- #1
geft
- 148
- 0
For r = 3, 0 < theta < pi/2, 0 < phi < pi/3
[tex]\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{\frac{\pi}{3}}r^2sin\theta d \theta d \phi[/tex]
[tex]=R^2 [-cos\theta]_{0}^{\frac{\pi}{2}} [\phi]_{0}^{\frac{\pi}{3}}[/tex]
[tex]=(3^2)(1)(\frac{\pi}{3}) = 3\pi[/tex]
The above is my working. The answer should be the same as the volume since 1/3R^3 (in the volume equation) = R^2 when R = 3 but the answer given is 9pi.
[tex]\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{\frac{\pi}{3}}r^2sin\theta d \theta d \phi[/tex]
[tex]=R^2 [-cos\theta]_{0}^{\frac{\pi}{2}} [\phi]_{0}^{\frac{\pi}{3}}[/tex]
[tex]=(3^2)(1)(\frac{\pi}{3}) = 3\pi[/tex]
The above is my working. The answer should be the same as the volume since 1/3R^3 (in the volume equation) = R^2 when R = 3 but the answer given is 9pi.