Volume of cylinder between planes Calculate the volume inside the cylinder $x^2+4y^2=4$, and between the two planes $z=12-3x-4y$ and $z=1$.
Converting to cylindrical coordinates gives $$r^2\cos^2\theta +4r^2\sin^2\theta=4\\ z=12-3r\cos\theta-4r\sin\theta\\ z=1$$.
$r$ goes from $0$ to $1$ and $2$. Can i use $0\le r\le 1.5$?
$0\le \theta \le 2\pi$
$1\le z \le 12-3r\cos\theta-4r\sin\theta$
$$\int_{0}^{2\pi}\int_{0}^{1.5}\int_{1}^{12-3r\cos\theta-4r\sin\theta} rdzdrd\theta $$
What have i done wrong?
 A: *

*$z=12-3x-4y\implies z=12-3\color{red}{r\cos\theta}-4\color{blue}{r\sin\theta}$.

*Note that the area over which you perform the integration is not a circle (it is an ellipse) so you will not be able to keep constant bounds for $r$. The upper bound of $r$ depends on $\theta$: for $\theta=0,r$ ranges till $2$ while for $\theta=\pi/2,r$ ranges till $1$. The upper bound on $r$ is given by:$$\frac{x^2}4+y^2=1\implies r^2(\cos^2(\theta)/4+\sin^2\theta)=1$$which gives $r_\max=(\cos^2(\theta)/4+\sin^2\theta)^{-1/2}$.

These are the two errors in your integral besides the typo on the bounds of $\theta$ (it should be from $0\to2\pi$ as you wrote earlier).

As an alternative, you could use the substitution $x=2r\cos\theta,y=r\sin\theta$ instead of the regular polar substitution. The upper bound of $r$ will become $1$, independent of $\theta$. This should be relatively easier to integrate.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
{\cal V} & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}\bracks{x^{2} + 4y^{2} < 4}
\bracks{z < 12 - 3x - 4y}}
\\[2mm] &\ \phantom{AAAAA}
\bbox[5px,#ffd]{\bracks{z > 1}\dd x\,\dd y\,\dd z}
\\[5mm] & \stackrel{x/2\ \mapsto\ x}{=}\,\,\,
2\int_{1}^{\infty}\iint_{\mathbb{R}^{2}}\bracks{x^{2} + y^{2} < 1}
\\[2mm] &\ \phantom{AAAAAAAAAAA}
\bracks{z < 12 - 6x - 4y}
\dd x\,\dd y\,\dd z
\end{align}
Lets use Cylindrical Coordinates:
\begin{align}
{\cal V} &
\,\,\,\stackrel{x/2\ \mapsto\ x}{=}\,\,\,
2\int_{1}^{\infty}\iint_{\mathbb{R}^{2}}\
\bracks{0 < \rho < 1}\ \times
\\[2mm] &\ \phantom{\stackrel{x/2\ \mapsto\ x}{=}\,\,\,\,\,\,\,}
\bracks{z < 12 - 6\rho\cos\pars{\theta} - 4\rho\sin\pars{\theta}}
\rho\,\dd \rho\,\dd\theta\,\dd z
\\[5mm] & =
2\int_{1}^{\infty}\int_{0}^{2\pi}\int_{0}^{1}
\\[2mm] &\ \phantom{2 =}
\bracks{z < 12 - 6\rho\braces{\cos\pars{\theta} +
{2 \over 3}\sin\pars{\theta}}}
\rho\,\dd \rho\,\dd\theta\,\dd z
\end{align}
With $\ds{\alpha \equiv \arctan\pars{2 \over 3}}$:
\begin{align}
{\cal V} & =
2\int_{1}^{\infty}\int_{0}^{2\pi}\int_{0}^{1}
\bracks{z < 12 -
2\root{13}\rho\cos\pars{\theta - \alpha}}
\\[2mm] &\
\phantom{AAAAA}\rho\,\dd \rho\,\dd\theta\,\dd z
\end{align}
Indeed, the last integral is $\ds{\alpha}$-independent because the integrand is a periodic function of period $\ds{2\pi}$ and it depends on the difference $\ds{\theta - \alpha}$. Namely,
\begin{align}
{\cal V} & =
2\int_{-\pi}^{\pi}\int_{0}^{1}\int_{1}^{\infty}
\bracks{z < 12 + 2\root{13}\rho\cos\pars{\theta}}
\\[5mm] &\
\phantom{AAAAA}\rho\,\dd z\,\dd \rho\,\dd\theta
\\[5mm] & =
2\int_{-\pi}^{\pi}\int_{0}^{1}\int_{1}^{12 + 2\root{13}\rho\cos\pars{\theta} \color{red}{\ >\ 1}}
\rho\,\dd z\,\dd \rho\,\dd\theta
\\[5mm] & =
4\int_{0}^{\pi}\int_{0}^{1}
\bracks{11 + 2\root{13}\rho\cos\pars{\theta}}
\rho\,\dd \rho\,\dd\theta
\\[5mm] & =
4\int_{-\pi/2}^{\pi/2}\int_{0}^{1}
\bracks{11 - 2\root{13}\rho\sin\pars{\theta}}
\rho\,\dd \rho\,\dd\theta
\\[5mm] & =
4\int_{-\pi/2}^{\pi/2}\int_{0}^{1}
11\rho\,\dd \rho\,\dd\theta = \bbx{22\pi}
\approx 69.1150 \\ &
\end{align}
