Change of Variables. (f T) JT. f = U

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1 Change of Variables The change of ariables formla for mltiple integrals is like -sbstittion for single-ariable integrals. I ll gie the general change of ariables formla first, and consider specific cases in lower dimensions below. Theorem. Let U be an open set in n. Let T : U n be a fnction which satisfies: (a) T is injectie: If T(x) = T(y), then x = y. (b) T has continos partial deriaties. (c) The Jacobian JT = detdt(x) is nonzero for all x U. Sppose f : T(U) is an integrable fnction. Then T(U) f = U (f T) JT. The conditions on T ensre that the change of ariables is nicely behaed for example, we don t hae the region U dobling p on itself. The term f T is the composite of f and T: Yo change ariables in the fnction yo re integrating, sbstitting sing the transformation T. The term detdt is roghly the factor by which olmes are expanded by T. For instance, in conerting from rectanglar to polar in a doble integral, it is the factor of r in rdrdθ. In this corse, the important cases are in lower dimensions. Sppose T : is a coordinate transformation of the plane. Sppose T takes a region U in the - plane to a region in the x-y plane. y T x In component form, (x,y) = T(,) looks like x = g(,), y = h(,). As in the theorem, I need to assme make some technical assmptions to ensre that the transformation is reasonably nice. Specifically:. T is injectie.. T has continos first partial deriaties. 3. The Jacobian JT is nonzero on Q. Yo can sally get away with some exceptions to these conditions if the set of exceptional points is small. For example, if JT = on a cre (technically, on a set of content ) the reslt is still tre. To aoid technicalities, I ll only consider sitations in which it is okay to proceed.

2 Under the assmptions aboe, the change of ariables formla says that f(x,y)dxdy = To apply the formla to compte. eplace the limits for with the limits for U. U f [x(,),y(,)] JT dd. f(x,y)dxdy, yo need to:. Use the transformation eqations to sbstitte and for x and y. 3. Compte the absolte ale of the Jacobian JT and insert it in the integral. Steps and seem reasonable; what abot step 3? The idea is that for a small region in the - plane, the pictre looks like this: y d T dt d x For small increments of and, T is well approximated by the deriatie dt. Apply dt to the sides (d,), (,d) of the small rectangle in the - plane to get a small parallelogram in the x-y plane: ( Ths,, ) d and area is, d [ ] d = (, d,, d [ ] = d d. ) d are the adjacent sides of the parallelogram. The parallelogram s dd = JT dd. That is, the Jacobian represents the factor by which small areas are scaled p by the transformation. This explains its presence in the change of ariables formla.

3 Example. Compte the Jacobian of the transformation (x,y) = T(,) = (5, ). JT = detdt = det [ ] 3 6 = Example. Compte the Jacobian of the transformation (x,y,z) = T(,,w) = (5+3 +7w, +, ). JT = detdt = det ] = 7det[ = 8. Example. (a) Find a transformation (x,y) = (f(,),g(,)) which takes the nit sqare { } to the parallelogram with ertices A(, ), B(5, 3), C(7, ), D(4, 6). (The ertices are listed conterclockwise arond the parallelogram.) (b) Let be the parallelogram with ertices A(, ), B(5,3), C(7, ), D(4, 6). (The ertices are listed conterclockwise arond the parallelogram.) Use the transformation in (a) and change of ariables to compte (4x+y)dxdy. (a) AB = (3,4) and AC = (, 5), so the transformation is [ ] [ ][ ] [ ] x 3 = + y 4 5 In component form, this is + (b) The integrand is The Jacobian is Hence, (5x+y 7)dxdy = x = 3+ +, y = x+y = 4(3+ +)+(4 5 ) = +6. [ ] 3 det = 3 = ( +6) 3dd = 3 3 [ +6 ] d =

