# a double integral question

• Aug 1st 2007, 02:14 PM
kittycat
a double integral question
The sphere of radius a centered at the origin is expressed in rectangular coordinates as x^2 + y^2 +z^2 = a^2 , and hence its equation in cylindrical coordinates is r^2 + z^2 = a^2 . Use this equation and a polar integral to find the volume of the sphere.

Thank you very much.
• Aug 1st 2007, 02:21 PM
ThePerfectHacker
Quote:

Originally Posted by kittycat
The sphere of radius a centered at the origin is expressed in rectangular coordinates as x^2 + y^2 +z^2 = a^2 , and hence its equation in cylindrical coordinates is r^2 + z^2 = a^2 . Use this equation and a polar integral to find the volume of the sphere.

Thank you very much.

$\iiint_V \ dV = 2\int_0^{2\pi} \int_0^a \int_{0}^{\sqrt{a^2-r^2}} r \ dz \ dr \ d\theta$
• Aug 1st 2007, 02:31 PM
kittycat
hi perfecthacker,
Thank you very much for your reply. I haven't learnt triple integral. Could you please explain this question to me in double integral. Thanks.
• Aug 1st 2007, 02:39 PM
galactus
$\int_{0}^{2\pi}\int_{-a}^{a}r\cdot{r}drd{\theta}$

$\int_{0}^{2\pi}\int_{-a}^{a}r^{2}drd{\theta}$
• Aug 1st 2007, 02:43 PM
tukeywilliams
$V = 2 \int_{0}^{2 \pi} \int_{0}^{a} \sqrt{a^2-r^2} r \ dr \ d \theta$

$V = 2 \int_{0}^{2 \pi} \left[\frac{1}{3}(a^2-r^2)^{3/2} \right]$ from $0$ to $a$.

$V = \frac{2}{3} \int_{0}^{2 \pi} a^3 \ d \theta$

$V = \frac{4 \pi}{3} a^3$