Consider the region with $x,y,z>0$ bounded by $$\left(\frac{x}{a}+\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=\frac{x}{p}+\frac{y}{q}$$
I've tried compute this with a triple integral and Jacboian (|J|) using the coordinates $$x=ar\cos(\phi)\sin(\theta) \qquad y=br \sin(\phi)\sin(\theta) \qquad z=cr\cos(\theta)$$
Then $$r=\sin(\theta)\left(\frac a p\cos(\phi)+\frac b q\sin(\phi)\right) \\ |J|=abcr^{2} \sin(\theta)$$
I don't know what to do now or how to determine the limits of integration. Any ideas? Thank you in advance!
Having investigated this surface a bit, I feel I really need to ask whether there is a typo in the equation. Were it
$$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 + \left( \frac{z}{c} \right)^2 \ = \ \frac{x}{p} + \frac{y}{q} \ , $$
then we would simply have an ellipsoid with all three axes parallel to the coordinate axes, but with its center offset from the origin. Still, it would be not overly difficult to characterize the region in the first octant.
As written, however, the given equation takes the form
$$ \frac{( x - \frac{a^2}{2p} )^2}{a^2} \ + \ \frac{( y - \frac{b^2}{2q} )^2}{b^2} \ + \ \left( \frac{z}{c} \right)^2 \ + \ \frac{2xy}{ab} \ = \ \frac{1}{4} \ \left( \frac{a^2}{p^2} + \frac{b^2}{q^2} \right) \ , $$
which appears to be a rotated elliptic paraboloid which is tangent to the origin. With all of $ \ a, \ b,\ c, \ p, \ \text{and} \ q \ $ positive, the vertex of this surface lies outside the first octant, but there is a finite region within said octant. So the volume integral is at least finite, but characterizing the integration limits looks rather messy.
For $ \ a = 2 , \ b = 3 , \ c = 5 , \ p = 3 , \ q = 2 , \ $ I have attached graphs of the cross-sections of this surface in the $ \ xy- \ $ and $ \ xz-$planes, as well as two views of the surface which should provide an idea of the portion of the volume located in the first octant. I don't see much in the way of convenient symmetries to exploit, beyond the one about the $ \ xy-$plane...