. T (u; v) hu; 4vi. T (u; v) hu + 3; vi [; ] [; ] [; ] [; ] 3. T (u; v) hu; u + vi 4. T (u; v) hu v; u + vi [; ] [; ] [; ] [; ] 5. T (u; v) u v ; uv 6. T (u; v) hu v; uvi [; ] [; ] [; ] [; ] 7. T (u; v) h4u cos (v) ; 3u sin (v)i ; 8. T (u; v) hu cosh (v) ; u sinh (v)i ; [; ] [; ] [; ] [; ] Use the given transformation to evaluate the given iterated integral. 9. xy x + dydx. xy sin y dydx T (u; v) h p u; vi T (u; v) hu; p vi. 3. ex cos (e x ) dxdy. T (u; v) hln (u) ; vi x cos x dydx 4. T (u; v) hv; ui cos (y) esin(y) dydx T (u; v) u; sin (v) x cos y dydx T (u; v) hv; ui 5. 7. x+ x y 4x dydx 6. x+ p x y x dydx T (u; v) hu; u + vi T (u; v) hu; u + vi y+ y sin (y x) dxdy 8. y+ y T (u; v) hu + v; vi pdxdy xy y T (u; v) hu + v; vi 9.. y sin x dxdy. T (u; v) hu; uvi x dydx x ex T (u; v) hu; uvi p p xyda;. xy3 da; T (u; v) u v ; uv T (u; v) u v ; uv bounded by xy ; xy 9 bounded by xy ; xy 9 y x; y 4x y x; y 4x Use the given transformation to transform the given iterated integral. Then 9

reduce to a single integral and approximate numerically. h x 3. x cos (x y) i dydx 4. x x cos h (x + y) i dydx T (u; v) hv + u; v ui T (u; v) hv + u; v ui 5. p x sin(x +y ) x +y dydx 6. T (u; v) hv cos (u) ; v sin (u)i p :5 x sin(x +4y ) x +4y dydx T (u; v) hv cos (u) ; v sin (u)i 7. y 4 y 4 ln 6x 8xy + y 4 y dxdy 8. 4 8xy +y 4 dxdy y 4 e6x T (u; v) u v ; uv T (u; v) u v ; uv 9. Evaluate the following iterated integral in two ways: Z Z 3 x cos x dxdy. (a) By letting w x ; dw xdx and noting that x implies w while x 3 implies w 9: (b) Using the coordinate transformation T (u; v) p u; v 3. Evaluate the following iterated integral in two ways: Z Z 4 p x cos p x dxdy. (a) By letting w p x; dw dx ( p x) and noting that x implies w while x 4 implies w : (b) Using the coordinate transformation T (u; v) u ; v 3. In two di erent ways, we show that if g is a di erentiable function, then Z b Z g(d) a g(c) (x; y) dydx Z b Z d a c (u; g (v)) g (v) dvdu. (a) By letting y g (v) ; dy g (v) dv in the rightmost iterated integral. (b) By considering the coordinate transformation T (u; v) hu; g (v)i : 3. In two di erent ways, we show that if f is a di erentiable function, then Z f(b) Z g(d) f(a) g(c) (x; y) dydx Z b Z d a c (f (u) ; g (v)) f (u) g (v) dvdu

. (a) By letting x f (u) ; dx f (u) du and y g (v) ; dy g (v) dv in the rightmost iterated integral. (b) By considering the coordinate transformation T (u; v) hf (u) ; g (v)i 33. The parabolic coordinate system on the xy-plane is given by T (u; v) u v ; uv If T maps a region in the uv-plane to a region in the xy-plane, then what does the change of variable formula imply that (x; y) da will become when integrated over in the uv-coordinate system? 34. The tangent coordinate system on the xy-plane is given by u T (u; v) u + v ; v u + v If T maps a region in the uv-plane to a region in the xy-plane, then what does the change of variable formula imply that (x; y) da will become when integrated over in the uv-coordinate system? 35. The elliptic coordinate system on the xy-plane is given by T (u; v) hcosh (u) cos (v) ; sinh (u) sin (v)i If T maps a region in the uv-plane to a region in the xy-plane, then what does the change of variable formula imply that (x; y) da will become when integrated over in the uv-coordinate system? 36. The bipolar coordinate system on the xy-plane is given by sinh (v) T (u; v) cosh (v) cos (u) ; sin (u) cosh (v) cos (u) If T maps a region in the uv-plane to a region in the xy-plane, then what does the change of variable formula imply that (x; y) da will become when integrated over in the uv-coordinate system?

37. Write to Learn: The translation T (u; v) hu + a; v + bi translates a region in the uv-plane to a region in the xy-plane which is translated a units horizontally and b units vertically. Write a short essay which uses the change of coordinate formula to show that the area of is the same as the area of? 38. Write to Learn: A rotation T (u; v) hcos () u + sin () v; sin () u + cos () vi maps a region in the uv-plane to a region in the xy-plane which is rotated about the origin through an angle. Write a short essay which uses the change of coordinate formula to show that the area of is the same as the area of?