Let $\rho$, $\theta$, $\phi$, be spherical coordinates in $R^3$ and suppose that a surface surrounding the origin is described by a continuous positive function $\rho = f(\theta, \phi)$.
Complete the following passage regarding the volume enclosed by the surface.
The volume $V$ of the bound region is a triple integral over the volume element. The bounds of integration for $\theta$ are from $0$ to $2\pi$, for $\phi$ are from $0$ to $\pi$, and for $\rho$ are from $0$ to $f(\theta, \phi)$. In spherical coordinates, the volume element is $\rho^2 \sin(\phi) d\rho d\phi d\theta$.
Thus, the volume integral simplifies from a triple integral to a double integral in the following manner.
$V = \int_0^{2\pi} \int_0^{\pi} \int_0^{f(\theta, \phi)} \rho^2 \sin(\phi) d\rho d\phi d\theta$
$= \int_0^{2\pi} \int_0^{\pi} \frac{1}{3} [f(\theta, \phi)]^3 \sin(\phi) d\phi d\theta$
$= \frac{1}{3} \int_0^{2\pi} \int_0^{\pi} [f(\theta, \phi)]^3 \sin(\phi) d\phi d\theta$
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