Calculating the volume of a solid Determine the volume of the solid described by
$$x^2 + y^2 + 2z^2 \leq 1; \, x + 2y + 3z \geq 0.$$
I am pretty sure that I will need to be doing a triple integral here, but I'm not quite sure how to set up the integrals. I am mostly unsure about the bounds of integration of each variable. I have also tried converting to spherical coordinates but it didn't seem to be helpful.
I was able to discern that the region in question would be an ellipsoid intersected by a plane and was able to verify my idea with Mathematica.
This is the region in question.
 A: Your solid is precisely half the unit ball. Its volume is $\frac{2\pi}{3}$. One way to see it is to take the normal to the plane $x+2y+3z=0$, and rotate it to the vector $(0,0,1)$. The half-space $x+2y+3z\geq 0$ will then be rotated to the half-space $z\geq 0$, but the ball will remain unchanged, and it is evident that the intersection of $z\geq 0$ with the unit ball is half a ball.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson\ Bracket}$.
\begin{align}
V & \equiv
\bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}
\bracks{x^{2} + y^{2} + 2z^{2} < 1}\bracks{x + 2y + 3z > 0}
\dd x\,\dd y\,\dd z}
\\[5mm] & =
{\root{2} \over 2}\iiint_{\mathbb{R}^{3}}
\bracks{x^{2} + y^{2} + z^{2} < 1}
\bracks{x + 2y + {3\root{2} \over 2}\,z > 0}\dd x\,\dd y\,\dd z
\\[5mm] & =
\left.{\root{2} \over 2}\iiint_{r\ <\ 1}
\bracks{\vec{r}\cdot\vec{n} > 0}\dd^{3}\vec{r}
\,\right\vert_{\ \vec{\,\large n}\ \equiv\ \pars{1,2,3\root{2}/2}}
\\[5mm] & =
{\root{2} \over 2}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}
\bracks{rn\cos\pars{\theta} > 0}r^{2}\sin\pars{\theta}
\,\dd r\,\dd\theta\,\dd\phi
\\[5mm] & =
{\root{2} \over 2}\,2\pi\int_{0}^{\pi}\int_{0}^{1}
\bracks{\cos\pars{\theta} > 0}r^{2}\sin\pars{\theta}
\,\dd r\,\dd\theta
\\[2mm] & =
\root{2}\pi\int_{0}^{\pi}
\bracks{0 < \theta < {\pi \over 2}}\sin\pars{\theta}
\pars{1 \over 3}\,\dd\theta =
{\root{2} \over 3}\pi\
\overbrace{\int_{0}^{\pi/2}\sin\pars{\theta}\,\dd\theta}^{\ds{= 1}}
\\[5mm] & = \bbx{{\root{2} \over 3}\,\pi} \approx 1.4810\\ &
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson\ Bracket}$.
\begin{align}
V & \equiv
\bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}
\bracks{x^{2} + y^{2} + 2z^{2} < 1}\bracks{x + 2y + 3z > 0}
\dd x\,\dd y\,\dd z}
\\[5mm] & =
{\root{2} \over 2}\iiint_{\mathbb{R}^{3}}
\bracks{x^{2} + y^{2} + z^{2} < 1}
\bracks{x + 2y + {3\root{2} \over 2}\,z > 0}\dd x\,\dd y\,\dd z
\\[5mm] & =
\left.{\root{2} \over 2}\iiint_{r\ <\ 1}
\bracks{\vec{r}\cdot\vec{n} > 0}\dd^{3}\vec{r}
\,\right\vert_{\ \vec{\,\large n}\ \equiv\ \pars{1,2,3\root{2}/2}} =
{\root{2} \over 2}\iiint_{r\ <\ 1}
\int_{-\infty}^{\infty}{\expo{\ic k\,\vec{r}\cdot\vec{n}} \over
k - \ic 0^{+}}\,{\dd k \over 2\pi\ic}\,\dd^{3}\vec{r}
\\[5mm] & =
{\root{2} \over 2}\int_{-\infty}^{\infty}{1 \over
k - \ic 0^{+}}\iiint_{r\ <\ 1}\expo{\ic k\,\vec{r}\cdot\vec{n}}
\,\dd^{3}\vec{r}\,{\dd k \over 2\pi\ic}
\\[5mm] & =
{\root{2} \over 2}\int_{-\infty}^{\infty}{1 \over
k - \ic 0^{+}}\int_{0}^{1}4\pi r^{2}
\underbrace{\int_{\Omega_{\,\vec{r}}}\expo{\ic k\,\vec{r}\cdot\vec{n}}
\,{\dd\Omega_{\vec{r}} \over 4\pi}}_{\ds{\sin\pars{knr} \over knr}}
\,\dd r\,{\dd k \over 2\pi\ic}
\\[5mm] & =
2\root{2}\pi\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}}
{1 \over \pars{kn}^{3}}\int_{0}^{kn}r\sin\pars{r}\,\dd r\,{\dd k \over 2\pi\ic}
\\[5mm] & =
2\root{2}\pi\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}}
{\sin\pars{kn} - kn\cos\pars{kn} \over \pars{kn}^{3}}\,{\dd k \over 2\pi\ic}
\\[5mm] & =
2\root{2}\pi\,\ \underbrace{\lim_{k \to 0}
\braces{\ic\pi\,{\sin\pars{kn} - kn\cos\pars{kn} \over \pars{kn}^{3}}\,{1 \over 2\pi\ic}}}_{\ds{1 \over 6}} =
\bbx{{\root{2} \over 3}\,\pi} \approx 1.4810
\end{align}
