Discussion
The integral \( \displaystyle\iint e^{-(x^{2}+y^{2})}\,dA \) over the whole plane is most easily done by
Answer: B
In polar form the exponent becomes \( -r^{2} \) and the area element supplies the factor \( r \) that the substitution \( u = r^{2} \) needs:
\[ \int_{0}^{2\pi}\!\!\int_{0}^{\infty} e^{-r^{2}}r\,dr\,d\theta \;=\; 2\pi \cdot \frac{1}{2} \;=\; \pi. \]
This is the standard route to \( \int_{-\infty}^{\infty} e^{-x^{2}}dx = \sqrt{\pi} \), which has no elementary antiderivative in one variable
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