Answer
The required volume is given by:
$$
\begin{aligned}
V &=\iint_{x^{2}+y^{2} \leq 1 / 2}(\sqrt{1-x^{2}-y^{2}}-\sqrt{x^{2}+y^{2}}) d A \\
&=\frac{\pi}{3}(2-\sqrt{2}).
\end{aligned}
$$
Work Step by Step
The solid is bounded above by the cone $z=\sqrt {x^{2}+y^{2}} $ and below by the sphere $x^{2}+y^{2}+z^{2} = 1$.
If we put $z = \sqrt {x^{2}+y^{2}}$ in the equation of the sphere, we get $ x^{2}+y^{2}+(\sqrt {x^{2}+y^{2}})^{2} = 1 \text { or } x^{2}+y^{2} = \frac{1}{2} $. This means that the the cone intersects the sphere in the circle $x^{2}+y^{2} = \frac{1}{2} $, so the volume is
$$
\begin{aligned}
V& =\iint_{D}(Z_{1}-Z_{2}) d A\\
&=\iint_{x^{2}+y^{2} \leq 1 / 2}(\sqrt{1-x^{2}-y^{2}}-\sqrt{x^{2}+y^{2}}) d A\\
\end{aligned}
$$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $; thus the disk D is given by:
$$ D=\left\{(r, \theta) | 0 \leq r \leq \sqrt {\frac{1}{2}}, \quad 0 \leq \theta \leq 2\pi \right\} $$
and, the required volume is given by:
$$
\begin{aligned}
V &=\iint_{x^{2}+y^{2} \leq 1 / 2}(\sqrt{1-x^{2}-y^{2}}-\sqrt{x^{2}+y^{2}}) d A \\
&=\int_{0}^{2 \pi} \int_{0}^{1 / \sqrt{2}}(\sqrt{1-r^{2}}-r) r d r d \theta\\
&=\int_{0}^{2 \pi} d \theta \int_{0}^{1 / \sqrt{2}}\left(r \sqrt{1-r^{2}}-r^{2}\right) d r \\
&=[\theta]_{0}^{2 \pi}\left[-\frac{1}{3}\left(1-r^{2}\right)^{3 / 2}-\frac{1}{3} r^{3}\right]_{0}^{1 / \sqrt{2}}\\
& =2 \pi\left(-\frac{1}{3}\right)\left(\frac{1}{\sqrt{2}}-1\right)\\
&=\frac{\pi}{3}(2-\sqrt{2}).
\end{aligned}
$$
