I am trying to determine the volume of the blue shaded area in the first figure. This volume is essentially the volume between two tilted ellipses (one up and one down). Since I am not well versed with double integrals, I will include comments on what my thinking process is so that responders can correct me and I can think about this the correct way.
The volume is given by:
$$V = \int^{2\pi}_0\int^r_0h(\phi)rdrd\phi$$
where $h$ is the height, $r$ is the distance from the left focus and $\phi$ is the angle measured from the focus (see Fig. 2). In Fig. 2 "P" is the left focus and "O" is the center of the ellipse. Now, I guess the way I am seeing this is that there is a slice in the $r$ direction with a height of $h$ (z-axis). The volume is then determined by sweeping this slice along the $\phi$ direction. It turns out that the equation for this slice is given by,
$$h(\phi)r = \frac{R^2\sin^3\beta}{(1-\cos\beta\cos\phi)^2}$$
where $\beta$ is an angle that is constant. $R$ is the semi-minor axis of the ellipse and $\frac{R}{\sin\beta}$ is the semi-major axis of the ellipse as shown in the second figure. From Wikipedia, the polar form relative to the focus is:
$$r(\phi)=\frac{a(1-e^2)}{1\pm e\cos\phi}$$
where $a$ is the semi-major axis and
$$e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{R^2}{\frac{R^2}{\sin^2\beta}}} = \sqrt{1-\sin^2\beta} = \cos\beta $$
where $b$ is the semi-minor axis.
So, continuing on, this is what I think the double integral becomes
$$V = \int^{2\pi}_0\int^{\frac{a(1-e^2)}{1\pm e\cos\phi}}_0\frac{R^2\sin^3\beta}{(1-\cos\beta\cos\phi)^2}drd\phi$$
or
$$V = 2{R^2\sin^3\beta}\int^{\pi}_0\int^{\frac{a(1-e^2)}{1\pm e\cos\phi}}_0\frac{1}{(1-\cos\beta\cos\phi)^2}drd\phi$$
Since the volume is symmetrical about the $\phi$ direction, the outer integral upper bound is replaced by $\pi$ and the entire integral is multiplied by 2.
The upper bound for the inside integral can be simplified to
$$\frac{a(1-e^2)}{1\pm e\cos\phi} = \frac{\frac{R}{\sin\beta}(1-\cos^2\beta)}{1\pm \cos\beta\cos\phi} = \frac{R\sin\beta}{1\pm \cos\beta\cos\phi} $$
Since the upper bound for the integral on $r$ is a function of $\phi$, it seems to me that the inside integral has to be the one over $r$ and not $\phi$ (please confirm). So after evaluating the inside integral, I get:
$$V = 2{R^2\sin^3\beta}\int^{\pi}_0\frac{1}{(1-\cos\beta\cos\phi)^2}{\frac{R\sin\beta}{1\pm \cos\beta\cos\phi}}d\phi$$
Now before this integral even gets evaluated, I'm starting to get bogged down. From Fig. 2, I think the portion of the ellipse where the second half of the denominator is equal to $1-\cos\beta\cos\phi$ is to the right of the focus "P" and to the left of the focus "P", the right half of the denominator is equal to $1+\cos\beta\cos\phi$.
So should I split the integral up into two portions as in,
$$V = 2{R^3\sin^4\beta}\int_{0}^{\frac{\pi}{2}}\frac{1}{(1-\cos\beta\cos\phi)^2}{\frac{1}{1 - \cos\beta\cos\phi}}d\phi + 2{R^3\sin^4\beta}\int_{\frac{\pi}{2}}^{\pi} \frac{1}{(1-\cos\beta\cos\phi)^2}{\frac{1}{1 + \cos\beta\cos\phi}}d\phi$$
Any help including coming up with the final solution is welcome.
EDIT: Drawing of the volume in x and z plane only has been added for clarification.
