Discussion
Let \(D \subset \mathbb{R}^{2}\) denote the closed disc with center at the origin and radius 2. Then
\[ \iint_{D} e^{-(x^{2}+y^{2})}\, dx\,dy = \]
Answer: A
The integral equals \(\pi(1 - e^{-4})\).
Use polar coordinates, \(x^{2} + y^{2} = r^{2}\) and \(dx\,dy = r\,dr\,d\theta\), with \(0 \le r \le 2\) and \(0 \le \theta \le 2\pi\):
\[ \begin{aligned} I &= \int_{0}^{2\pi}\int_{0}^{2} e^{-r^{2}} r\,dr\,d\theta \\ &= 2\pi \left[-\frac{1}{2}e^{-r^{2}}\right]_{0}^{2} \\ &= 2\pi \cdot \frac{1}{2}(1 - e^{-4}) = \pi(1 - e^{-4}) \end{aligned} \]
The exponent is \(-r^{2} = -4\) at the edge of the disc of radius 2, so the options with \(e^{-2}\) use the radius instead of its square. The options with \(\frac{\pi}{2}\) lose the \(2\pi\) from the angle, or integrate only over a half turn.
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