Triple Integral Problem Using Spherical Coordinates

**dynas7y** · December 3rd 2009, 01:45 PM

Given that:

Integral[-a, 0] Integral[-Sqrt(a^2-y^2), 0] Integral[a-Sqrt(a^2-x^2-y^2), a] dz dx dy

http://tinypic.com/view.php?pic=4ibucl&s=6

Please write an equivalent integral in spherical coordinates.

**dynas7y** · December 4th 2009, 05:06 AM

bump

**shawsend** · December 4th 2009, 11:25 AM

Hi Dynas, Looks to me your integral:

$\int_{-a}^0 \int_{-\sqrt{a^2-y^2}}^0 \int_{a-\sqrt{a^2-x^2-y^2}}^a dz dx dy$

is the integral from the bottom of the sphere surface in the plot below, to the plane z=a shown in blue. Now, just because the integrand $f(x,y,z)=1$ , that's the same as just the volume of 1/8 of a sphere of radius $a$ in spherical coordinates:

$\int_{0}^{\pi/2}\int_0^{\pi/2} \int_0^a \rho^2 \sin(\phi) d\rho d\theta d\phi=(1/8) 4/3 \pi a^3$

**dynas7y** · December 4th 2009, 12:37 PM

Thanks a lot. That was a big help.

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