I need to find the surface area of a shell of radius R. to do so I assumed a circular ring of radius R being rotated by an angle π. How do I go about finding the area. My initial attempt was 2πr.rdθ where θ is the angle by which the ring rotates, but I have understood that this method is wrong as thickness rdθ isn't constant for every point on the circumference. How should I find the area?
I know there exist other methods too but please tell how do I go about finding area using this method only.
thanks
Consider a quarter circle $y=\sqrt{1-x^2}$ on $0<x<1$. Let's rotate this around the $x$-axis to get half of a unit spherical shell. We expect the surface area to be $2\pi$ (since we're taking $R=1$, and only half of the whole sphere).
If we slice the shell into thin slices, one may be tempted to simply integrate $2\pi y\,dx$ from $0$ to $1$, reasoning that we're adding up the surface areas of a bunch of rings the shell is made from. Unfortunately, this doesn't quite work. This slicing may work for finding the volume, but apparently this 'stair-stepping' of the surface (see (b) in the picture) messes up when calculating the surface area. So let's try refining this a bit.
Instead of approximating the curve of the sphere by making straight-across slices, let's use diagonal line segments (see (c) in the picture). This should make it more accurate.
(I bet you're so jealous of my drawing skills :P)
Now, $$ds^2=dx^2+dy^2\implies ds=\sqrt{dx^2+dy^2}=dx\sqrt{1+(\frac{dy}{dx})^2}$$
Since $y=\sqrt{1-x^2}$, we end up with, uh... $$ds=\frac{dx}{\sqrt{1-x^2}}$$
Then, the surface area of (half of) the shell is $$\int_0^1 2\pi\,y\,ds=\int_0^1 2\pi\,\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}\,dx=2\pi$$
Well, that was kind of underwhelming. But I believe that is how it goes.