Find limits of integration for region under sphere $x^2+y^2+z^2=a^2$ inside cone $x^2+y^2=z^2$ and above $0xy$

Question

I am asked to find the limits of integration for region under sphere $x^2+y^2+z^2=a^2$ inside cone $x^2+y^2=z^2$ and above $0xy$.

Should I use spherical coordinates or cylindrical coordinates? Is it possible to use cylindrical coordinates for this case?

Thank you.

Answer 1

Spherical coordinates seem to be a good choice.

$r = \lvert a \rvert$, $\theta \in [0, \theta_m]$ with $\theta_m = \pi/4$.

Answer 2

Cylindrical coordinates is probably easiest. On top, you have $$x^2+y^2+z^2=r^2+z^2=a^2$$ So there $$z=\sqrt{a^2-r^2}$$ On the bottom, $$x^2+y^2=r^2=z^2$$ So in that case $r^2=z^2$, or $r=z$. Where the two surfaces intersect, $$r=\sqrt{a^2-r^2}$$ So $r=a/\sqrt2$. Thus you get $$\int_0^{2\pi}\int_0^{\frac a{\sqrt2}}\int_r^{\sqrt{a^2-r^2}}f(r\cos\theta,r\sin\theta,z)dz\,r\,dr\,d\theta$$ But spherical coordinates is also feasible. Then on top, $$x^2+y^2+z^2=\rho^2=a^2$$ So $\rho=a$. Then on the bottom, $$x^2+y^2=\rho^2\sin^2\phi=z^2=\rho^2\cos^2\phi$$ So $\tan\phi=1$ and $\phi=\frac{\pi}4$ Then you will have $$\int_0^{2\pi}\int_0^{\frac{\pi}4}\int_0^af(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\theta)\rho^2d\rho\sin\phi\,d\phi\,d\theta$$ So either way of setting up the problem seem relatively straightforward. Actually, I would look at the integrand $f(x,y,z)$ to see whether it's more simply expressed in one coordinate system or another.