# Volume enclosed by a cone in spherical coordinates

• Mar 12th 2009, 07:30 PM
s7b
Volume enclosed by a cone in spherical coordinates
Hi,

I really don't understand how to set up this integral;

Find the volume of the solid enclosed by the cone z=sqrt(x^2 + y^2) between the planes z=1 and z=2

The only thing I did so far was convert the first z equation into phi=pi/4
I'm not sure where to go from here.
• Mar 12th 2009, 09:15 PM
ThePerfectHacker
Quote:

Originally Posted by s7b
Hi,

I really don't understand how to set up this integral;

Find the volume of the solid enclosed by the cone z=sqrt(x^2 + y^2) between the planes z=1 and z=2

The only thing I did so far was convert the first z equation into phi=pi/4
I'm not sure where to go from here.

The $\displaystyle \phi$ angle opens up from $\displaystyle 0$ to $\displaystyle \tfrac{\pi}{4}$.
The $\displaystyle \theta$ angles travels a complete revolution.
What is trickier is finding the bounds on $\displaystyle \rho$.

If $\displaystyle z=1$ then the plane cut is $\displaystyle 1$ unit above the $\displaystyle xy$ plane, but the side of the cone makes an angle of $\displaystyle \tfrac{\pi}{4}$, therefore the length of the side of the cone is $\displaystyle \cos \tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}$. For $\displaystyle z=2$ by similar argument we get $\displaystyle \rho = 2\cdot \tfrac{\sqrt{2}}{2} = \sqrt{2}$. Therefore, the integral becomes,
$\displaystyle \int_{\sqrt{2}/2}^{\sqrt{2}} \int_0^{2\pi} \int_0^{\pi/4}\rho^2 \sin \phi ~ d\phi ~ d\theta ~ d\rho$
• Apr 11th 2009, 06:07 AM
fisix
Quote:

Originally Posted by ThePerfectHacker
The $\displaystyle \phi$ angle opens up from $\displaystyle 0$ to $\displaystyle \tfrac{\pi}{4}$.
The $\displaystyle \theta$ angles travels a complete revolution.
What is trickier is finding the bounds on $\displaystyle \rho$.

If $\displaystyle z=1$ then the plane cut is $\displaystyle 1$ unit above the $\displaystyle xy$ plane, but the side of the cone makes an angle of $\displaystyle \tfrac{\pi}{4}$, therefore the length of the side of the cone is $\displaystyle \cos \tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}$. For $\displaystyle z=2$ by similar argument we get $\displaystyle \rho = 2\cdot \tfrac{\sqrt{2}}{2} = \sqrt{2}$. Therefore, the integral becomes,
$\displaystyle \int_{\sqrt{2}/2}^{\sqrt{2}} \int_0^{2\pi} \int_0^{\pi/4}\rho^2 \sin \phi ~ d\phi ~ d\theta ~ d\rho$

Surely the limits for r would be from $\displaystyle r=\sqrt{2}$ to $\displaystyle r=2\sqrt{2}$. No? Using the lengths as $\displaystyle \tfrac{1}{\cos \tfrac{\pi}{4}}$ and $\displaystyle \tfrac{2}{\cos \tfrac{\pi}{4}}$.