Solution using double integrals:
I'm trying to calculate this using triple integrals in spherical coordinates.
I'm using the method described here Triple Integrals in spherical coordinates .
Is this integral and especially the limits correct? Doubt mainly regarding limits for phi.Isnt phi varying from pi/3 to 5pi/3?
Best Answer
At intersection of the plane $z = 1$ and sphere $\rho = 2$,
$z = \rho \cos\phi \implies \cos\phi = \frac12$ i.e $\phi = \frac{\pi}{3}$
Now for $0 \leq \phi \leq \pi/3$, $\rho$ is bound above by the plane $z = \rho \cos\phi = 1$ and for $\phi \geq \pi/3$, $\rho$ is bound above by the sphere.
So the triple integral to find volume should be,
$ \displaystyle V_1 = \int_0^{2\pi} \int_0^{\pi/3} \int_0^{\sec\phi} \rho^2 \sin \phi ~ d\rho ~ d\phi ~ d\theta = \pi$
$ \displaystyle V_2 = \int_0^{2\pi} \int_{\pi/3}^{\pi} \int_0^2 \rho^2 \sin \phi ~ d\rho ~ d\phi ~ d\theta = 8 \pi$
$V = V_1 + V_2 = 9 \pi$
The other option is to find the volume of the smaller region and then subtract from sphere volume $\dfrac{32 \pi}{3}$.
$ \displaystyle V = \frac{32 \pi}{3} - \int_0^{2\pi} \int_0^{\pi/3} \int_{\sec\phi}^2 \rho^2 \sin \phi ~ d\rho ~ d\phi ~ d\theta = 9 \pi$