Having trouble deriving the volume of an elliptical ring toroid

Like the title and the summary suggest, I can derive the volume ##V=2\,\pi^{2}\,r^{2}\,R## for a ring torus - a doughnut-style toroid (one such that the major radius ##R## > the minor radius ##r##, and it therefore has a hole at the center) that is of circular cross-section. But I want to be able to prove the general case, that is, derive the volume of any elliptical ring toroid, including those of non-circular cross-section (those with an elliptical cross-section of non-zero eccentricity, and therefore have semi-major and semi-minor axes ##a\neq b##).

I run into a road block at a point in my integration where I have to make a trigonometric substitution. Specifically, I'm having trouble getting the upper bound of integration I want while making this change of variables from ##x## to ##\theta##. I had to do a trig substitution while deriving the volume of the ring torus (the one w/ circular cross-section) as well. ##\sin\theta## ended up being equal to 1, so ##\theta## (and therefore my upper bound) ended up being equal to ##\frac{\pi}{2}##, and everything worked out beautifully...

...not so with the one of elliptical cross-section. Unfortunately I don't really know how to show you where I'm hitting a roadblock without simply starting at the beginning. So I'll start by posting my successful derivation of the volume of the ring torus of circular cross-section first, and then post what I have so far for the derivation of the volume of a ring toroid of non-circular cross-section up to the point where I get stuck. I apologize in advance for the lengthiness of this post .


DERIVATION OF THE VOLUME OF AN ELLIPTICAL RING TORUS (A RING TOROID OF CIRCULAR CROSS-SECTION):

I started with a circle of radius ##r## in the ##xy##-plane whose center is at the point ##(0,R,0)##, where ##R>r## so that there is a hole at the center of the torus, doughnut-style:


Since the implicit equation of a circle becomes two functions when y is solved for x explicitly (the upper and lower halves of the circle), what I did was revolve the upper half of the circle about the x-axis first, giving me the volume I labeled ##V_{1}##. I then did the same for the lower half, called it ##V_{2}##, and subtracted it from ##V_{1}## to get ##V_{torus}##. Starting with the equation of a circle and solving for ##y##, we have:

$$x^2+(y-R)^2=r^2\rightarrow (y-R)^2=r^2-x^2\rightarrow y-R=\pm\sqrt{r^2-x^2}\rightarrow y=+\sqrt{r^2-x^2}+R $$

Substituting in the function ##\sqrt{r^2-x^2}+R## for ##y## in the "solid of revolution" formula and evaluating, we have:

$$V_1=\int_{a}^{b}{\pi y^2\;dx}=\pi\int_{-r}^{r}{\left(\sqrt{r^2-x^2}+R\right)^2\;dx}$$ $$=2\pi\int_{0}^{r}{\left(\sqrt{r^2-x^2}+R\right)^2\;dx}=2\pi\int_{0}^{r}{\left(r^2-x^2+2R\sqrt{r^2-x^2}+R^2\right)\;dx}$$ $$=2\pi\int_{0}^{r}{r^2\mathrm{\ } dx}-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx+2\pi\int_{0}^{r}{2R\sqrt{r^2-x^2}\mathrm{\ } dx}}+2\pi\int_{0}^{r}{R^2\mathrm{\ } dx}$$ $$=2\pi\cdot r^2\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx+2\pi\cdot2R\int_{0}^{r}{\sqrt{r^2-x^2}\mathrm{\ } dx}}+2\pi\cdot R^2\int_{0}^{r}dx$$ $$=2\pi\left(r^2+R^2\right)\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx+4\pi R\int_{0}^{r}{\sqrt{r^2-x^2}\mathrm{\ } dx}}$$

At this point I grouped the first two integrals together and evaluated third integral separately because it required a trigonometric substitution. I dealt with the first two integrals first:

$$2\pi\left(r^2+R^2\right)\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx=2\pi\left(r^2+R^2\right)\left[x\right]{r\atop0}-2\pi\left[\frac{x^3}{3}\right]{r\atop0}}$$ $$=2\pi\left(r^2+R^2\right)\left(r-0\right)-2\pi\left(\frac{r^3}{3}-\frac{0^3}{3}\right)=2\pi r\left(r^2+R^2\right)-\frac{2\pi r^3}{3}$$ $$=2\pi r^3+2\pi rR^2-\frac{2\pi r^3}{3}=\mathbf{2}{\pi r}{R}^\mathbf{2}-\frac{\mathbf{4}{\pi}{r}^\mathbf{3}}{\mathbf{3}}$$

