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In our first example, the limits of integration for $r$ do not depend on $\theta$.
When our regions are more complicated than circles, we have to be more careful.
The function $e^{-x^2}$ does not have an antiderivative that can be expressed in closed form, and it is impossible to compute $\int_{0}^1 e^{-x^2} dx$ exactly. However, the integral $\displaystyle{\int_{-\infty}^\infty e^{-x^2} dx}$ turns out to equal $\sqrt{\pi}.$ In the following video, we use double integrals and polar coordinates to explain this surprising result. |