triple integral - ecliptic coordinate I need to find the $V$ by triple  integral.
the limits from up is (1) - ecliptic cone.
and from dwon - (2) - elepsoide.  
$$(1) \ \ \ \ z=-\sqrt{3x^2+5y^2}$$
$$(2) \ \ \ \ {3 \over 10}x^2+5y^2+{z^2 \over 3}=8$$

I try to use:
$x=5r\cos\theta$
$y=3r\sin\theta$
$z=z$  
but I'm not sure it's good coordinate and I can't find the limit.
 A: This one is obviously a little messy, but there is a simple methodology by which the volume may be reduced to a single integral.  The idea is to compute the cross sectional area $A(z)$ for each value of $z$ in the volume, and then integrate over $z$ to get the volume.  
First, we needs the bounds in $z$.  This is done by observing that the relevant $z$ vertex of the ellipsoid is at $z=-2 \sqrt{6}$.  In between, it should be clear that each cross section is an intersection of two ellipses centered at the origin.  The first has equation
$$\frac{x^2}{z^2/3} + \frac{y^2}{z^2/5} = 1$$
the other has equation
$$\frac{x^2}{\frac{80}{3} \left(1-\frac{z^2}{24}\right)} + \frac{y^2}{\frac{8}{5} \left(1-\frac{z^2}{24}\right)}=1$$
There are three different scenarios: one in which the first ellipse is entirely within the second, one in which the second is entirely within the first, and one in which they stick out of each other.  The intervals in $z$ where these different scenarios occur may be determined by equating the vertices in $x$ and $y$, respectively, in the first and second ellipses.  The result is three intervals in $z$: 


*

*I: $z \in \left[-2 \sqrt{6},-\sqrt{\frac{240}{13}}\right]$

*II: $z \in \left[-\sqrt{\frac{240}{13}},-\sqrt{6}\right]$

*III: $z \in \left[-\sqrt{6},0\right]$


In interval I, the area is simply the area of the ellipsoid cross section:
$$A(z) = \frac{8 \sqrt{6}}{3} \pi \left(1-\frac{z^2}{24}\right)$$
(Note that I used the formula $A=\pi a b$ as the area of the ellipse $(x^2/a^2)+(y^2/b^2)=1$.)
In interval III, the area is simply the area of the conical cross section:
$$A(z) = \frac{\pi}{\sqrt{15}} z^2$$
That leaves interval II, in which there is an intersection:

The cross sectional area is broken up into 2 pieces: one bounded by the first ellipse at the sides, and another bounded by the second ellipse in the center.  The area is then
$$4 \int_{x_0(z)}^{z/\sqrt{3}} dx \: \frac{1}{\sqrt{5}}\sqrt{z^2-3 y^2} + 2 \int_{-x_0(z)}^{x_0(z)} dx \: \sqrt{\frac{8}{5}} \sqrt{1-\frac{z^2}{24}-3 \frac{x^2}{80}}$$
where 
$$x_0(z)=\frac{\sqrt{z (z+3)-24}}{3 \sqrt{\frac{1}{z}-\frac{1}{10}}}$$
