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Consider a transformation T ( u,v ) = ( x ( u,v ) , y ( u,v )) from R 2 to R 2 .

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 D T ( u,v ) =.  x  x — —  u  v =  y  y — —  u  v. a b c d.

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Consider a transformation T ( u,v ) = ( x ( u,v ) , y ( u,v )) from R 2 to R 2 .

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  1. Consider a transformation T(u,v) = (x(u,v) , y(u,v)) from R2 to R2. Suppose T is a linear transformation T(u,v) = (au + bv , cu + dv) . Then the derivative matrix of T is DT(u,v) = x x — — u v = y y — — u v abcd We now explore what the transformation does to the rectangle defined by points (vectors) (0 , 0), (u0 , 0), (u0 , v0), (0 , v0). v y (0 , v0) (u0 , v0) u x (0 , 0) (u0 , 0) (0 , 0)

  2. v y (au0 + bv0 , cu0 + dv0) (bv0 , dv0) (0 , v0) (u0 , v0) T(D) D (au0 , cu0) u x (0 , 0) (u0 , 0) (0 , 0) T(0,0) = (a(0) + b(0) , c(0) + d(0)) = (0,0) T(0 , v0) = (a(0) + b(v0) , c(0) + d(v0)) = (bv0 , dv0) T(u0 , v0) = (au0 + bv0 , cu0 + dv0) T(u0 , 0) = (a(u0) + b(0) , c(u0) + d(0)) = (au0 , cu0) The area of the rectangle D in the uv plane is u0v0 . The area of the parallelogram T(D) in the xy plane is

  3. T(0,0) = (a(0) + b(0) , c(0) + d(0)) = (0,0) T(0 , v0) = (a(0) + b(v0) , c(0) + d(v0)) = (bv0 , dv0) T(u0 , v0) = (au0 + bv0 , cu0 + dv0) T(u0 , 0) = (a(u0) + b(0) , c(u0) + d(0)) = (au0 , cu0) The area of the rectangle D in the uv plane is u0v0 . The area of the parallelogram T(D) in the xy plane is ||(au0 , cu0 , 0)(bv0 , dv0 , 0)|| = ||(0 , 0 , adu0v0–bcu0v0)|| = u0v0|ad–bc| . du dv = u0v0 and dx dy = u0v0|ad–bc|. Now we know that D T(D) Consequently, dx dy = |ad – bc| du dv = |det[DT]| du dv . T(D) D D

  4. This is a special case of a more general result. Suppose transformation T(u,v) = (x(u,v) , y(u,v)) from R2 to R2 is one-to-one. Then f(x,y) dx dy = f(x(u,v),y(u,v)) |det[DT]| du dv = T(D) D (x,y) f(x(u,v),y(u,v)) ——– du dv . (u,v) D This notation is used to represent x x — — u v y y — — u v det 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.

  5. Example Consider the integral of xy over the parallelogram P formed by the lines y = 2x , y = 2x– 2 , y = x , and y = x + 1 . (a) v u–— 2 u–v —— , 2 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. (3,4) u (2,0) D (1,2) (2,2) T(D) = P x (0,–4) (2,–4) y v (x,y) = (0,0)  (u,v) = (0,0) (x,y) = (1,2)  (u,v) = (2,0) 2(y–x) u = v = (x,y) = (3,4)  (u,v) = (2,–4) x = (u–v)/2  y = u–v/2 2y– 4x (x,y) = (2,2)  (u,v) = (0,–4)

  6. x = (u – v)/2 y = u – v/2 (b) Evaluate by making the change of variables xydx dy P x x — — u v = y y — — u v 1/2 – 1/2 (x,y) ——– = (u,v) det det = 1/4 1 – 1/2 (u– v)(2u – v) (x,y) —————— ——– du dv = 4 (u,v) xydx dy = P D 0 2 (u – v)(2u – v) —————– du dv = 16 (2u2– 3uv + v2) ——————– du dv = 16 7 D –4 0

