91Ó°ÊÓ

We are given that the region \(S\) in the \(uv\)-plane is defined by the inequalities $$v \leq 1-u, \ \ u \geq 0, \ \ v \geq 0$$

We need to apply the given transformation \(T: x=u, y=v^2\) to the inequalities of \(S\). This means that we need to rewrite each inequality in terms of \(x\) and \(y\). Inequality 1: $$v \leq 1-u \implies v^2 \leq (1-u)^2 \implies y \leq (1-x)^2$$ Inequality 2: $$u \geq 0 \implies x \geq 0$$ Inequality 3: $$v \geq 0 \implies v^2 \geq 0 \implies y \geq 0$$

Replacing the inequalities in terms of \(x\) and \(y\) gives us the region \(R\) in the \(xy\)-plane: $$R=\\{(x, y): y \leq (1-x)^2, x \geq 0, y \geq 0\\}$$

We will sketch both the region \(S\) in the \(uv\)-plane and the region \(R\) in the \(xy\)-plane. Region S: 1. First, sketch the curve \(v=1-u\). 2. Shade the area below the curve \(v=1-u\), since \(v \leq 1-u\). 3. Shade the area to the right of the line \(u=0\), since \(u \geq 0\). 4. Shade the area above the line \(v=0\), since \(v \geq 0\). 5. Note that the shaded regions overlap in a triangular shape. Region R: 1. First, sketch the curve \(y=(1-x)^2\), which is a parabola with its vertex at \((1,0)\), opening downward. 2. Shade the area below the curve \(y=(1-x)^2\), since \(y \leq (1-x)^2\). 3. Shade the area to the right of the line \(x=0\), since \(x \geq 0\). 4. Shade the area above the line \(y=0\), since \(y \geq 0\). 5. Note that the shaded regions overlap in a parabolic shape starting from the point \((0,0)\) to the vertex of the parabola \((1,0)\). Now, we have successfully found and sketched the regions \(S\) and \(R\).