<s>[INST] Finding volume of a solid over region E, bounded below by cone $z = \sqrt{x^2 + y^2}$ and above by sphere $ z^2 = 1- x^2 - y^2$ I set up the triple integral as follows:
$v=\int^1_0\int^{\sqrt{x-1}}_{-\sqrt{x}}\int^{\sqrt{1-x^2-y^2}}_{\sqrt{x^2+y^2}}(1)dzdydx$ Then went to polar coords getting $$\int^{2\pi}_0\int_0^1(\sqrt{1-r^2}r=r^2)drd\theta$$$$2\pi(\frac{-1}{2})[\frac{2}{3}(1-r^2)^{3/2}-1]|^1_0=-\pi[(0-1)-(\frac{2}{3}-1)]=\pi+\frac{1}{3}$$
The answer should be $\frac{\pi}{3}(2-\sqrt{2})$
 [/INST] Note that the solid is of revolution, formed by rotating the circular sector between $z=\sqrt {1-x^2}$ and $z=x$ (about the $z$-axis).
From cylinder method, we have that
\begin{align}V&=2\pi \int_{0}^{\sqrt2 \over 2}x|\sqrt {1-x^2}-x|dx\\
&=2\pi({1\over3} - {\sqrt2\over6})
\end{align}
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