Triple Integrals

Question

Find the volume of the region bounded by $(x^2+y^2+z^2)^2=x.$

I'm having issues setting my bounds (specifically $\theta$)

so far I have $0<r< [\sin(\phi) \cos(\theta)] ^{(\frac{1}{3})}$

$0<\phi<\pi$

Answer 1

Hint: Inasmuch as $(x^2+y^2+z^2)^2\ge 0$ then if $x=(x^2+y^2+z^2)^2$, $x\ge0$ also.

This restricts the region in spherical coordinates, given by

$$x=r\sin \theta \cos \phi$$ $$y= r\sin \theta \sin \phi$$ $$z=r\cos \theta $$

to

$$-\pi/2\le \phi\le \pi/2$$

The volume is then

$$\int_{-\pi/2}^{\pi/2} \int_0^{\pi} \int_0^{(\sin \theta \cos \phi)^{1/3}} r^2 \sin \theta dr d\theta d\phi$$