# Finding Volume using spherical coordinates

• March 26th 2010, 04:39 PM
Belowzero78
Finding Volume using spherical coordinates
Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere $\rho=4 cos\phi$ and the hemisphere $\rho=6, z\geq0$. Then evaluate the integral.

I found the integral with the correct limits for the solid in spherical coordinates as the following: $V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta$

I would like to check my volume, the answer which i got is $\frac{26\pi}{3}$.
• March 26th 2010, 07:04 PM
dwsmith
As long as all your bounds are correct for the figure, your answer is off. I obtained 400*pi/3 for that integral as is.
• March 26th 2010, 07:11 PM
Belowzero78
Alright, do you mind to show a few steps?
• March 26th 2010, 07:13 PM
dwsmith
If you show me your steps, it will be quicker for me to find your error.
• March 26th 2010, 07:35 PM
11rdc11
Quote:

Originally Posted by Belowzero78
Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere $\rho=4 cos\phi$ and the hemisphere $\rho=6, z\geq0$. Then evaluate the integral.

I found the integral with the correct limits for the solid in spherical coordinates as the following: $V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta$

I would like to check my volume, the answer which i got is $\frac{26\pi}{3}$.

$2\pi\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi$

= $2\pi\int_{0}^{\pi/2} \frac{\rho^3}{3}\sin\phi \bigg|^{6}_{4\cos\phi} d\phi$

= $2\pi\int_{0}^{\pi/2} 72\sin\phi -\frac{64\cos^3\phi\sin\phi}{3}\ d\phi$

= $2\pi( -72\cos\phi +\frac{64\cos^4\phi}{12}) \bigg|^\frac{\pi}{2}_{0}$

= $2\pi(72-\frac{16}{3}) = \frac{400\pi}{3}$