I need to compute the following volume in spherical coordinates:
where the angle between the base of the cylinder of radius $R$ and the plane is $30ยบ$.
I already computed the equation of the plane with 3 points: $x+ \sqrt3 z =0$. After that, I decided to use cylindrical coordinates and it was pretty straightfoward:
$$V= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{0}^{R} \int_{0}^{\frac{-r \cos(\theta)}{\sqrt3}} r \mathrm{dz}\mathrm{dr}\mathrm{d \theta} = \frac{2R^3}{3 \sqrt3}.$$
However, I need to get the result using spherical coordinates. When I tried to get the volume in spherical coordinates I got the following:
$$V =\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{0}^{R\csc(\phi)} r^2 \sin(\phi) \mathrm{dr}\mathrm{d \phi}\mathrm{d \theta} = \frac{\pi R^3}{3 \sqrt3}.$$
I am pretty sure that the result in the cylindrical coordinates is correct, but I don't know why my result in spherical coordinates is wrong. Any hints about the integration limits are appreciated.
Best Answer
Please note that $\phi$ and $\theta$ are not independent of each other.
$x = \rho \cos\theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos\phi$
$\phi$ and $\theta$ are bound by the plane $x + \sqrt3 z = 0 \implies \rho (\cos\theta \sin\phi + \sqrt3 \cos\phi) = 0$
$tan \phi = - \sqrt3 \sec \theta$
So the volume integral should be,
$V = \displaystyle \int_{\pi/2}^{3\pi/2} \int_{\arctan(- \sqrt3 \sec \theta)}^{\pi/2} \int_0^{R \csc \phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta$