Volume above cone φ = π/3 inside torus ρ = 4sin φ

[eok20]

Hi, I need to find the volume of the solid that lies above the cone with equation (in spherical coordinates) $$\phi = \frac{\Pi}{3}$$ and inside the torus with equation $$\rho = 4\sin\phi$$. I thought that the bounds are: $$0\leq\rho\leq4\sin\phi$$, $$\frac{\Pi}{3}\leq\phi\leq\frac{\Pi}{2}$$, and $$0\leq\theta\leq2\Pi$$ but when I evaluated the integral (using Mathematica) of $$\rho^2\sin\phi$$ (the Jacobian) using these bounds I got the wrong answer. Any help would be greatly appreciated.

Thanks.

 

[whozum]

$$\frac{\Pi}{3}\leq\phi\leq\frac{\Pi}{2}$$

Between Pi/3 and Pi/2 you ahve the volume outside the cone, if you want the volume inside the cone (like water ina coneshaped cup) then you'd want to go from phi = 0 to Pi/3.

 

[saltydog]

The attached plot shows the cross-section of the donut intersecting with the cone. This is what I come up with:

$$\int_0^{2\pi}\int_0^{\pi/3}\int_0^{4Sin(\phi)}\rho^2 Sin(\phi)d\rho d\phi d\theta$$

Mathematica reports approx 19.99. What did you get?

 

[eok20]

Thanks a lot for both of your help. After fixing the bounds of phi i got the correct answer of around 19.99