Consider the following integral, $$ {1 \over 4\pi^{2}}\int_{0}^{2\pi}\int_{0}^{2\pi} \sqrt{\, 9 \sin^{2}\left(\theta_{1} \over 2\right) \sin^{2}\left(\theta_{2} \over 2\right)\,} \,{\rm d}\theta_{1}\,d\theta_{2} $$ This integral comes up in computing the volume of $3$dimensional special orthogonal matrices of Hessenberg form, i.e., the bottom left entry is $0$. Mathematica isn't able to produce close form solution. Numerically it's about $2.95$.
Assuming that $$I=\int_0^{2\pi} \int_0^{2\pi}\sqrt{9\sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$ is correct, $$I=3\int_0^{2\pi} \int_0^{2\pi}\sqrt{1\frac19 \sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$ then, $$I=12\int_0^{2\pi} \int_0^{\frac{\pi}{2}}\sqrt{1\frac19 \sin^2 \theta_1 \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$ $$I=12\int_0^{2\pi}E\left(\frac19 \sin^2 \frac{\theta_2 }{2}\right)\mathrm{d}\theta_2$$ ($E(m)$ is the complete elliptic integral of the second kind, with parameter $m$; for the Maple people, what you have is $E(k)$ where $k^2=m$) $$I=48\int_0^{\frac{\pi}{2}}E\left(\frac19 \sin^2 \theta_2\right)\mathrm{d}\theta_2$$ and letting $m=\sin^2 \theta_2$, $$I=24\int_0^1 \frac1{\sqrt{m}\sqrt{1m}} E\left(\frac{m}{9}\right)\mathrm{d}m$$ which Mathematica evaluates to $$12\pi^2 {}_3 F_2\left(\frac12,\frac12,\frac12 ; 1,1 ; \frac19\right)$$ where ${}_3 F_2$ is a hypergeometric function; further "simplification" can be done using the formula here. 


12 NIntegrate[EllipticE[(Sin[th/2]/3)^2], {th, 0, 2Pi}]
– J. M. Sep 23 '10 at 3:43