The volume below by $z=\sqrt{x^2+y^2}$ and above by $x^2+y^2+z^2=1$.
My Solution: Wrote the integral. Converted it into cylindrical coordinates. But keep getting $0$ as my answer. Can someone help me by writing the solution. Will be able to find my mistake.
On
I can't tell where in your work your mistake is unless you post your work.
However, let me suggest using spherical coordinates instead of cylindrical coordinates.
The cone $z = \sqrt{x^2+y^2}$ becomes $\phi = \dfrac{\pi}{4}$ and the sphere $x^2+y^2+z^2 = 1$ becomes $\rho = 1$.
Therefore, the bounds are $0 \le \theta \le 2\pi$, $0 \le \phi \le \dfrac{\pi}{4}$, $0 \le \rho \le 1$.
This will give you a triple integral which is easy to compute.
Observe that the intersection between this two surfaces is the circle $x^2+y^2=1/2$. Since you are asking for volume below $z=\sqrt{x^2+y^2}$ and above the unit sphere, the integration region is the set $D=\{(x,y):1/2\le x^2+y^2\le 1\}$. Using cylindrical coordinates, you have \begin{align}V&=\int_{0}^{2\pi}\int_{ \frac{1}{\sqrt{2}}} ^{1} \left(r-\sqrt{1-r^2}\right) r\ drd\theta \\ &= \dfrac{2 \pi}{3} \left[r^3+(1-r^2)^{3/2} \right]_{ 1/\sqrt{2}}^{1}\\ &= \left( \dfrac{2-\sqrt{2}}{3} \right) \pi\approx 0.6134 \end{align}