...and now for the integral that required a trig substitution:

• ##cos\ \theta=\frac{\sqrt{r^2-x^2}}{r}\rightarrow\sqrt{r^2-x^2}=r\ cos\ \theta##
• Also, ##sin\ \theta=\frac{x}{r}\rightarrow x=r\ sin\ \theta\rightarrow dx=r\ cos\ \theta\ d\theta##
• Also, when ##x=0##, ##sin\ \theta=\frac{0}{r}=0\rightarrow\theta=0## when ##sin\ \theta=0## (upper limit)
• Also, when ##x=r##, ##sin\ \theta=\frac{r}{r}=1\rightarrow\theta=\frac{\pi}{2}## when ##sin\ \theta=1## (lower limit)

Substituting these quantities and these upper and lower bounds back into our integral, we have:


Here, I had to make another substitution: let ##u=2\theta\rightarrow du=2\ d\theta\rightarrow d\theta=\frac{1}{2}du##

$$\rightarrow4\pi r^2R\int_{0}^{\frac{\pi}{2}}{\frac{1}{2}\ d\theta}+4\pi r^2R\int_{0}^{\frac{\pi}{2}}{\frac{1}{2}cos\ 2\theta\ d\theta}=4\pi r^2R\cdot\frac{1}{2}\int_{0}^{\frac{\pi}{2}}d\theta+4\pi r^2R\cdot\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{cos\ u\cdot\frac{1}{2}du}$$ $$=2\pi r^2R\int_{0}^{\frac{\pi}{2}}d\theta+4\pi r^2R\cdot\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{cos\ u\ du}$$ $$=2\pi r^2R\left[\theta\right]{\frac{\pi}{2}\atop0}+\pi r^2R\left[sin\ u\right]{\frac{\pi}{2}\atop0}=2\pi r^2R\left[\theta\right]{\frac{\pi}{2}\atop0}+\pi r^2R\left[sin\ 2\theta\right]{\frac{\pi}{2}\atop0}$$ $$=2\pi r^2R\left(\frac{\pi}{2}-0\right)+\pi r^2R\left(sin\ \pi-sin\ 0\right)=2\pi r^2R\left(\frac{\pi}{2}\right)+\pi r^2R\left(0-0\right)=\pi^2r^2R+\pi r^2R\left(0\right)={\pi}^\mathbf{2}{r}^\mathbf{2}{R}$$

Putting things back together, we have:

$$\left(2\pi\left(r^2+R^2\right)\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx}\right)+\left(4\pi R\int_{0}^{r}{\sqrt{r^2-x^2}\mathrm{\ } dx}\right)$$ $$=2\pi{rR}^2-\frac{4\pi r^3}{3}+\pi^2r^2R$$ $$Thus, {V}_\mathbf{1}=\mathbf{2}{\pi}{{rR}}^\mathbf{2}-\frac{\mathbf{4}{\pi}{r}^\mathbf{3}}{\mathbf{3}}+{\pi}^\mathbf{2}{r}^\mathbf{2}{R}$$

Then I had to calculate the volume beneath the lower half of the circle ##y=-\sqrt{r^2-x^2}+R##:

