Use Polar Coordinates to evaluate $\int_{-4}^4\int_0^{\sqrt{16-x^2}} e^{-(x^2+y^2)}$

Question

Here is what I have so far: $ x = r\cos(\theta) \ y=r\sin(\theta)\ dydx = rdrd\theta$ $$ \int_{-4}^4\int_0^{\sqrt{16-x^2}} e^{-((r\cos(\theta))^2+(r\sin(\theta))^2)} \\ \int_{-4}^4\int_0^{4} e^{-r^3} \\ \int_{-4}^4 3r^2e^{-64}$$

Answer 1

If you're going to use polar coordinates, the variable $\;r\;$ is always non negative, so it can be $\;-4\le r\le4\;$ .

You're first integral's limits make it sure you're integrating over the upper half circle $\;x^2+y^2=16\;$ in the $\;xy\,-$ plane, from which you get in polar coordinates

$$\begin{cases}0\le r\le 4\\{}\\0\le\theta\le\pi\end{cases}\;\;\implies\;\text{your integral is}\;\;\int_0^4\int_0^\pi re^{-r^2}\,d\theta\,dr$$

which is an almost immediate integral (can you see the Jacobian there?)