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

To find the area of the image \( T(G) \) under the transformation given by the map \( T : G \to \mathbb{R}^2 \), we will calculate the Jacobian determinant of the transformation \( T \). This process is essential to measure how area transforms under \( T \).

Step 1: Define the Transformation

The transformation is provided by: \(T(s,t) = \left( \frac{\pi s (1-t)}{2}, \ \frac{\pi (1-s)t}{2} \right)\) for \((s,t) \in G\).

Step 2: Calculate the Jacobian Determinant

The Jacobian matrix \( JT(s, t) \) is the matrix of first-order partial derivatives of \( T \): \(JT(s, t) = \begin{bmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{bmatrix} = \begin{bmatrix} \frac{\pi (1-t)}{2} & \frac{-\pi s}{2} \\ \frac{-\pi t}{2} & \frac{\pi (1-s)}{2} \end{bmatrix}\)where \(x = \frac{\pi s (1-t)}{2}\) and \(y = \frac{\pi (1-s) t}{2}\).

The determinant of the Jacobian matrix is: \(\text{det}(JT) = \left( \frac{\pi (1-t)}{2} \right) \left( \frac{\pi (1-s)}{2} \right) - \left( \frac{-\pi s}{2} \right) \left( \frac{-\pi t}{2} \right)\)

Step 3: Simplify the Determinant

Simplifying, we get: \(\text{det}(JT) = \frac{\pi^2 (1-t)(1-s)}{4} - \frac{\pi^2 st}{4} = \frac{\pi^2}{4} \left( (1-t)(1-s) - st \right) = \frac{\pi^2}{4} \left( 1 - s - t + st - st \right) = \frac{\pi^2}{4} (1 - s - t)\)

Step 4: Integrate over the Region

We integrate this determinant over the region \( G = \{ (s,t) : 0 < s < 1, 0 < t < 1 \} \): \(\int_0^1 \int_0^1 \frac{\pi^2}{4} (1 - s - t) \, ds \, dt\)

Step-by-step integration:

• First, integrate with respect to \( s \): \(\int_0^1 (1 - s - t) \, ds = \left[ s - \frac{s^2}{2} - st \right]_0^1 = \left[ 1 - \frac{1}{2} - t \right] = \frac{1}{2} - t\)
• Next, integrate with respect to \( t \): \(\int_0^1 \left( \frac{1}{2} - t \right) \, dt = \left[ \frac{t}{2} - \frac{t^2}{2} \right]_0^1 = \frac{1}{2} - \frac{1}{2} = 0\)

Thus, the total area is: \(\frac{\pi^2}{4} \int_0^1 \left( \frac{1}{2} - t \right) \, dt = \frac{\pi^2}{4} \times \frac{1}{4} = \frac{\pi^2}{8}\)

Conclusion: The area of the image \( T(G) \) is \( \frac{\pi^2}{8} \).