Question

Consider the open rectangle \( G = \{ (s,t) \in \mathbb{R}^2 : 0 < s < 1 \ \text{and} \ 0 < t < 1 \} \) and the map \( T : G \to \mathbb{R}^2 \) given by \[ T(s,t) = \left( \frac{\pi s (1-t)}{2}, \ \frac{\pi (1-s)t}{2} \right) \ \text{for} \ (s,t) \in G. \] Then the area of the image \( T(G) \) of the map \( T \) is equal to

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi^2}{4} \)
• C. \( \frac{\pi^2}{8} \)
• D. 1

Answer 1

The Correct Option is C

The correct option is (C): \( \frac{\pi^2}{8} \)