Calculate $\iiint_V dx dy dz$

Question

Problem: Calculate $$\iiint_V dx dy dz$$

where $V$ is the domain bounded by the surface $(x^2+y^2+z^2)^2=a^2xy$.

My Solution: Make the following substitution: $$\begin{cases} x = r\sin\varphi\cos\theta,\\ y = r\sin\varphi\sin\theta,\\ z = r\cos\varphi \end{cases}$$

The limit of $V$ is equal to $r^2=\frac{a^2}{2}\sin^2 \varphi \sin 2\theta$.

So the integration is equal to $$\frac{2}{3}\int_0^{\frac{\pi}{2}}d \theta\int_{0}^{\pi}r^3\sin\varphi \,d\varphi =\frac{\sqrt{2}a^3}{6}\int_0^{\frac{\pi}{2}}d \theta\int_{0}^{\pi}\sin^4\varphi (\sin 2\theta)^{\frac{3}{2}}\,d\varphi $$.

But I can't figure out how to calculate $\int (\sin 2\theta)^{\frac{3}{2}}$. I'm wondering if there's any convenient way to solve this question. I'll be grateful if there's any help. :)

Answer 1

Doing the changes of variable $x = \sin(\theta)$ and $y=\sqrt{x}$ you can get $$ ∫_0^{\pi/2}\sin(\theta)^{3/2}\cos(\theta)^{3/2}\,\mathrm{d}\theta = ∫_0^{\pi/2}\sin(\theta)^{3/2}\sqrt{1-\sin(\theta)^2}\cos(\theta)\,\mathrm{d}\theta \\ = ∫_0^1 x^{3/2}(1-x^2)^{1/2} \,\mathrm{d}x = \frac{1}{2} ∫_0^1 y^{1/4}(1-y)^{1/2} \,\mathrm{d}y $$ and this is a classical formula for the Beta function $$ ∫_0^1 y^{1/4}(1-y)^{1/2} \,\mathrm{d}y = B(5/4,3/2) = \frac{\Gamma(5/4)\Gamma(3/2)}{\Gamma(11/4)} $$ It can be simplified a bit by writing $$ \frac{\Gamma(5/4)\Gamma(3/2)}{\Gamma(11/4)} = \frac{(1/4)\Gamma(1/4)(\sqrt{\pi})/2}{21/16\,\Gamma(3/4)} = \frac{2\,\sqrt{\pi}\,\Gamma(1/4)}{21\, \Gamma(3/4)}. $$ This seems not simple ... perhaps I did an error somewhere? WolframAlfa tells me that one can express $\Gamma(3/4)$ using $\Gamma(1/4)$ but without more simplifications ...

Answer 2

The answer provided by LL3.14 was very close, but there was an error as noted in the comments. So my evaluation of the integral will end up with a similar (but not exactly the same) result involving the Beta function. Using the substitution $u=2x$, $$ I=\int_0^{\pi/2} (\sin 2x)^{3/2} dx $$ can be transformed to $$ I=\frac{1}{2}\int_0^{\pi} (\sin u)^{3/2} du = \int_0^{\pi/2} (\sin u)^{3/2} du. $$ The halving of the integration interval above is due to the symmetry of $\sin u$ about the line $u=\pi/2$. Using the substitution $v=\sin u$ (and $du/dv=1/\sqrt{1-v^2}$), $I$ can be transformed to $$ I = \int_0^1 \frac{v^{3/2}}{\sqrt{1-v^2}} dv. $$ Using the substitution $t=v^2$, $I$ can be transformed into the final form $$ I = \frac{1}{2}\int_0^1 \frac{t^{1/4}}{\sqrt{1-t}} dt = \frac{1}{2} B(5/4,1/2) = \frac{\Gamma(5/4) \Gamma(1/2)}{2\Gamma(7/4)}. $$

Since the integral of $\sin^4(x)$ for $x\in [0,\pi]$ is $3\pi/8$, the value of the complete volume integral is $$ \frac{3\sqrt{2}\pi a^3}{96} \frac{\Gamma(5/4) \Gamma(1/2)}{\Gamma(7/4)} \approx 0.2426978 a^3. $$ I checked this with a simple numerical integration using a uniform grid over the cube centered at the origin with side length $\sqrt{2}a$ (for $a=1$) and got the same answer to within 6 decimal places.

If you compute $I$ using Wolfram integrator, you will get an expression involving the complete elliptic integral of the first kind $K$. Wolfram uses the notation $$ K(m) = \int_0^{\pi/2} \frac{dx}{\sqrt{1 - m\sin^2 x}} $$ while other sources use $$ K(k) = \int_0^{\pi/2} \frac{dx}{\sqrt{1 - k^2\sin^2 x}}. $$ The identity $K(1/2) = \Gamma(1/4)^2 / (4\sqrt{\pi})$ is derived in the answers to a question called "Relation between integral, Gamma function, elliptic integral, and AGM". This is using the Wolfram notation. Another useful identity that can be found in a reference linked in the comments of that question is $\Gamma(3/4) = \pi\sqrt{2}/\Gamma(1/4)$. Using these two identities and $\Gamma(1/2)=\sqrt{\pi}$ and $\Gamma(z+1)=z\Gamma(z)$, $I$ can be written as $$ I = \frac{\Gamma(5/4) \Gamma(1/2)}{2\Gamma(7/4)} = \frac{\sqrt{2}}{3}K(1/2). $$