MATH 200 WEEK 10 - WEDNESDAY THE JACOBIAN & CHANGE OF VARIABLES

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1 WEEK - WEDNESDAY THE JACOBIAN & CHANGE OF VARIABLES

2 GOALS Be able to convert integrals in rectangular coordinates to integrals in alternate coordinate systems

3 DEFINITION Transformation: A transformation, T, from the uv-plane to the xy-plane is a function that maps (u,v) points to (x,y) points. x = x(u,v); y = y(u,v)

4 EXAMPLE Consider the region in the rθ-plane bounded between r=, r=2, θ=, and θ=π/2 The transformation T:x=rcosθ, y=rsinθ maps the region in the rθ-plane into this region in the xy-plane Transformations need to be one-to-one and must have continuous partial derivatives

5 JACOBIAN The area of a cross section in the xy-plane may not be exactly the same as the area of a cross section in the uvplane. We want to determine the relationship; that is, we want to determine the scaling factor that is needed so that the areas are equal.

6 IMAGE OF S UNDER T Suppose that we start with a tiny rectangle as a cross section in the uv-plane with dimensions u and v. The image will be roughly a parallelogram (as long as our partition is small enough). T: x=x(u,v), y=y(u,v) x and y are functions of u and v

7 Let s label the corners of S as follows A(u,v ) B(u + Δu,v ): B is a a little to the right of A C(u,v + Δv): C is a little higher that A D(u + Δu,v + Δv): D is a little higher and to the right of A

8 What happens to these points under the transformation T? Well, we have transformation functions x(u,v) and y(u,v), so we can plug the coordinates of the points A, B, C, and D to get the corresponding points in the xy-plane For the image of a point P under T, we ll write T(P) = P, so we get: A (x(u,v ),y(u,v )) B (x(u + Δu,v ), y(u + Δu,v )) C (x(u,v + Δv), y(u,v + Δv)) D (x(u + Δu,v + Δv), y(u + Δu,v + Δv))

9 We want to know how the area of R compares to the area of S. R is a parallelogram, so its area is the cross-product of the vectors a and b: a b a = A B = x(u + u, v ) x(u,v ),y(u + u, v ) y(u,v ), b = A C = x(u,v + v) x(u,v ),y(u,v + v) y(u,v ), WE VE ADDED THE Z- COMPONENT BECAUSE CROSS PRODUCTS ARE ONLY DEFINED FOR VECTORS IN 3-SPACE

10 To simplify these vectors down to something more manageable, we look to the limit definition of the partial derivative: df dx = lim x f(x + x) f(x) x FROM CALC f x = lim x f(x + x, y) f(x, y) x Notice that the first component of the vector a looks like the numerator of the limit definition of partial derivative of x with respect to u x x(u + u, v ) x(u,v ) u u IT S APPROXIMATE BECAUSE WE RE MISSING THE LIMIT

11 Applying this idea to both vectors, we get a = x(u + u, v ) x(u,v ),y(u + u, v ) y(u,v ), x u u, y u u, x u u, y u u, b = x(u,v + v) x(u,v ),y(u,v + v) y(u,v ), x y v, v v v, x v v, y v v,

12 Now we can take the cross product a b = i j k x u u y u u x v v y v v =[(x u u)(y v v) (x v v)(y u u)] k =(x u y v x v y u ) u v k For the area of R, we get a b = x u y v x v y u u v A R = x u y v x v y u u v

13 CONCLUSIONS We ve shown that the area of R is The area of S is A R = x u y v x v y u u v A S = u v So the scaling factor we were looking for is x u y v - x v y u We can write this as the determinate of a special matrix called the Jacobian Matrix: (x, y) (u, v) = ( ) xu x v y u y v

14 EXAMPLE Compute the Jacobian (i.e., the determinate of the Jacobian Matrix) for the transformation from the rθ-plane to the xy-plane { x(r, θ) =r cos θ T : y(r, θ) =r sin θ

15 T : { x(r, θ) =r cos θ y(r, θ) =r sin θ x r = cos θ x θ = r sin θ y r =sinθ y θ = r cos θ (x, y) (r, ) = x r x y r y v = cos r sin sin r cos = cos (r cos ) ( r sin )(sin ) = r cos 2 + r 2 sin 2 = r(cos 2 +sin 2 ) = r

