Triple Integral In a Sphere Outside of a Cone Find the volume of the solid that lies within the hemisphere $x^2 + y^2 + z^2 =6$, z ≤ 0, and outside the cone $z=-\sqrt{x^2+y^2}$. I am not sure what to do for this question, I tried subtracting the area of the sphere from the cone, I also tried spherical but a unsure of what the limits for phi would be.
 A: Due to symmetry, the solid is identical to the one which lies within the hemisphere $x^2 + y^2 + z^2 = 6$, $z \geq 0$ and outside the cone $z = \sqrt{x^2 + y^2}$, the only difference being that one is the mirror image of the other across $xy$-plane. (This is done just to avoid negative signs. This is a personal choice, and you can very well work with either of the solids). We will work with the solid which lies above $xy$-plane.
Notice that this solid is identical to the solid of revolution if we revolve $\Omega = \{(x,y)| \; x,y \geq 0; \; x^2 + y^2 \leq 6; \; x \geq y \}$ around $y$-axis.
Using Disk Method, the volume of this solid of revolution is given by:
$V = \pi \int \limits_{y = 0}^{\sqrt{3}} [(6 - y^2) - y^2] dy = 4 \sqrt{3} \pi$.
Alternatively, using triple integration,
$V = \int \limits_{\phi = 0}^{2 \pi} \; \; \int \limits_{\theta = \pi/4}^{\pi/2} \; \; \int \limits_{\rho = 0}^{\sqrt{6}} %
\rho^2 \; \sin{\theta} \; d\rho \; d\theta \; d\phi = %
4 \sqrt{3} \pi$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, $\ds{\bracks{\cdots}}$ are Iverson Brackets. Namely,
  $\ds{\bracks{P} = 1}$ whenever $\ds{P}$ is true and $\ds{0}$ otherwise.

$$
\bbox[15px,#ffe,border:1px dotted navy]{\mbox{You'll see how the calculation is done without worrying with pictures}}
$$
The problem symmetry is claiming for cylindrical coordinates. The volume $\ds{V}$ is given by:
\begin{align}
V & = \iiint_{\large\mathbb{R}^{3}}\bracks{x^{2} + y^{2} + z^{2} < 6}\bracks{-\root{x^{2} + y^{2}} < z < 0}\dd x\,\dd y\,\dd z
\\[5mm] & =
\int_{0}^{2\pi}\int_{-\infty}^{0}\int_{0}^{\infty}
\bracks{r^{2} + z^{2} < 6}\bracks{-r < z}r\,\dd r\,\dd z\,\dd\phi
\\[5mm] & =
2\pi\int_{0}^{\infty}\int_{0}^{\infty}
\bracks{r^{2} + z^{2} < 6}\bracks{r > z}r\,\dd r\,\dd z
\\[5mm] & =
2\pi\int_{0}^{\infty}\bracks{\verts{z} < \root{6}}\int_{0}^{\infty}
\bracks{z < r < \root{6 - z^{2}}}r\,\dd r\,\dd z
\\[5mm] & =
2\pi\int_{0}^{\root{6}}\bracks{z < \root{6 - z^{2}}}
\int_{z}^{\root{6 - z^{2}}}r\,\dd r\,\dd z
\\[5mm] & =
2\pi\int_{0}^{\root{6}}\bracks{z < \root{3}}
{\pars{6 - z^{2}} - z^{2} \over 2}\,\dd z
=
2\pi\int_{0}^{\root{3}}\pars{3 - z^{2}}\dd z = \bbx{4\root{3}\pi}
\end{align}
