Consider a transformation T(u,v) = (x(u,v), y(u,v)) from R 2 to R 2. Suppose T is a linear transformation T(u,v) = (au + bv, cu + dv). Then the derivative matrix of T is DT(u,v) =  x —  u  v =  y —  u  v a b c d We now explore what the transformation does to the rectangle defined by points (vectors) (0, 0), (u 0, 0), (u 0, v 0 ), (0, v 0 ). u v (0, 0) (0, v 0 )(u 0, v 0 ) (u 0, 0) x y (0, 0)

u v (0, v 0 )(u 0, v 0 ) (u 0, 0) x y T(0,0) =(a(0) + b(0), c(0) + d(0)) = (0,0) T(0, v 0 ) = (0, 0) (bv 0, dv 0 ) (au 0 + bv 0, cu 0 + dv 0 ) (au 0, cu 0 ) (a(0) + b(v 0 ), c(0) + d(v 0 )) = (bv 0, dv 0 ) T(u 0, v 0 ) =(au 0 + bv 0, cu 0 + dv 0 ) T(u 0, 0) =(a(u 0 ) + b(0), c(u 0 ) + d(0)) = (au 0, cu 0 ) The area of the rectangle D in the uv plane is u 0 v 0. The area of the parallelogram T(D) in the xy plane is D T(D)T(D)

T(0,0) =(a(0) + b(0), c(0) + d(0)) = (0,0) T(0, v 0 ) =(a(0) + b(v 0 ), c(0) + d(v 0 )) = (bv 0, dv 0 ) T(u 0, v 0 ) =(au 0 + bv 0, cu 0 + dv 0 ) T(u 0, 0) =(a(u 0 ) + b(0), c(u 0 ) + d(0)) = (au 0, cu 0 ) The area of the rectangle D in the uv plane is u 0 v 0. The area of the parallelogram T(D) in the xy plane is ||(au 0, cu 0, 0)  (bv 0, dv 0, 0)|| = ||(0, 0, adu 0 v 0 – bcu 0 v 0 )|| = u 0 v 0 |ad – bc|. D du dv = Now we know that u0v0u0v0 T(D)T(D) dx dy =u 0 v 0 |ad – bc|.and Consequently, T(D)T(D) dx dy = D |ad – bc| du dv = D |det[DT]| du dv.

This is a special case of a more general result. Suppose transformation T(u,v) = (x(u,v), y(u,v)) from R 2 to R 2 is one-to-one. Then T(D)T(D) f(x,y) dx dy = D f(x(u,v),y(u,v)) |det[DT]| du dv = D  (x,y) f(x(u,v),y(u,v)) ——– du dv.  (u,v) This notation is used to represent xx——uvyy——uvxx——uvyy——uv—— which is called the Jacobian determinant of T. One useful fact about one-to-one linear transformations is that a triangle is always mapped to a triangle, and a parallelogram is always mapped to a parallelogram. det

ExampleConsider the integral of xy over the parallelogram P formed by the lines y = 2x, y = 2x – 2, y = x, and y = x + 1. (a) Sketch the region P in the xy plane, and with T(u,v) = find the region D in the uv plane so that T(D) = P. x y (3,4) (2,2) (1,2) u v T(D) = P x = (u – v)/2  y = u – v/2 u = v = 2(y – x) 2y – 4x (x,y) = (0,0)  (u,v) = (0,0) (x,y) = (1,2)  (u,v) = (2,0) (x,y) = (3,4)  (u,v) = (2,–4) (x,y) = (2,2)  (u,v) = (0,–4) (2,0) (2,–4)(0,–4) D u – v ——, 2 v u – — 2

(b)Evaluateby making the change of variables P xy dx dy x = (u – v)/2 y = u – v/2  (x,y) ——– =  (u,v)  x —  u  v =  y —  u  v det 1/2– 1/2 1 =1/4 P xy dx dy = D (u – v)(2u – v)  (x,y) —————— ——– du dv = 4  (u,v) D (u – v)(2u – v) —————– du dv = 16 (2u 2 – 3uv + v 2 ) ——————– du dv = –4 7

ExampleConsider the following integral: (a) D*D* xy 3 ——— dx dy (1 – x) 4 y x D*D* (3/4, 1/4)(0, 1/4) (0, 1/2) (1/2, 1/2) where D is the region displayed in the graph. With the transformation T(u,v) = (u/v, v – u), there will be a trapezoidal region D in the uv plane so that T(D) = D*. Find this trapezoidal region D in the uv plane. u x = —u = v  y = v – uv = xy —— 1 – x y —— 1 – x v u (0, 1/4) (3/4, 1) D (0, 1/2) (1/2, 1)

(b)Evaluate by making the change of variables D*D* x = u / v y = v – u  (x,y) ——– =  (u.v)  x —  u  v =  y —  u  v det 1/v– u/v 2 –1 1 = v – u —— v 2 D*D* xy 3 ——— dx dy (1 – x) 4 xy 3 ——— dx dy = (1 – x) 4 D uv du dv = Since v > u on D, we see that this is equal to its absolute value.

