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Differential Equations Question on Differential Equations

Consider the open rectangle G=(s,t)∈R2:0<s<1 and 0<t<1G = \\{ (s,t) \in \mathbb{R}^2 : 0 < s < 1 \ \text{and} \ 0 < t < 1 \\} and the map T:G→R2T : G \to \mathbb{R}^2 given by T(s,t)=(πs(1−t)2, π(1−s)t2) for (s,t)∈G.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)T(G) of the map TT is equal to

A

π4\frac{\pi}{4}

B

π24\frac{\pi^2}{4}

C

π28\frac{\pi^2}{8}

D

1

Answer

π28\frac{\pi^2}{8}

Explanation

Solution

The correct option is (C): π28\frac{\pi^2}{8}