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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $3$.

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