  7. Example Consider the following integral: (a) xy3 ——— dx dy (1 – x)4 (1/2 , 1/2) (0 , 1/2) D* D* (0 , 1/4) (3/4 , 1/4) where D is the region displayed in the graph. x y 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. (1/2 , 1) (3/4 , 1) u x = — u = v  y = v – uv = xy —— 1 – x (0 , 1/2) D y —— 1 – x (0 , 1/4) u v

  8. xy3 ——— dx dy (1 – x)4 (b) Evaluate by making the change of variables x = u/v y = v–u D* x x — — u v = y y — — u v 1/v – u/v2 (x,y) ——– = (u.v) v – u —— v2 det det = –1 1 xy3 ——— dx dy = (1 – x)4 Since v > u on D, we see that this is equal to its absolute value. uv du dv = D D*

  9. (1/2 , 1) (3/4 , 1) (0 , 1/2) D xy3 ——— dx dy = (1 – x)4 (0 , 1/4) uv du dv = u D v D* 1/2 v– 1/4 1 v – 1/4 239 —— 6144 uv du dv + uv du dv = 1/4 0 1/2 v – 1/2 1/2 1/2 + u 3/4 1 239 —— 6144 uv dv du + uv dv du = 0 1/4 + u 1/2 1/4 + u

  10. Recall: For each point (x,y) in R2, the polar coordinates (r,) are defined by x = r cos  and y = r sin  , where r =  x2 + y2 is the length of the vector (x,y) , and  = the angle that the vector (x,y) makes with the positive x axis . y (x,y)  (r,) r  x We have that 0  r and 0   < 2 .

  11. Example Describe the region x2 + y2 36 in terms of rectangular coordinates and in terms of polar coordinates.  6  x  6 –  6  x2  y   6  x2 or  6  y  6 –  6  y2  x   6  y2 r   < 6 0 2

  12. 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. 5/3    r 2 (2,0) 2 / cos  ( or –/3    0 r 2 / cos  ) (2,–23) 0  x  2 –23  y  0 or – (3)x  y  0 – y / 3  x  2

  13.  / 2     r  3 / 2 Example Describe the region (x + 5)2 + y2 25 in terms of rectangular coordinates and in terms of polar coordinates. 0 – 10 cos

  14. Example Describe the region interior to both circles x2 + y2= 1 and x2 + (y + 1)2= 1 in terms of rectangular coordinates and in terms of polar coordinates. r = 1 r = 1  = r = 1  = 7 — 6 11 —— 6 r = – 2 sin 11 —— 6 7 — 6 7 — 6 11 —— 6     r      r      r   2   – 2 sin – 2 sin 0 0 0 1

  15. Example Suppose 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 (x,y) f( x(r,) , y(r,) ) ——— dr d (r,) f(x,y) dx dy = D* D y  (x,y) Find ——— . (r,) D D* r x cos – r sin (x,y) ——— = (r,) det = | r cos2 + r sin2 | = r sin r cos

  16. We see then how we can change from rectangular coordinates to polar coordinates: f(x,y) dx dy = f( r cos  , r sin ) r dr d . D* D x2 + y2 Example Consider 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 x2 + y2 = 4 and x2 + y2 = 25. (a) Sketch and describe the region of integration D* in the xy plane and the corresponding region D in the r plane. y  x2 + y2 = 25 (2 , /2) (5 , /2) D* x2 + y2 = 4 D r x (2 , 0) (5 , 0) 0     r   / 2 2 5

  17. x2 + y2 (b) Evaluate by making the change of variables edx dy D* x = r cos y = r sin /2 5 r2 r2 x2 + y2 er dr d = er dr d = edx dy = D* D 0 2 5 /2 /2 r2 e — d = 2 e25– e4 ——— d = 2 (e25– e4) ———— 4 0 0 r = 2

  18. –x2 Example Find edx by squaring and using polar coordinates. –     –x2 –y2 – x2 –y2 e dx e dy = e dx e dy = – – – –   2  2  –r2 –(x2 + y2) –r2 edx dy = e rdr d = e – —— d = 2 – – 0 0 0 r = 0 2  1 — d = 2 –x2  We see then that e dx =  0 –

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