D*D* xy 3 ——— dx dy = (1 – x) 4 D uv du dv = 239 —— /2 0 1/2 + u uv dv du + 1/21/4 + u 3/4 1/4 + u 1 uv dv du = v u (0, 1/4) (3/4, 1) D (0, 1/2) (1/2, 1) 239 —— /2 1/4 v – 1/4 uv du dv + 1/20 1 v – 1/2 v – 1/4 uv du dv =

For each point (x,y) in R 2, the polar coordinates (r,  ) are defined by x = r cos  and y = r sin , where r =  x 2 + y 2 is the length of the vector (x,y), and  = the angle that the vector (x,y) makes with the positive x axis. x y (x,y)  (r,  ) r  We have that 0  r and 0   < 2 . Recall:

Example Describe the region x 2 + y 2  36 in terms of rectangular coordinates and in terms of polar coordinates. r    < 6 0 22  x  x   y  y   6 6 –  6  x 2  6  x 2 or  y  y   x  x   6 6 –  6  y 2  6  y 2

Example Describe the region inside the triangle with vertices (0,0), (2,0), and (2,–2  3) in terms of rectangular coordinates and in terms of polar coordinates.    r  ( or –  /3    0 r  2 / cos  )  x  x   y  y  02 – (  3)x 0  y  y   x  x  –23–23 0 – y /  3 2 or (2,0) (2,–2  3) 5  /322 2 / cos 

Example Describe the region (x + 5) 2 + y 2  25 in terms of rectangular coordinates and in terms of polar coordinates.    r    r   / 23  / 2 0 – 10 cos 

Example Describe the region interior to both circles x 2 + y 2 = 1 and x 2 + (y + 1) 2 = 1 in terms of rectangular coordinates and in terms of polar coordinates. r = 1 r = – 2 sin  r = 1  = 7  — 6 r = 1  = 11  —— 6    r    r     r    r     r    r   7  — 6 0 – 2 sin  7  — 6 11  ——  —— 6 22 0 – 2 sin   1

ExampleSuppose we want to integrate the function f(x,y) over the region D* = T(D) in the xy plane, where D is the same region described in the r  plane and T(r,  ) = (r cos , r sin  ), that is, T is the polar coordinates transformation. Then, we know that D f(x,y) dx dy= D*D*  (x,y) f( x(r,  ), y(r,  ) ) ——— dr d   (r,  )  (x,y) Find———.  (r,  ) r  x y D  (x,y) ——— =  (r,  ) det cos  – r sin  sin  r cos  = | r cos 2  + r sin 2  | = r D*D*

D*D* f(x,y) dx dy= D f( r cos , r sin  ) r dr d . We see then how we can change from rectangular coordinates to polar coordinates: ExampleConsider the integral of the function f(x,y) = e over D* defined to be the region in the first quadrant of the xy plane between the two circles x 2 + y 2 = 4 and x 2 + y 2 = 25. (a) x 2 + y 2 Sketch and describe the region of integration D* in the xy plane and the corresponding region D in the r  plane. r  x y x 2 + y 2 = 4 x 2 + y 2 = 25 (2, 0)(5, 0) (2,  /2)(5,  /2)    r    r  0  / 2 25 D*D* D

(b)Evaluate by making the change of variables D*D* e dx dy x = r cos  y = r sin  x 2 + y 2 D*D* e dx dy = x 2 + y 2 D e r dr d  = r2r2 r2r  /2 r2r2 e — d  = 2 0  /2 r = 2 5 e 25 – e 4 ——— d  = 2 0  /2  (e 25 – e 4 ) ———— 4

ExampleFinde dx by squaring and using polar coordinates. ––  –x 2 ––  ––  –y 2 e dx e dy = – x 2 e dxe dy=  –– –– – y2– y2 –(x 2 + y 2 ) e dx dy =  –– –– –r 2 e r dr d  = 22 0  0 –r 2 e – —— d  = 2 0 22 r = 0  0 22 1 — d  = 2  We see then that ––  –x 2 e dx = 