$$V_2=\int_{a}^{b}{\pi y^2\;dx}=\pi\int_{-r}^{r}{\left(-\sqrt{r^2-x^2}+R\right)^2\;dx}=2\pi\int_{0}^{r}{\left(-\sqrt{r^2-x^2}+R\right)^2\;dx}$$ $$=2\pi\int_{0}^{r}{\left(r^2-x^2-2R\sqrt{r^2-x^2}+R^2\right)\;dx}$$ $$=2\pi\int_{0}^{r}{r^2\mathrm{\ } dx}-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx-2\pi\int_{0}^{r}{2R\sqrt{r^2-x^2}\mathrm{\ } dx}}+2\pi\int_{0}^{r}{R^2\mathrm{\ } dx}$$ $$=2\pi\cdot r^2\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx-2\pi\cdot2R\int_{0}^{r}{\sqrt{r^2-x^2}\mathrm{\ } dx}}+2\pi\cdot R^2\int_{0}^{r}dx$$ $$=2\pi\left(r^2+R^2\right)\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx-4\pi R\int_{0}^{r}{\sqrt{r^2-x^2}\mathrm{\ } dx}}$$ ##\rightarrow## I already calculated these integrals in the previous step, so I already know that ##2\pi\left(r^2+R^2\right)\int_{0}^{r}dx-2\pi\int_{0}^{r}{x^2\mathrm{\ } dx}=2\pi rR^2-\frac{4\pi r^3}{3} ##, and I know that ##4\pi R\int_{0}^{r}{\sqrt{r^2-x^2}\mathrm{\ } dx}=\pi^2r^2R##, but whereas I added these values together in the previous step, I have to subtract the latter from the former here:

$$\left(2\pi r R^2-\frac{4\pi r^3}{3}\right)-\left(\pi^2r^2R\right)$$


...and finally, I calculated ##V1-V2= Vtorus## and got:

$$2\pi{rR}^2-\frac{4\pi r^3}{3}+\pi^2r^2R-\left(2\pi r R^2-\frac{4\pi r^3}{3}-\pi^2r^2R\right)$$ $$=2\pi{rR}^2-\frac{4\pi r^3}{3}+\pi^2r^2R-2\pi{rR}^2+\frac{4\pi r^3}{3}+\pi^2r^2R=2\pi^2r^2R$$


OK...hopefully you're not so bored yet that that I've lost your attention . Now I'll post what I have so far for the derivation of the volume of a ring toroid of non-circular cross-section up to the point that I get stuck...


DERIVATION OF THE VOLUME OF AN ELLIPTICAL PARABOLOID OF NON-CIRCULAR CROSS-SECTION:

I started with an ellipse of semi-major axis ##a## and semi-minor axis ##b## in the ##xy##-plane whose center is at the point ##(0,R,0)##, where ##R>a## (or ##b##, whichever semi-axis is parallel to the major axis) so that there is a hole at the center of the torus, doughnut-style. We have the same setup as before, the only difference being the shape of the toroid's cross-section:



Since the implicit equation of an ellipse becomes two functions when y is solved for x explicitly (its upper and lower halves), what I did was revolve the upper half of the ellipse about the x-axis first, giving me the volume I labeled ##V_{1}##. I then did the same for the lower half, called it ##V_{2}##, and subtracted it from ##V_{1}## to get ##V_{toroid}##. Starting with the equation of an ellipse and solving for ##y##, we have:

$$\frac{x^2}{a^2}+\frac{\left(y-R\right)^2}{b^2}=1\rightarrow\frac{\left(y-R\right)^2}{b^2}=1-\frac{x^2}{a^2}\rightarrow\left(y-R\right)^2=b^2\left(1-\frac{x^2}{a^2}\right)$$ $$\rightarrow y-R=\pm\sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}\rightarrow y=\pm b\sqrt{1-\frac{x^2}{a^2}}+R$$

Substituting in the function ##y=b\sqrt{1-\frac{x^2}{a^2}}+R## for ##y## in the "solid of revolution" formula and evaluating, we have:

