It is often advantageous to evaluate òò_R f(x,y) dA in a coordinate system other than the xy-coordinate system. In this section, we develop a method for converting double integrals into iterated integrals in other coordinate systems.

Let's suppose that T( u,v) = á f( u,v),g( u,v) ñ maps a region S in the uv-plane to a region R in the xy-plane, and let's suppose that S is contained in a rectangle [ a,b] ×[ c,d] on which both f and g are differentiable. A partition of [ a,b] into n subintervals of width Du_i and [ c,d] into m subintervals of width Dv_j covers S with a collection of rectangles. If we require Du_i < h and Dv_j < h for some sufficiently small h > 0, then the image of the rectangles under T is a collection of regions covering R which are approximately parallelograms. In chapter 3, we learned that the area of the image of a rectangle is approximately

DAij = ê ê  ¶( x,y) ¶(u,v) ê ê ( ui*,vj*)   DuiDvj

where ( u_i^*,v_j^*) is a point in the ij-th rectangle. Thus, the sum over the rectangles intersecting S leads to an approximation of the area of R of the form

Area = å å DAij = å å ê ê  ¶(x,y) ¶( u,v) ê ê (ui*,vj*)   DuiDvj

and in the limit as h approaches 0, the double sum leads to a double integral. Thus, if T( u,v) maps a region S to a region R, then

Area  of  R = ó õ ó õ R  dA = ó õ ó õ S  ê ê  ¶(x,y) ¶( u,v) ê ê dudv

If S is type 1, however, then the differential is instead dvdu.       

EXAMPLE 1    Find and describe the image of [ 0,2p] ×[ 0,1] under the transformation

T( u,v) = á 2v cos( u) ,v sin(u) ñ

and then find the area of that image.       

Solution: Since x = 2vcos( u) and y = vsin(u) , we have

x2 4 +  y2 1 = v2cos2( u) +v2sin2( u) = v2

which because u ranges from 0 to 2p is an ellipse centered at the origin. Moreover, these ellipses completely cover the region inside the ellipse corresponding to v = 1, which is the ellipse with semi-major axis 2 and semi-minor axis 1. Moreover, the Jacobian determinant is

¶( x,y) ¶( u,v) = æ è  ¶ ¶u 2vcos( u) ö ø æ è  ¶ ¶v vsin( u) ö ø - æ è  ¶ ¶v 2vcos( u) ö ø æ è  ¶ ¶u vsin( u) ö ø = -2vsin2( u) -2vcos2( u) = -2v

Thus, dA = 2vdvdu and the area of the image region is

Area  of  R = ó õ ó õ S  ê ê  ¶( x,y) ¶( u,v) ê ê dvdu = ó õ 2p 0  ó õ 1 0  2vdvdu = ó õ 2p 0  v2  1 0 du = ó õ 2p 0  du = 2p