Part 2: Change of Variable in Double Integrals   

Let f( x,y) be continuous on a region R that is the image under T( u,v) of a region S in the uv-plane. Then the double integral over R is of the form

ó õ ó õ R  f( x,y) dA = lim h® 0  å å f( xi*,yj*) DAij

where dA = dA_xy is the area differential in the xy-plane.  If we choose ( u_i*,v_j*) such that (x_i*,y_j*) = ( f( u_i*,v_j*),g( u_i*,v_j*) ) , then similar to the discussion in part 1, we have

ó õ ó õ R  f( x,y) dAxy = lim h® 0  å å f( xi*,yj*) DAij = lim h® 0  å å f( f(ui*,vj*) ,g( ui*,vj*) ) ê ê  ¶( x,y) ¶( u,v) ê ê ( ui*,vj*)   DuiDvj = ó õ ó õ S  f( f( u,v) ,g( u,v) ) ê ê  ¶( x,y) ¶( u,v) ê ê dAuv

where dA_uv is the area differential in the uv-plane. That is, is f is continuous on R which is the image under T(u,v) = á f( u,v) ,g( u,v) ñ of a type II region S in the uv-plane, then

ó õ ó õ R  f( x,y) dAxy = ó õ ó õ S  f( f(u,v) ,g( u,v) ) ê ê  ¶(x,y) ¶( u,v) ê ê dAuv

(1)

The formula (1) is called the change of variable formula for double integrals, and the region S is called the pullback of R under T. In order to make the change of variables formula more usuable, let us notice that implementing (1) requires 3 steps:

1. Compute the pullback S of R
2. Find the Jacobian and substitute for dA_xy
3. Replace x and y by f( u,v) and g(u,v) , respectively.

Finally, evaluate the result double integral over S, using dA_uv = dv du if S is type I and dA_uv = du dv if S is type II.

EXAMPLE 2    Evaluate òò_R( x+y) dA where R is the region with boundaries y = x, y = 3x, and x+y = 4 Use the transformation T( u,v) = áu-v,u+v ñ .       

Solution: To begin with, T( u,v) = áu-v,u+v ñ  is equivalent to x = u-v, y = u+v. Thus, the pullback of the boundaries is as follows:

y = x    Þ     u+v = u-v    Þ    2v = 0    Þ     v = 0 y = 3x    Þ     u+v = 3u-3v    Þ    4v = 2u    Þ     v =  u 2 x+y = 4    Þ     u-v+u+v = 4    Þ    2u = 4    Þ     u = 2

That is, S is the region bounded by v = 0, v = u/2, u = 2. The Jacobian of the transformation is

¶( x,y) ¶( y,v) =  ¶x ¶u  ¶y ¶v  -   ¶x ¶v  ¶y ¶u = 1·1 -  (-1) ·1 = 2

Thus, dA = 2dvdu and (1) implies that

òòR ( x+y) dAxy =   òòS  ( u-v+u+v) 2 dAuv = ó õ 2 0  ó õ u/2 0  4udvdu = 16/3