$$V_1=\int_{a}^{b}{\pi y^2\;dx}=\pi\int_{-b}^{b}{\left(y=b\sqrt{1-\frac{x^2}{a^2}}+R\right)^2\;dx}=2\pi\int_{0}^{b}{\left(b\sqrt{1-\frac{x^2}{a^2}}+R\ \right)^2\;dx}$$ $$=2\pi\int_{0}^{b}{\left(b^2\left(1-\frac{x^2}{a^2}\right)+2b\sqrt{1-\frac{x^2}{a^2}}+R^2\right)\;dx}$$ $$=2\pi\int_{0}^{b}{b^2\left(1-\frac{x^2}{a^2}\right)dx}+2\pi\int_{0}^{b}{2b\sqrt{1-\frac{x^2}{a^2}}\mathrm{\ } dx}+2\pi\int_{0}^{b}{R^2\mathrm{\ } dx}$$ $$=2\pi\cdot b^2\int_{0}^{b}{\left(1-\frac{x^2}{a^2}\right)dx+2\pi\cdot2b\int_{0}^{b}{\sqrt{1-\frac{x^2}{a^2}}\mathrm{\ } dx}}+2\pi\cdot R^2\int_{0}^{b}dx$$ $$=2\pi b^2\int_{0}^{b}{1\ dx-2\pi b^2\int_{0}^{b}\frac{x^2}{a^2}\ dx+4\pi b\int_{0}^{b}{\sqrt{1-\frac{x^2}{a^2}}\mathrm{\ } dx}}+2\pi R^2\int_{0}^{b}dx$$ $$=\left(2\pi b^2+2\pi R^2\right)\int_{0}^{b}{dx-2\pi b^2\cdot\frac{1}{a^2}\int_{0}^{b}x^2dx+4\pi b\int_{0}^{b}{\sqrt{1-\frac{x^2}{a^2}}\mathrm{\ } dx}}$$ $$=2\pi\left(b^2+R^2\right)\int_{0}^{b}{dx-\frac{2\pi b^2}{a^2}\int_{0}^{b}x^2dx+4\pi b\int_{0}^{b}{\sqrt{1-\frac{x^2}{a^2}}\mathrm{\ } dx}}$$

At this point I grouped the first two integrals together and evaluated third integral separately because it required a trigonometric substitution. I dealt with the first two integrals first:

$$2\pi\left(b^2+R^2\right)\int_{0}^{b}dx-\frac{2\pi b^2}{a^2}\int_{0}^{b}{x^2\mathrm{\ } dx=2\pi\left(b^2+R^2\right)\left[x\right]{b\atop0}-\frac{2\pi b^2}{a^2}\left[\frac{x^3}{3}\right]{b\atop0}}$$ $$=2\pi\left(b^2+R^2\right)\left(b-0\right)-\frac{2\pi b^2}{a^2}\left(\frac{b^3}{3}-\frac{0^3}{3}\right)=2\pi\left(b^2+R^2\right)\left(b\right)-\frac{2\pi b^2}{a^2}\left(\frac{b^3}{3}\right)$$ $$=2\pi b\left(b^2+R^2\right)-\frac{2\pi b^5}{{3a}^2}=2\pi b^3+2\pi bR^2-\frac{2\pi b^5}{{3a}^2}=\frac{6\pi{a^2b}^3}{{3a}^2}+2\pi bR^2-\frac{2\pi b^5}{{3a}^2}$$ $$=2\pi bR^2+\frac{6\pi{a^2b}^3-2\pi b^5}{{3a}^2}=\mathbf{2}{\pi b}{R}^\mathbf{2}+\mathbf{2}{\pi}{b}^\mathbf{3}\frac{\mathbf{3}{a}^\mathbf{2}-{b}^\mathbf{2}}{{\mathbf{3}{a}}^\mathbf{2}}$$

...and now for the integral that required a trig substitution:

• ##cos\ \theta=\frac{\sqrt{1-\frac{x^2}{a^2}}}{1}\rightarrow\sqrt{1-\frac{x^2}{a^2}}=cos\ \theta##
• Also, ##sin\ \theta=\frac{\frac{x}{a}}{1}=\frac{x}{a}\rightarrow x=a\ sin\ \theta\rightarrow dx=a\ cos\ \theta\ d\theta##
• Also, when ##x=0, sin\ \theta=\frac{0}{a}=0\rightarrow\theta=0## when ##sin\ \theta=0## (upper limit)
• Also, when ##x=b, sin\ \theta=\frac{b}{a}=?\rightarrow\theta=?## when ##sin\ \theta=\frac{b}{a}## (lower limit)