16 THEOREM Given a transformation T from the uv-plane to the xy-plane, if f is continuous on R and the Jacobian is nonzero, we have R f(x, y) da xy = S f(x(u, v),y(u, v)) (x, y) (u, v) da uv For example, when converting from rectangular to polar, we have f(x, y) da xy = f(x(r, ),y(r, ))rda r R S

17 EXAMPLE Consider the following double integral: R x y da, R : region enclosed by y = x, y = x, y= x +,y= x +3 x + y R IT WOULD BE NICE IF WE COULD TRANSFORM THIS REGION INTO AN UPRIGHT RECTANGLE

18 Let s take the transformation T to be { u = x y T : v = x + y To convert our xy-integral to a uv-integral, we want the Jacobian (x, y) (u, v) = ( ) xu x v We could solve for u and v in our transformations OR we could use the convenient fact that y u y v (x, y) (u, v) = (u,v) (x,y)

19 First, we need the partial derivatives: u x =,u y =, v x =,v y = Plug it all in to the Jacobian Matrix (u, v) (x, y) = u x u y v x v y =2 Take the reciprocal to get = (x, y) (u, v) = 2

20 For the bounds, we have y = x = x y == u = y = x = x y == u = y = x += x + y == v = y = x +3= x + y =3= v =3 Finally, our integral becomes x y x + y da = R 3 u v 2 dudv

21 R x y 3 x + y da = = 2 = 2 = u v u 2 v u v dv = 4 ln v 3 2 dudv v dudv dv = 4 ln 3

22 EXAMPLE 2 Use the transformation x = v/u, y=v to evaluate the integral x y 2 Let s first convert the region from the xy-plane to the uv-plane The region extends from the line y= to the line y=x From x= to x= x 2 ey/x dydx

23 Apply the conversion formulas x=v/u and y=v y= becomes v= y=x becomes v=v/u u= x= becomes v/u= v= x= becomes v/u= v=u In summary, the region extends from the line v= to the line v=u Bounded by u= SO THE REGION LOOKS THE SAME IN THE UV-PLANE

24 For the Jacobian, we need the partials of x=v/u and y=v x u = -v/u 2 x v = /u y u = y v = (x, y) (u, v) = x u y u x v y v = v/u2 /u = v/u 2 Lastly, we need to convert the integrand (the function we re integrating) y 2 x 2 ey/x = y 2 (v)2 (v/u) 2 ev/(v/u) x 2 ey/x = u 2 e u Now we have all the pieces we need to set up the integral

25 x y 2 x 2 ey/x dydx = = = = u u u 2 e u 2 v2 e u v u 2 ve u dvdu u du 2 u2 e u du dvdu At this point we need to do integration by parts twice to finish the problem

26 2 u2 e u du = 2 u2 e u = 2 e 2 = 2 e 2 2 2ue u 2ue u du 2e 2e u 2e u du = 2 e e + 2 (2e 2) = 2 e

27 EXAMPLE 3 Let R be the region enclosed by xy =, xy = 2, xy 2 =, and xy 2 = 2. Evaluate the following integral. y 2 da R R

28 Let u=xy and v=xy 2 This gives us a nice rectangular region in the uv-plane u= to u=2 and v= to v=2 To find the Jacobian we ll need to solve for x and y in terms of u and v Solve each equation for x to get x = u/y and x = v/y 2 Setting those equal: u/y = v/y 2 Solve for y: y = v/u Plug into the u-equation: u=x(v/u) x = u 2 /v

29 Jacobian: (x, y) (u, v) = x u y u x v y v = 2u/v u2 /v 2 v/u 2 /u = 2 v v = v Setup: R y 2 da = 2 2 v 2 u 2 v dvdu

30 R y 2 da = = = = = = 3 2u 2 2 v 2 2u 2 v 2 u 2 v dvdu v u 2 dvdu 2 du 4 2u 2 2u 2 du 3 2u 2 du 2 = = 3 4