Use Spherical Coordinates to Evaluate the Triple Integral

Question

Use spherical coordinates to evaluate : $$ \int_{0}^{4} \int_{0}^{\sqrt{16 - x^2}} \int_{\sqrt{x^2 + y^2}}^{\sqrt{32 - (x^2 + y^2)}} \sqrt{x^2 + y^2 + z^2 \space } \, dz\,dy\,dx$$ Here is what I have tried so far $$ = \int_0^4 \int_0^{\sqrt{16 - {p^2\sin^2\phi \cos^2\theta}^2}} \left. \frac{p^2} 2 \right|_{ \sqrt{p^2\sin^2\phi \cos^2\theta + p^2\sin^2 \phi \sin^2 \theta}}^{\sqrt{p^2\sin^2\phi \cos^2\theta + p^2\sin^2 \phi \sin^2 \theta + 32}} \space d \phi \, d\theta $$ $$ = \int_0^4 \int_0^{\sqrt{16 -p^2\sin^2\phi \cos^2\theta}^{\,2}} \frac{\sqrt{p^2\sin^2\phi \cos^2\theta + p^2\sin^2 \phi \sin^2 \theta + 32}}{2}^2 - \frac{\sqrt{p^2\sin^2\phi \cos^2\theta + p^2\sin^2 \phi \sin^2 \theta}}{2}^2 \, d\phi \, d\theta $$ $$ = \int_0^4 \int_0^{\sqrt{16 - p^2\sin^2\phi \cos^2\theta}} 16 $$ $$ = \int_0^4 \left. 8\phi \vphantom{\frac11} \right|_0^{\sqrt{16 - p^2\sin^2\phi \cos^2\theta}} $$ $$ = \int_0^4 8 \sqrt{16 - p^2\sin^2\phi \cos^2\theta} $$ At this point how do i get the equation to only have one varaiable ?

Answer 1

The bounds on the $z$ coordinate suggest you are integrating an ice cream cone shape: at the lower bound you have $$ z^2=x^2+y^2 $$ the equation of a cone, and above you have a spherical cap, where the sphere has radius $4\sqrt{2}$. Note also your integrand is the radius. This all suggests spherical coordinates.

Fortunately, the bounds on $\theta$ and $\rho$ are not too hard to discern; the bounds on the $x,y$ coordinates are exactly the equation of a half circle of radius $4$, telling us that $$ 0\leq \theta\leq \pi $$ These bounds however further restrict your bounds on $\phi$, try drawing a triangle in a cross section of a plane parallel to the $x,z$ plane, you will see that when you want to stop your integration of the shape (when the projection onto the $xy$ plane is about to leave the allowed region), you have a distance from the origin in the $xy$ plane of $4$, and thus a height of $4$, using the equation $z=\sqrt{32-(x^2+y^2)}=4$, and so using some basic trig, $$ 0\leq \phi \leq \pi/4 $$ Recalling the jacobian $\rho^2\sin\phi$ and that your integrand is $\rho$ yields the integral $$ \int_0^{\pi}\int_0^{4\sqrt{2}} \int_0^{\pi/4}\rho^3\sin\phi \, \mathrm d\phi \, \mathrm\ d\rho \, \mathrm d\theta\\ = 128 (2 -\sqrt{2}) \pi $$