...and this is where I get stuck. Actually, I'm able to carry on with calculating the integral itself, I just can't evaluate it over an interval because I don't know what my upper bound is. In the previous derivation, when the cross-section was circular and I just had a radius ##r## to contend with, you can see that ##sin\ \theta=\frac{x}{r}##, and that ##sin\ \theta=1## (and therefore ##\theta=\frac{\pi}{2}##) when ##x=r##, and everything worked out nicely from there. Now, I feel like the upper bound has to equal ##\frac{\pi}{2}## in this derivation just like in the previous one in order for things to work out in the end. But this time ##sin\ \theta=\frac{x}{a}##, not ##sin\ \theta=\frac{x}{r}##, and substituting in my upper limit ##b## gives me ##sin\ \theta=\frac{b}{a}## which, unless ##a=b## (##=r##), cannot equal 1. So if ##sin\ \theta\neq 1## then ##\theta\neq \frac{\pi}{2}##, and I don't have an upper limit that makes the integral work out nicely. In fact, I don't have an exact value at all...just the formula ##sin\ \theta=\frac{b}{a}## (or ##\theta=arcsin\ \frac{b}{a}##, but I have my doubts about going down that road and substituting in ##arcsin\ \frac{b}{a}## for my upper limit...I can't see that resulting in an integral that isn't messier than a pair of soiled shorts LOL).

So that's it - I can't come up with a value that makes sense for the upper limit of my integral during a trig substitution and the associated change of variables from ##x## to ##\theta##. Any ideas on where to go from here? Apparently the volume of an elliptical ring toroid is ##{V}_{{toroid}}=2\pi^{2}abR## - almost the same as ##{V}_{{torus}}##, only ##r^2## gets replaced w/ ##ab##. I feel like I'm so close, yet so far...

Thanks in advance,

Eric

These calculations are much simpler if you use an appropriate parametrization.

For the circular torus (rotating a circle of radius [itex]r[/itex] in the [itex]xz[/itex] plane with center [itex](R,0,0)[/itex] about the [itex]z[/itex] axis), we can set [tex]\begin{split}
x &= (R + \rho \cos v)\cos u \\
y &= (R + \rho \cos v) \sin u \\
z &= \rho \sin v\end{split}[/tex] for [itex]0 \leq \rho \leq r[/itex], [itex]0 \leq u < 2\pi[/itex], [itex]0 \leq v < 2\pi[/itex]. Rather than calculate the jacobian by hand, I did it using SymPy and obtained [itex]\rho(R + \rho\cos v)[/itex]. The volume is then [tex]
\int_0^{2\pi} \int_0^{2\pi} \int_0^r \rho(R + \rho \cos v)\,d\rho\,du\,dv = 2\pi^2 Rr^2.[/tex] The 2D ellipse is generally parametrized as [tex]\begin{split}
x = a\rho\cos u \\
y = b \rho \sin u \end{split}[/tex] with [itex]\rho = 1[/itex] being the ellipse. Adapting this, we can parametrize the elliptical toroid as [tex]\begin{split}
x &= (R + a\rho \cos v)\cos u \\
y &= (R + a\rho \cos v) \sin u \\
z &= b\rho \sin v\end{split}[/tex] where now [itex]0 \leq \rho \leq 1[/itex] and the jacobian is (again using SymPy) [tex]ab\rho(R + a\rho \cos v).[/tex]

pasmith said:

These calculations are much simpler if you use an appropriate parametrization.

For the circular torus (rotating a circle of radius [itex]r[/itex] in the [itex]xz[/itex] plane with center [itex](R,0,0)[/itex] about the [itex]z[/itex] axis), we can set [tex]\begin{split}
x &= (R + \rho \cos v)\cos u \\
y &= (R + \rho \cos v) \sin u \\
z &= \rho \sin v\end{split}[/tex] for [itex]0 \leq \rho \leq r[/itex], [itex]0 \leq u < 2\pi[/itex], [itex]0 \leq v < 2\pi[/itex]. Rather than calculate the jacobian by hand, I did it using SymPy and obtained [itex]\rho(R + \rho\cos v)[/itex]. The volume is then [tex]
\int_0^{2\pi} \int_0^{2\pi} \int_0^r \rho(R + \rho \cos v)\,d\rho\,du\,dv = 2\pi^2 Rr^2.[/tex] The 2D ellipse is generally parametrized as [tex]\begin{split}
x = a\rho\cos u \\
y = b \rho \sin u \end{split}[/tex] with [itex]\rho = 1[/itex] being the ellipse. Adapting this, we can parametrize the elliptical toroid as [tex]\begin{split}
x &= (R + a\rho \cos v)\cos u \\
y &= (R + a\rho \cos v) \sin u \\
z &= b\rho \sin v\end{split}[/tex] where now [itex]0 \leq \rho \leq 1[/itex] and the jacobian is (again using SymPy) [tex]ab\rho(R + a\rho \cos v).[/tex]

