0 Video Example 1) EXAMPLE 1 A transformation is dened by the equations X=u2v2 y=8uv. Find the image of the square 5 = {(u, v), 0 s u s 1, 0 s v s 1}. SOLUTION The transformation maps the boundary of S into the boundary of the image. So we begin by finding the images of the sides of S. The first side, 51, is given by V = 0 (0 s u s 1). (See the bottom gure.) From the given equations we have X = , y = 0, and so 0 s x s 1. Thus 51 is mapped into the line segment from (0, 0) to (1, 0) in the xy-plane. The second side, 52, is u = 1 (0 S V S 1) and, putting u = 1 in the equations, we get Eliminating v, we obtain 2 X = 1 y S X S 1 64' O which is part ofa parabola. Similarly, S3, is given by v = 1 (O s u 5 1), whose image is the parabolic arc 2 x=%1, ISXSO. Finally, 54 is given by u=0 (OS vs 1) whose image is X: , y= O, that is, 1 s X s 0. (Notice that as we move around the square in the counterclockwise direction, we also move around the parabolic region in the counterclockwise direction.) The image of Sis the region R (shown in the gure) bounded by the X-axis and the parabolas given by the equations above