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? 

My solution is another way of stating @Jack D'Aurizio's solution. We want to rotate coordinates so that the plane $x+4z=a$ is horizontal. This can be achieved through
$$\begin{array}{rl}x^{\prime}&=\frac4{\sqrt{17}}x-\frac1{\sqrt{17}}z\\
z^{\prime}&=\frac1{\sqrt{17}}x+\frac4{\sqrt{17}}z\end{array}$$
The inverse transformation is
$$\begin{array}{rl}x&=\frac4{\sqrt{17}}x^{\prime}+\frac1{\sqrt{17}}z^{\prime}\\
z&=-\frac1{\sqrt{17}}x^{\prime}+\frac4{\sqrt{17}}z^{\prime}\end{array}$$
In terms of these coordinates, the plane is $z^{\prime}=\frac a{\sqrt{17}}$. Then the equation of the boundary reads
$$\frac{(-x^{\prime}+4z^{\prime})^2}{17}=\frac{(4x^{\prime}+z^{\prime})^2}{4\cdot17}+y^2$$
Or
$$(x^{\prime})^2-8x^{\prime}z^{\prime}+16(z^{\prime})^2=\frac14\left(16(x^{\prime})^2+8x^{\prime}z^{\prime}+(z^{\prime})^2\right)+17y^2$$
$$0=3(x^{\prime})^2-6x^{\prime}z^{\prime}-\frac{63}4(z^{\prime})^2+17y^2=3(x^{\prime}-z^{\prime})^2-\frac{75}4(z^{\prime})^2+17y^2$$
EDIT: Whoops, mistake on the above line. Should be
$$0=3(x^{\prime})^2+10x^{\prime}z^{\prime}-\frac{63}4(z^{\prime})^2+17y^2=3\left(x^{\prime}+\frac53z^{\prime}\right)^2-\frac{289}{12}(z^{\prime})^2+17y^2$$
Now we can write this as the equation of an ellipse
$$\frac{\left(x^{\prime}+\frac53z^{\prime}\right)^2}{\frac{289}{36}(z^{\prime})^2}+\frac{y^2}{\frac{17}{12}(z^{\prime})^2}=\frac{\left(x^{\prime}+\frac53z^{\prime}\right)^2}{\bar a^2}+\frac{y^2}{\bar b^2}=1$$
This is the equation of an ellipse with semi-major axis $\bar a=\frac{17}6z^{\prime}$, semi-minor axis $\bar b=\frac12\sqrt{\frac{17}3}z^{\prime}$ parallel to the $x^{\prime}y^{\prime}$-plane and centered at $\langle x_c^{\prime},y_c,z_c^{\prime}\rangle=\langle-\frac53z^{\prime},0,z^{\prime}\rangle$, so we know its area is $A(z^{\prime})=\pi\bar a\bar b=\pi\frac{17}{12}\sqrt{\frac{17}3}(z^{\prime})^2$, so we can add up the areas of the ellipses to get the volume
$$V=\int_0^{\frac a{\sqrt{17}}}A(z^{\prime})dz^{\prime}=\int_0^{\frac a{\sqrt{17}}}\pi\frac{17}{12}\sqrt{\frac{17}3}(z^{\prime})^2dz^{\prime}=\left.\pi\frac{17}{12}\sqrt{\frac{17}3}\frac{(z^{\prime})^3}3\right|_0^{\frac a{\sqrt{17}}}=\frac{\pi a^3}{36\sqrt3}$$
So with the edit, my solution is now consistent with the others.
