Finding the volume of the region bounded by $z=\sqrt{\frac{x^2}{4}+y^2}$and $x+4z=a$. Cylindrical coordinates. I would like the answer to preferably be done using either using a surface integral, or an integral with substitutions. But anything other than this is alright, if nothing else exists.
I have to find the volume of the region bounded by $z=\sqrt{\frac{x^2}{4}+y^2}$and $x+4z=a$.
So, here we have a cone and a plane "cutting" it. I definitely must do this using some some of coordinate substitution. When doing a problem in class, that is, the area to be found being bounded in between $(z-1)^2=x^2+y^2$ and $z=0.$ the substitution was made, (which would be logical here to do aswell): $$x=r\cos\varphi \\ y=r \sin\varphi \\ z=z$$ and I also understand that the bounderies being $0\leq r\leq 1,0\leq\varphi\leq2\pi,0\leq z\leq 1-r.$ But in the problem I give that is a lot more difficult to do, I know that the bounderies for $\varphi$ should be the same,but with $z$ and $r$ I find it impossible. Should I find these conditional extreme points on the cone with the plane equation or am I not seeing something quite obvious here? 

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With
$\ds{\mathcal{V} =
 \braces{\pars{x,y,z}\ \left.\vphantom{A^A}\right\vert\
         \root{{x^{2} \over 4} + y^{2}}\ <\ z\ <\ {a - x \over 4}}}$:
\begin{align}
\color{#f00}{\iiint_{\mathcal{V}}\dd x\,\dd y\,\dd z} & =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\Theta\pars{z - \root{{x^{2} \over 4} + y^{2}}}\Theta\pars{{a - x \over 4} - z}
\,\dd x\,\dd y\,\dd z
\end{align}
Where $\Theta$ is the Heaviside Step function. Then,
\begin{align}
\color{#f00}{\iiint_{\mathcal{V}}\dd x\,\dd y\,\dd z} & =
2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\Theta\pars{z - \root{x^{2} + y^{2}}}\Theta\pars{{a - 2x \over 4} - z}
\,\dd x\,\dd y\,\dd z
\\[3mm] & =
2\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}
\Theta\pars{z - \rho}\Theta\pars{{a - 2\rho\cos\pars{\phi} \over 4} - z}
\rho\,\dd\rho\,\dd\phi\,\dd z
\\[3mm] & =
2\int_{0}^{2\pi}\int_{0}^{\infty}
\bracks{{a - 2\rho\cos\pars{\phi} \over 4} - \rho}
\Theta\pars{{a - 2\rho\cos\pars{\phi} \over 4} - \rho}\rho\,\dd\rho\,\dd\phi
\\[3mm] & =
\color{#88f}{2\int_{0}^{2\pi}\int_{0}^{\infty}
\braces{{1 \over 4}\,a\rho - \half\bracks{2 + \cos\pars{\phi}}\rho^{2}}
\Theta\pars{{a/2 \over 2 + \cos\pars{\phi}} - \rho}\,\dd\rho\,\dd\phi}
\\[3mm] & =
2\int_{0}^{2\pi}
\braces{{1 \over 8}\,a\,\bracks{a/2 \over 2 + \cos\pars{\phi}}^{2} -
\half\bracks{2 + \cos\pars{\phi}}
{1 \over 3}\bracks{{a/2 \over 2 + \cos\pars{\phi}}}^{3}}\,\dd\phi
\\[3mm] & =
{1 \over 48}\,a^{3}\
\overbrace{\int_{0}^{2\pi}{\dd\phi \over \bracks{2 + \cos\pars{\phi}}^{\,2}}}
^{\ds{{4\root{3} \over 9}\,\pi}}\ =\
\color{#f00}{{\root{3} \over 108}\,\pi a^{3}}
\end{align}
Note that the $\color{#88f}{\mbox{"blue integration"}}$ requires $a > 0$ as we
can see from the $\Theta$ argument. Otherwise it vanishes out.
