# Volume of sphere

• Nov 15th 2009, 03:45 PM
Volume of sphere
Find the volume of the sphere:

f(x,y,z)=e^(x^2+y^2+z^2) over x^2+y^2+z^2≤9 using spherical coordinates

After working trough I get this enormous number 228965, when I know its 113.097 using the simple volume formula for a sphere. I think I am having trouble with the integration of e^(rho^2)*rho... Any tips guys?
• Nov 15th 2009, 03:57 PM
Jhevon
Quote:

Find the volume of the sphere:

f(x,y,z)=e^(x^2+y^2+z^2) over x^2+y^2+z^2≤9 using spherical coordinates

After working trough I get this enormous number 228965, when I know its 113.097 using the simple volume formula for a sphere. I think I am having trouble with the integration of e^(rho^2)*rho... Any tips guys?

the function given is not a sphere...

as for your integral, a substitution of $\displaystyle u = \rho^2$ would work nicely.
• Nov 15th 2009, 03:58 PM
redsoxfan325
Quote:

Find the volume of the sphere:

f(x,y,z)=e^(x^2+y^2+z^2) over x^2+y^2+z^2≤9 using spherical coordinates

After working trough I get this enormous number 228965, when I know its 113.097 using the simple volume formula for a sphere. I think I am having trouble with the integration of e^(rho^2)*rho... Any tips guys?

Integrating $\displaystyle e^{\rho^2}\cdot\rho$ is just a u-sub. Let $\displaystyle u=\rho^2$ so $\displaystyle \frac{du}{2}=\rho\,d\rho$

$\displaystyle \int e^{\rho^2}\cdot\rho\,d\rho=\int\frac{e^u}{2}\,du=\ frac{e^u}{2}=\frac{e^{\rho^2}}{2}$
• Nov 15th 2009, 04:08 PM
EDIT: I'm sorry I meant the integration is e^(rho^2)*(rho^2) that is giving me trouble I forgot to that rho is getting squared. My apologies
• Nov 15th 2009, 04:14 PM
redsoxfan325
That has no closed form expression.
• Nov 15th 2009, 04:16 PM
Jhevon
Quote:

make sure your integral is set up correctly. perhaps a change of integration order will solve the problem. that is, integrating with respect to another variable other than $\displaystyle \rho$ first may give you a factor you need to get an integrable expression.