I have made an attempt at this question:
Find the volume below $z = \sqrt{1 - r^{2}}$ and above the top half of the cone $z = r$
I solved for $r$ first, then I calculated: \begin{equation} \int_{0}^{2\pi}\int_{0}^{\frac{1}{2}\sqrt{2}}r\sqrt{1-r^{2}}drd\theta = -\frac{1}{6}\pi(\sqrt{2} - 4) \end{equation} I wasn't sure of my answer and after plotting the surfaces in Geogebra I'm sure that my inner integral bounds are incorrect, can anybody provide any guidance? Thanks in advance!