I saw that in order to compute the volume of a surface of revolution, we can use $\int_a^b\pi f^2\left(x\right)dx$, where $f$ is the curve to be rotated. This seemed really intuitive: for each "slice" (i.e: for each value of $x$) we look at the curve's value, $f(x)$, as the radius of a circle (since we're going to rotate it, so that area is exactly our eventual solid's intersection with the plane $x=$whichever value we're currently looking at), and then compute its area with $\pi f(x)^2$, and perform integration over all of those slices to end up with our volume. I'm aware this is not remotely rigorous, but I thought it was a relatively safe intuitive explanation for why it works.
I then tried to apply this same logic to compute the solid's surface area, but here it came apart. Specifically, what I attempted to claim was that if instead of integrating over the areas of the discs, I just integrated over their perimeters, I'd end up with the surface; so:
$\int_a^b 2 \pi f(x) dx$
Surprisingly for me but probably not for anyone else reading this, that didn't work! 🙁
I was hoping for an explanation as to why, exactly, and if there's some tweak I could perform to make it work. I looked online and couldn't see anything close to this; I did find another formula with derivatives under a square root, but I failed to find something to help me reason about the fault in my logic here.
In case this is relevant: what I'd attempted to do is compute the surface area of half a sphere; so I integrated with $f(x)=\sqrt{r^2-x^2}$ from $x=0$ to $x=r$.
This gives $\int_0^r \pi (1-x^2)dx = \frac{4\pi r^3}{3}$ as the volume, which is awesome!
But $\int_0^r \sqrt{r^2-x^2}dx = \frac{\pi r^2}{4}$ as the surface area, which is unfortunate.
Best Answer
I can hopefully help you to see the fault in your logic by using some "physicist's thinking" — not very rigorous, but perhaps more illustrative (and hopefully the real mathematicians here won't tear me apart for this).
Here's a super-rough sketch by hand:
This is a sketch of one of your "slices" enclosed between two very close planes in heights $z$ and $z + \mathrm{d}z$, where by $\mathrm{d}z$ I mean an infinitesimally small increment of $z$. The distance of the curve from the axis of rotation is $r$ in the lower plane and slightly different $r + \mathrm{d} r$ in the upper one.
So what happens when you calculate the volume of the slice? You just calculate the volume of a small cylinder with radius $r$ and height $\mathrm{d} z$, getting $\pi r^2\,\mathrm{d} z$. Notice that in the sketch, this is the rectangle with two black horizontal and two blue vertical lines. So by counting like that, we're essentially "throwing away" the little colorful triangles.
Why do we get the correct answer? It's because the triangles are way too small and they give zero volume when rotated. You can intuitively see that just by looking at the areas in the sketch: you count the volume that corresponds to the black/blue rectangle with area $r\,\mathrm{d} z$, and throw away the triangle with area $\tfrac12 \mathrm{d} r\,\mathrm{d}z$. So the volume of the slice is infinitesimally small, and the volume of the triangle is infinitesimally small with respect to that infinitesimally small slice. It's "infinitesimally small squared"; no wonder it doesn't contribute.
However: when you calculate the surface area with your method, you're still counting the surface area of the little cylinders (i. e. the blue bits in the sketch). Sadly, that won't do, because the "real surface" is the orange one. And you're using $\mathrm{d} z$ where you should really use the length of the sloped orange line, which is $\sqrt{\mathrm{d} z^2 + \mathrm{d} r^2}$ (just apply the Pythagorean theorem to the colorful triangle). $\mathrm{d} r$ and $\mathrm{d} z$ are pretty similar in magnitude, so you can't just throw one of them out and expect to get away with it — it's like saying that $\sqrt2$ is pretty much equal to $1$ because $\sqrt2 = \sqrt{1 + 1}$ and we can neglect one of the $1$'s!
So, taking your half-sphere $r = \sqrt{R^2 - z^2}$, we first need to calculate
$$ \mathrm{d} r = \frac{\mathrm d (R^2 - z^2)}{2\sqrt{R^2-z^2}} = \frac{- z\,\mathrm d z}{\sqrt{R^2 - z^2}}. $$
And the surface area of the slice is essentially the surface of a cylinder with radius $r$ and height $\sqrt{\mathrm d z^2 + \mathrm d r^2}$, so it needs to be
$$ \mathrm{d} S = 2\pi r \sqrt{\mathrm d z^2 + \mathrm d r^2} = 2\pi \sqrt{R^2 - z^2} \times \sqrt{\mathrm d z^2 + \frac{z^2}{R^2 - z^2} \mathrm d z^2} = 2\pi R |\mathrm d z|. $$
Integrate this from $z = 0$ to $z = R$ to get the $2\pi R^2$ that we know is correct.