2. Find the image of the square bounded by the lines \( u=0, u=1, v=0, v=1 \) under the transformation \( x=v, y=u\left(1+v^{2}\right) \).

To find the image of the square defined by \( (u,v) \) coordinates between \( 0 \) and \( 1 \) under the transformation given by \( x = v \) and \( y = u(1 + v^2) \), let's analyze how the corners of this square transform. 1. For the corner \( (u, v) = (0, 0) \): \( x = 0 \), \( y = 0\) - Image: \( (0,0) \) 2. For the corner \( (u, v) = (0, 1) \): \( x = 1 \), \( y = 0\) - Image: \( (1,0) \) 3. For the corner \( (u, v) = (1, 0) \): \( x = 0 \), \( y = 1\) - Image: \( (0,1) \) 4. For the corner \( (u, v) = (1, 1) \): \( x = 1 \), \( y = 2\) - Image: \( (1,2) \) Now, we can analyze the transformed region. The lower boundary occurs when \( u = 0 \) (mapping \( y \) to 0), which gives us the segment from \( (0, 0) \) to \( (1, 0) \). The upper boundary occurs when \( u = 1 \), generating the line segment from \( (0, 1) \) to \( (1, 2) \). For fixed values of \( v \), \( y \) varies linearly from \( 0 \) to \( 1+v^2 \), with the maximum occurring at the edge corners. Thus, the image of the square is a quadrilateral with vertices at \( (0, 0) \), \( (1, 0) \), \( (1, 2) \), and \( (0, 1) \).