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.
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.
Hi Dynas, Looks to me your integral:
$\displaystyle \int_{-a}^0 \int_{-\sqrt{a^2-y^2}}^0 \int_{a-\sqrt{a^2-x^2-y^2}}^a dz dx dy$
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 $\displaystyle f(x,y,z)=1$, that's the same as just the volume of 1/8 of a sphere of radius $\displaystyle a$ in spherical coordinates:
$\displaystyle \int_{0}^{\pi/2}\int_0^{\pi/2} \int_0^a \rho^2 \sin(\phi) d\rho d\theta d\phi=(1/8) 4/3 \pi a^3$