Question 9.17.1: Image of a Region Find the image of the region S shown in FI......
Image of a Region
Find the image of the region S shown in FIGURE 9.17.2(a) under the ransformation x=u^{2}+v^{2}, y=u^{2}-v^{2}.
We begin by finding the images of the sides of S that we have indicated by S_{1}, S_{2}, and S_{3}.
S_{1} : On this side v=0 so that x=u^{2}, y=u^{2}. Eliminating u then gives y=x. Now imagine moving along the boundary from (1,0) to (2,0) (that is, 1 \leq u \leq 2). The equations x=u^{2}, y=u^{2} then indicate that x ranges from x=1 to x=4 and y ranges simultaneously from y=1 to y=4. In other words, in the x y-plane the image of S_{1} is the line segment y=x from (1,1) to (4,4).
S_{2} : On this boundary u^{2}+v^{2}=4 and so x=4. Now as we move from the point (2,0) to \left(\sqrt{\frac{5}{2}}, \sqrt{\frac{3}{2}}\right), the remaining equation y=u^{2}-v^{2} indicates that y ranges from y=2^{2}-0^{2}=4 to y=\left(\sqrt{\frac{5}{2}}\right)^{2}-\left(\sqrt{\frac{3}{2}}\right)^{2}=1. In this case the image of S_{2} is the vertical line segment x=4 starting at (4,4) and going down to (4,1).
S_{3} : Since u^{2}-v^{2}=1, we get y=1. But as we move on this boundary from \left(\sqrt{\frac{5}{2}}, \sqrt{\frac{3}{2}}\right), to (1,0), the equation x=u^{2}+v^{2} indicates that x ranges from x=4 to x=1. The image of S_{3} is the horizontal line segment y=1 starting at (4,1) and ending at (1,1).
The image of S is the region R given in Figure 9.17.2(b).