I have accepted that the equation of a sphere in spherical coordinates is ${\rho^2\sin\phi}$. The triple integral is just to nice. What I don't understand is what happened to ${\theta}$. How can you evaluate the integral through ${d\rho d\phi d\theta}$ if the polar (xy) plane is not in the derivative?
Author's note: After much conversation on the topic I realize the above was a gross misinterpretation. But it did start me on the path to the right answer.
You want to find the volume element $dV$ in various coordinate systems. We know that in the cartesian system it takes the form $dV = dx \, dy \, dz$. How does this change over a different system? Answer: using the Jacobian. In the case of spherical coordinates, you make the following substitutions:
$$\begin{cases} x = r \cos \theta \sin \varphi, \\ y = r \sin \theta \sin \varphi, \\ z = r \cos \varphi, \end{cases}$$
where I am assuming that $\theta$ is the angle in the $xy$ plane and $\varphi$ is the angle with the $z$ axis (also known as azimuthal angle, I believe).
The Jacobian is then the determinant of the matrix composed by the partial derivatives of $x,y,z$ with respect to $r, \theta, \varphi$. We write it as
$$J = \frac{\partial (x,y,z)}{\partial (r, \theta, \varphi)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \varphi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \varphi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \varphi} \end{vmatrix}.$$
By performing the calculations you'll find $J = - r^2 \sin \varphi.$ Since we are interesting in the unsigned volume, the volume element becomes
$$dV = dx \, dy \, dz = \left\vert \frac{\partial (x,y,z)}{\partial (r,\theta,\varphi)} \right\vert \, dr \, d \theta \, d \varphi = r^2 \sin \varphi \, dr \, d \theta \, d \varphi,$$
that is, we take the absolute value of the Jacobian. The sign in it means that this coordinate switch reversed the orientation, but we are not interested in that.