$$ \int_{0}^{6}\int_{0}^{\sqrt{36-x^2}}e^{-x^2-y^2}dydx $$
I tried converting it into polar coordinates using the following equations
$y = \sqrt{36-x^2} \implies y^2 + x^2 = 36 \implies r = 6$
$0=6\cos\theta, 6=6\cos\theta \therefore \theta = \frac{pi}{2}, 0$
$$ \int_{\frac{\pi}{2}}^{0}\int_{0}^{6}re^{-r^2}drd\theta \\ = \int_{\frac{\pi}{2}}^{0} \frac{1}{2}(1-e^{-36})d\theta \\ = \frac{\pi}{4}(e^{-36}-1) $$
But I keep getting the answer wrong, I am not sure what I am doing wrong
The region where you are integrating can be described as: \begin{align*} R = \{(x,y)\in\mathbb{R}^{2} \mid (x^{2} + y^{2} \leq 36)\wedge(0\leq y\leq 6)\wedge(0\leq x\leq 6)\} \end{align*}
If you make the change of variable $x = r\cos(\theta)$ and $y = r\sin(\theta)$, the region $R$ can be described as \begin{align*} R = \{(r,\theta)\in\mathbb{R}_{\geq0}\times[0,2\pi) \mid (0\leq r\leq 6)\wedge(0\leq \theta \leq \pi/2)\}. \end{align*}
Given that the correction factor in this case is given by $r$, one arrives at the desired integral: \begin{align*} \int_{0}^{6}\int_{0}^{\sqrt{36 - x^{2}}}e^{-x^{2} - y^{2}}\mathrm{d}y\mathrm{d}x = \int_{0}^{\pi/2}\int_{0}^{6}re^{-r^{2}}\mathrm{d}r\mathrm{d}\theta \end{align*}
From now on, I think you can handle it.
EDIT
Here I provide a picture of the integration region: