I am stuck with the following xy-plane integral:
$$ \int_{x^2+y^2\leq3^2}\sqrt{25-y^2}dxdy. $$
Putting in in polar coordinates, one finds
$$ \begin{align} \int_{0}^{2\pi}\Bigg(\int_{0}^{3}\sqrt{25-\rho^2\sin^2(\theta)}\rho\,d\rho\Bigg)\,d\theta&=\int_{0}^{2\pi}\Bigg(\int_{0}^{3}\sqrt{25-\rho^2\sin^2(\theta)}\rho\,d\rho\Bigg)\,d\theta\\ &=\int_{0}^{2\pi}-\frac{1}{2\sin^2(\theta)}\Bigg(\int_{0}^{3}-2\rho\sin^2(\theta)\sqrt{25-\rho^2\sin^2(\theta)}\,d\rho\Bigg)\,d\theta\\ &=\int_{0}^{2\pi}-\frac{1}{2\sin^2(\theta)}\Bigg[\frac{2}{3}\big(25-\rho^2\sin^2(\theta)\big)^{\frac{3}{2}}|^{3}_{0}\Bigg]\,d\theta\\ &=\int_{0}^{2\pi}-\frac{1}{2\sin^2(\theta)}\Bigg[\frac{2}{3}\big(25-9\sin^2(\theta)\big)^{\frac{3}{2}}-\frac{2}{3}\big(25\big)^{\frac{3}{2}}\Bigg]\,d\theta\\ &=\int_{0}^{2\pi}\Bigg(-\frac{\big(25-9\sin^2(\theta)\big)^{\frac{3}{2}}}{3\sin^2(\theta)}+\frac{125}{3\sin^2(\theta)}\Bigg)\,d\theta \end{align} $$ and there I cannot come up with anything encouraging (by the way, I don't know if it is so legit to multiply and divide by something like $\sin(\theta)$ since $\frac{1}{\sin(\theta)}$ has a singularity in $[0,2\pi]$).
Your result in the last equation is correct. Although there is singularity, the integral is converging.
It may be a bit encouraging that your result could be translated into a cleaner form below, with the variable change $t=\cot^2\theta$,
$$ \int_{0}^{2\pi}\Bigg[\frac{125}{3\sin^2\theta}-\frac{\big(25-9\sin^2\theta\big)^{\frac{3}{2}}}{3\sin^2\theta}\Bigg]\,d\theta =\frac{500}{3}\int_0^{\infty} \left[ 1-\left( \frac{a^2+t^2}{1+t^2}\right)^{\frac{3}{2}}\right]dt $$
where $a=4/5$. It converges as $t\rightarrow \infty$. But, unfortunately, it could not be integrated with elementary calculus, which has to resort to special elliptic functions.