Volume of surface using triple integrals

Question

Using triple integrals I need to find the volume of surface bounded with $$(x^2+y^2+z^2)^2=a^3x$$. First of all I'm not sure what surface is this. I even used wolfram and geogebra but they did not sketched it as well. From equation it is obvious that if $a>0$ that $x$ can't be negative so four octants are of. Also what i did is converting to spherical coordinates and got that $r=a\sqrt[3]{\cos\phi\sin\theta}$. Then, since $\phi$ is the angle in $xOy$ is is easy to see that $-\frac{\pi}{2}\le\phi\le\frac{\pi}{2}$. But for $\theta$ not sure, looking at solution $\frac{a^3\pi}{3}$ it should have same bounds as $\phi$ but since i don't know to graph this surface I am not sure how to get these bounds.

Answer 1

Your surface is actually this: in the range of $x$ such that $x^4\le a^3x$ (whatever that range is), you have $\sqrt{y^2+z^2}=\sqrt{\sqrt{a^3x}-x^2}$. Thus, the surface can be obtained by rotating the curve $y=\sqrt{\sqrt{a^3x}-x^2}$ around $x$-axis.

Now this implies that you can use cylindrical coordinates ($x=x, y=\rho\cos\theta, z=\rho\sin\theta$), where $x$ is in the above range, $\rho$ is in the range from $0$ to $\sqrt{\sqrt{a^3x}-x^2}$ and $\theta\in[0,2\pi]$.

The range for $x$ depends on $a$. If $a\ge 0$, it is $[0,a]$. If $a\le 0$, it is $[a,0]$; this is easily checked and so I will leave the details to you.

Answer 2

It would be much easier to use spherical coordinates centered on the $x$ axis rather than the $z$ axis, giving

$r^3 = a^3\cos\theta$

Then from here set up the bounds with the angular integral first

$$\int_0^{2\pi}\int_0^a\int_0^{\cos^{-1}\left(\frac{r^3}{a^3}\right)}r^2\sin\theta\:d\theta dr d\phi = 2\pi\int_0^ar^2-\frac{r^5}{a^3}\:dr=\frac{a^3\pi}{3}$$