Suppose we have a smooth, bounded, closed surface in $\mathbb{R}^3$ which can be parametrized by giving the distance from the origin as a function $r(\varphi,\theta)$ of spherical angles $\varphi,\theta$. (So for a sphere of radius $R$, we would $r(\varphi,\theta) \equiv R$.)
Is there a simple way to express the area and volume of this surface in terms of integrals over the function $r$ and its derivatives?
Use $\varphi$ and $\theta$ as parameters, write $x, y, z$ as functions of $\varphi$ and $\theta$, using spherical coordinate formula, calculate the partial derivatives $T_\varphi = \langle x_\varphi, y_\varphi, z_\varphi \rangle$ and $T_\theta = \langle x_\theta, y_\theta, z_\theta \rangle$ (which will be expressions in the partial derivatives of $r$), then the surface area is given by the standard integral $$ S = \int_{0}^{2\pi}\!\!\! \int_{0}^{\pi} \| T_\varphi \times T_\theta \| \, d\varphi \, d\theta $$ I am not sure how much this will simplify in the end...
For the volume it is a little easier, you can argue that the integral over a small coordinate patch is approximately a pyramid of height $r$ and base of area $r^2 \sin \varphi \, \Delta\varphi \, \Delta\theta$, so the volume is $\approx (r^3/3) \sin \varphi \, \Delta \varphi \, \Delta \theta$, and the total volume thus is $$ V = \frac13\int_{0}^{2\pi}\!\!\! \int_{0}^{\pi} r^3 \sin \varphi \, d\varphi \, d\theta $$