My objective:

Using spherical coordinates, set up and compute an integral to find the volume of the ice-cream-cone shaped solid lying above the cone $z = \sqrt{x^2 + y^2}$ and below the sphere $\rho = 6 \cos(\phi)$.

I think that $\rho = 6 \cos(\phi)$ is a sphere with points at $(0,0,0)$ and $(0,0,6)$, so the radius is 3

So I setup the integral to find the volume as follows: $\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{3} 1\rho\sin{\phi}d\rho d\phi d\theta$ which works out to $2\pi(1 - \frac{\sqrt{2}}{2})\frac{3^3}{3} = (\frac{18\pi}{3}) (2 - \sqrt{2})$ if I'm not mistaken, but this is not correct.

I'm thinking my bounds for $\rho$ are incorrect (because I'm fairly confident about the others), but how are they incorrect?

  • $\begingroup$ First draw a picture, draw it in $(z,r)$ for simplicity ($z$ being vertical and $r$ being radius, your $\sqrt{x^2+y^2}=r$, then post the pic $\endgroup$ – Alec Teal Nov 1 '14 at 16:01
  • $\begingroup$ Got it yet?.... $\endgroup$ – Alec Teal Nov 1 '14 at 16:07

First of all, the Jacobian of the spherical coordinat transformation is $\rho^2\sin\phi$ not $\rho\sin\phi$.

Also, the upper bound for $\rho$ should have been $6\cos\phi$ not $3$. The equation $\rho = 3$ describes a sphere of radius $3$ centered at the origin while the equation $\rho = 6\cos\phi$ describes a sphere of radius $3$ centered at $(x,y,z) = (0,0,3)$.

After fixing those two errors, you should get an integral which evaluates to the correct answer.

  • $\begingroup$ Thanks for the explanation. Just to check (it still says I'm getting an error), the evaluated integral is $-72\pi(\sqrt{2} - 2)$, right? $\endgroup$ – Zach Saucier Nov 1 '14 at 21:13

Just spotted it:

It should be $\rho^2\sin(\phi)$ and the integral along $\rho$ should be from $0$ to $6\cos\phi$, not $0$ to $3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.