Part 1: Area of the Image of a Region

Let's suppose that T( u,v) = á f( u,v),g( u,v) ñ is a 1-1 map of a region S in the uv-plane onto 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. If DA_ij is the area of the image of the ij^th rectangle, i=1,...,n and j = 1,...,n, then the area of R is approximately

Area of R  »     å å DAij

However, 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 of R  »     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, we have

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

where dA_uv = dv du if S is type I and dA_uv = du dv if S is type II.

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

Since S is type I, it follows that

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