4 3 (6 )d = 3 [ 6 ] = 345. Example. (a) Let be the region in the x-y plane bonded on the left by y = x, x, on the right by y = (x ), x, aboe by y =, and below by y =. Use horizontal segments to find a coordinate transformation which carries the nit sqare, to. (b) Compte (x +3y)dxdy. (a) Here is the region: y x Notice that it can be broken p into horizontal segments. I will parametrize the segment at height y =. ecall that if P and Q are points, the line segment from P to Q can be parametrized by (x,y) = ( ) P + Q,. Yo can derie this formla by finding the ector eqation of the line throgh P and Q. Howeer, it s easier to simply note that it has the right form for a line eqation, and it passes throgh P (set = ) and Q (set = ) so it mst be the right thing. The left hand cre is y = x. Set y = : Then x =, so I get x = I want the positie sqare root, since the region is in the first qadrant. The left hand intersection point is (,). (, ) y = ( +, ) The right hand cre is y = (x ). Set y = : Then (x ) =, so I get x = +. The right hand intersection point is (+,). Now I constrct the segment joining the intersection points. By the discssion aboe, it is (x,y) = ( )(,)+(+,) = (, )+(+,) = (+,). What are the limits? For the segment parametrization I m sing, always goes from to. is the height of the segment, and the region extends from to. Ths, the coordinate transformation is x = +, y =,,. 4

5 (b) We se the transformation from (a): The Jacobian is The integrand is Hence, (x +3y)dxdy = x = +, y =,,. J = det [ ] =. x +3y = (+ ) +3 = ( + +4) dd = ( ) [ d = 3 + ] 3 3/ + = 3. [ ] +4 d = Example. (a) Consider the coordinate transformation (x,y) = f(,) = (, ). Find the image in the x-y plane of the sqare,. (b) Is the transformation injectie (one-to-one) on the sqare? (c) Use the change of ariables formla to compte (x + y)dxdy. Check by compting the integral directly. Does it work? (a) In component form the transformation is x =, y =. A reasonable way to find the image is to look at where the for sides of the sqare go. C D B A Look at side A:, =. The eqations become x =, y =. Eliminating, I get x = y. So this side of the sqare is mapped to the parabola x = y. Since and y =, it is the part of the parabola which goes from (,) to (,). Look at side B: =,. The eqations become x =, y =. Then = y, so x = ( y) = y y. This is a parabola; since and y =, I hae y. Hence, I want the part of the parabola which goes from (,) to (,). 5

6 Look at side C: =,. The eqations become x =, y =. Now = y +, so x = (y +) = y +y. This is a parabola again; since and y =, I want the part of the parabola which satisfies y. It extends from (,) to (, ). Finally, look at side D: =,. The eqations become x =, y =. Then x = y, which is another parabola. Since and y =, I want the part of the parabola for which y. It extends from (,) to (, ). The region looks like two petals or leaes. I e drawn them below, with their edges labelled according to the sides of the sqare that mapped to them. A.5 B C -.5 D - (b) Do different impts always prodce different otpts? No if =, then x = and y =. That is, eerything on the diagonal line = is mapped to the origin in the x-y plane. It isn t one-to-one. (c) The Jacobian is [ ] J = det = ( ). There are two cases. If, then J = ( ) = ( ). If, then J = ( ) = ( ). These two cases correspond to the hales of the sqare aboe and below the line = : = The bottom half (grey) is, (J = ( )). The top half (white) is, (J = ( )). The integrand is x+y = +. 6

7 Therefore, the change of ariables formla gies (x+y)dxdy? = ( + ) (( )) dd+ ( + ) ( ( )) dd =. Now I ll check by doing the integral as is, in x-y coordinates. The following ineqalities describe the two petals: { } { } y y y x y y y +y x y So the integral is y y y (x+y)dxdy + y y+y (x+y)dxdy =. The transformation is not one-to-one on the line =, and the Jacobian anishes along that line as well, so I m not entitled to assme it will work bt it does. c 8 by Brce Ikenaga 7