Evaluating $\iiint z(x^2+y^2+z^2)^{−3/2}\,dx\,dy\,dz$ over the part of the ball $x^2+y^2+z^2\le 81$ defined by $z\ge 4.5$

Question

Spherical Coordinate Homework Question

Evaluate the triple integral of $f(x,y,z)=z(x^2+y^2+z^2)^{−3/2}$ over the part of the ball $x^2+y^2+z^2\le 81$ defined by $z\ge 4.5$.

I set up a triple integral with the bounds $$0<\theta<2\pi,\ \ 0<\phi<\pi/3, \ \ 4/5/\cos(\phi)<\rho<9 $$

and end up getting the integral with those bounds: $$I=\iiint \rho^{1/2}\cos(\phi)\sin(\phi)\,\mathrm d\rho \,\mathrm d\phi \,\mathrm d\theta.$$

I have tried to carry this out so many times using online calculators, and this question is driving me crazy so any help would be appreciated!

Answer 1

We have that

• $x=r\sin\phi \cos \theta$

• $y=r\sin \phi \sin \theta$

• $z=r\cos \phi $

• $dV=r^2 \sin \phi \,d\phi\, d\theta \,dz$

then we obtain

$$\iiint_R z(x^2+y^2+z^2)^{−3/2} dV =\int_0^{2\pi} \int_0^\frac{\pi}3 \cos \phi \sin \phi \int_{\frac{4.5}{\cos \phi}}^9 dr \,d\phi\,d\theta$$

Answer 2

Once you have the integrand fixed you can try the other integral order too

$$\int_0^{2\pi}\int_{\frac{9}{2}}^9\int_0^{\cos^{-1}\left(\frac{9}{2\rho}\right)}\cos\phi\sin\phi\:d\phi\:d\rho\:d\theta = \pi\int_{\frac{9}{2}}^9 1 - \frac{81}{4\rho^2}\:d\rho$$

in this case it's neither easier nor harder than the other order. Notice though, had your integrand had the square root factor inside, this integration order would have been the way to do it.