Differentiate the volume of a solid of revolution to find its surface area

Question

I know that for a solid of revolution generated by rotating a function $f(x)$ around the $x$-axis, the volume $V$ is given by:

$$ V = \int_a^b \pi f^2(x) \,\mathrm dx. $$

Now, for a sphere, the volume is given by $V = \frac{4}{3} \pi r^3$, and when we differentiate this with respect to the radius $r$, we get:

$$ \frac{\mathrm dV}{\mathrm dr} = 4\pi r^2, $$

which is exactly the surface area of the sphere.

My question is: Can we extend this idea of differentiating the volume of a solid of revolution with respect to its radius to solids of revolution? In other words, is it always valid to differentiate the volume formula for a solid of revolution to get the surface area, or are there cases where this doesn’t work? In particular, why does or doesn’t this work?

Can I reason through the surface area of a solid of revolution by differentiating the volume as I did with the sphere?

Now, I want to differentiate the volume formula with respect to $r = f(x)$. The volume of the solid of revolution is given by:

$$ V = \int_a^b \pi f(x)^2 \,\mathrm dx. $$

Differentiating both sides with respect to $r$, we apply the Leibniz rule:

$$ \frac{\mathrm dV}{\mathrm dr} = \int_a^b \pi \frac{\mathrm d}{\mathrm dr} \big( f(x)^2 \big) \,\mathrm dx, $$

$$ \frac{\mathrm d}{\mathrm dr} f(x)^2 = 2 f(x); $$

so the differentiation yields:

$$ \frac{\mathrm dV}{\mathrm dr} = \int_a^b 2\pi f(x) \,\mathrm dx. $$

Answer 1

The formula $$\int_{\mathbb{R}} 2\pi f(x) \,\mathrm dx$$ you arrive at does not compute the surface area. In fact, the formula also doesn’t correctly calculate the surface area in the case of spheres, as you can check — something else is going on.

Indeed, in the case of spheres we have a family of functions $f_r$ depending on $r>0$, namely $f_r = \sqrt{r^2-x^2}$. For this family of functions, we indeed have $$\frac{\mathrm d}{\mathrm dr}V_r = A_r.$$ But for a general family of functions, this of course doesn’t need to be true at all — for most families of functions you come up with, it will not hold. It is more challenging to find families of functions for which the relation does hold.