Consider the open rectangle G = {(s,t) ∈ R²: 0 < s < 1 and 0 < t < 1} and the map T: G → R² given by T(s, t) = (πs(1 - t)/2, π(1 - s)/2) for (s, t) ∈ G. Then the area of the image... Consider the open rectangle G = {(s,t) ∈ R²: 0 < s < 1 and 0 < t < 1} and the map T: G → R² given by T(s, t) = (πs(1 - t)/2, π(1 - s)/2) for (s, t) ∈ G. Then the area of the image T(G) of the map T is equal to? Read more

Steps to Solve

1. Identify the Transformation
   The map is given by ( T(s, t) = \left( \frac{\pi s(1 - t)}{2}, \frac{\pi(1 - s)}{2} \right) ).

2. Compute the Jacobian
   To find the area of the image ( T(G) ), we need to compute the Jacobian determinant.
   The Jacobian matrix ( J ) is computed by differentiating ( T ) with respect to ( s ) and ( t ): $$ J = \begin{bmatrix} \frac{\partial T_1}{\partial s} & \frac{\partial T_1}{\partial t} \ \frac{\partial T_2}{\partial s} & \frac{\partial T_2}{\partial t} \end{bmatrix} $$

Calculating the partial derivatives:

• For ( T_1 = \frac{\pi s(1 - t)}{2} ):

  • ( \frac{\partial T_1}{\partial s} = \frac{\pi(1 - t)}{2} )
  • ( \frac{\partial T_1}{\partial t} = -\frac{\pi s}{2} )

• For ( T_2 = \frac{\pi(1 - s)}{2} ):

  • ( \frac{\partial T_2}{\partial s} = -\frac{\pi}{2} )
  • ( \frac{\partial T_2}{\partial t} = 0 )

Thus, the Jacobian matrix is: $$ J = \begin{bmatrix} \frac{\pi(1 - t)}{2} & -\frac{\pi s}{2} \ -\frac{\pi}{2} & 0 \end{bmatrix} $$

3. Calculate the Jacobian Determinant
   The determinant of ( J ) is: $$ \text{det}(J) = \left(\frac{\pi(1-t)}{2}\right)(0) - \left(-\frac{\pi s}{2}\right)\left(-\frac{\pi}{2}\right) = -\frac{\pi^2 s(1 - t)}{4} $$ So, we consider the absolute value: $$ |\text{det}(J)| = \frac{\pi^2 s(1 - t)}{4} $$

4. Find the Area of the Image
   To find the area of ( T(G) ), we need to integrate the Jacobian determinant over the rectangle ( G ): $$ \text{Area} = \int_0^1 \int_0^1 |\text{det}(J)| , ds , dt = \int_0^1 \int_0^1 \frac{\pi^2 s(1 - t)}{4} , ds , dt $$

5. Compute the Double Integral
   First, integrate with respect to ( s ): $$ \int_0^1 s , ds = \left[ \frac{s^2}{2} \right]_0^1 = \frac{1}{2} $$ Now integrate with respect to ( t ): $$ \int_0^1 (1 - t) , dt = \left[ t - \frac{t^2}{2} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} $$ Thus, the area becomes: $$ \text{Area} = \frac{\pi^2}{4} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{\pi^2}{16} $$