Spoiler: SymPy code

Python:

from sympy import *

R = Dummy('R', positive=True)
a = Dummy('a', positive=True)
b = Dummy('b', positive=True)

r = Dummy('\\rho', positive=True)
u = Dummy('u', real=True)
v = Dummy('v', real=True)

x = (R + a*r*cos(v))*cos(u)
y = (R + a*r*cos(v))*sin(u)
z = b*r*sin(v)

jac = trigsimp(
    Matrix([
        [diff(x, r), diff(x, u), diff(x, v)],
        [diff(y, r), diff(y, u), diff(y, v)],
        [diff(z, r), diff(z, u), diff(z, v)],
    ]).det()
)

No doubt this problem is more easily solved using spherical coordinates than it is using Cartesian coordinates. I must admit I remember very little of what I learned about the Jacobian 20-something years ago, but I'm assuming you're using it here to transform from Cartesian to spherical coordinates? I also don't know any Python coding, but thanks for including it just in case it might be helpful/useful to someone who stumbles across this thread. I did however compute by hand your triple integral for my own edification, and sure enough it checks out.

If one were to try to proceed using Cartesian coordinates, do you see a way forward from the point at which I got stuck? Obviously there's no way to make ##sin\ \theta=\frac{a}{b}=1## unless ##a=b##, which defeats the purpose because that means that I'm deriving the volume of a toroid of circular cross-section all over again (and not one with an elliptical cross-section). Do you think using ##arcsin\ \frac{b}{a}## as the upper bound in the integral where I got stuck will lead to anything fruitful, or is this approach a dead end? My gut tells me that one should in theory be able to derive the volume of a ring torus of non-circular cross-section using the "solid of revolution" method, and do so entirely in Cartesian coordinates, as messy as it may get. Having said that, neither my derivation of the volume of a ring torus nor what I have thus far for the derivation of the volume of a ring toroid of non-circular cross-section get particularly messy or complicated...they're just time-consuming because there's a fair amount of combing/factoring/simplifying/substituting etc. to get from start to finish.

You are revolving around the wrong axis.

If you set up the cross-section as [tex]
\frac{x^2}{a^2} + \frac{(y-R)^2}{b^2} \leq 1[/tex] then you want to revolve around the [itex]x[/itex] axis, not the [itex]y[/itex] axis. The volume is then [tex]
V = 4\pi \int_{R- b}^{R + b} y x(y)\,dy = 4\pi a \int_{R-b}^{R+b} y \sqrt{1 - \frac{(y-R)^2}{b^2}}\,dy.[/tex] Now substitute [itex]y = R + bu[/itex] to obtain [tex]
V = 4\pi ab \int_{-1}^1 (R+ bu)\sqrt{1 - u^2}\,du.[/tex] The contribution from [itex]u\sqrt{1 - u^2}[/itex] is zero, because it is an odd function integrated over a symmetric interval; the contribution from [itex]\sqrt{1 - u^2}[/itex] is twice the integral over positive values of [itex]u[/itex], so the volume is [tex]
V = 8\pi abR \int_0^1 \sqrt{1 - u^2}\,du.[/tex]

Isn't the volume you seek equal to the cross-sectional area of the smaller ellipse multiplied by the length of the perimeter about the central axis of the bigger ellipse? Let ##a_1## and ##b_1## be the semi-major and semi-minor axis of the small ellipse and likewise ##a_2## and ##b_2## for the large ellipse. The cross-sectional area of the small ellipse is ##\pi a_1b_1##. We need to compute the perimeter of the larger ellipse which turns out to be an elliptic integral(see Perimeter of ellipse);
$$
P=4a_2\int_0^{\frac{\pi}{2}}\sqrt{1-e^2 \sin^2{\theta}} d\theta
$$
and the volume becomes
$$
4\pi a_2 a_1b_1 E(\frac{\pi}{2},e^2)
$$
$$
e^2=1- \frac{(b_1 + b_2)^2}{(a_1 +a_2)^2}
$$
I think your problem eludes a closed form solution because of broken radial symmetry.