Solution: If $(x,y) \in D$, then $x^2 + y^2 \leq 1$. We have \begin{align*} T(D) & = \left\{ \left( \frac{x+y}{\sqrt{2}} , \frac{x-y}{\sqrt{2} } \right) : x^2 + y^2 \leq 1 \right\}. \end{align*} Consider \begin{align*} \left( \frac{x+y}{\sqrt{2} } \right) ^2 + \left( \frac{x-y}{\sqrt{2} } \right) ^2 & = \frac{(x+y)^2}{2} + \frac{(x-y)^2}{2} = x^2 + y^2. \end{align*} Therefore, $T(D) = D$ and hence the area of $T(D)$ will be $\pi $.

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We can also solve this problem by using the following fact: the area of the image of a figure under a linear transformation is the area of the figure times the absolute value of determinant of the transformation.

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Let us first find the determinant of the given linear transformation. We take a basis of $\mathbb{R} ^2$ as $\{ (1,0), (0,1) \} $. With respect to this basis the matrix of $T$ will be \[ [T] = \begin{pmatrix} \frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} } \\[1ex] \frac{1}{\sqrt{2} } & -\frac{1}{\sqrt{2} } \\ \end{pmatrix} \implies \det [T] = -1. \] Therefore, \[ \text{area of $T(D)$} = \text{area of $D$} \cdot \vert \det [T] \vert = \pi \times 1 = \pi . \]