Using infinitesimals to find the volume of a sphere/surface

• B
I've always thought of dxat the end of an integral as a "full stop" or something to tell me what variable I'm integrating with respect to.
I looked up the derivation of the formula for volume of a sphere, and here, dx is taken as an infinitesimally small change which is multiplied by the area of a disc(pi r^2) giving $$\displaystyle V = 2\pi \int_0^r x^2 dy$$ which is the sum of these infinitesimals.

Now I'm really confused. Is it correct to think of it this way? Is there any other way to prove this result without using infinitesimals? Also, if I'm integrating from 0 to r, wouldn't this give me the area of only half the sphere?

Part two to my question:
Using this same logic of using infinitesimals, I tried to find the surface area of a sphere and looked at it as the sum of infinite rings.

$$\displaystyle A = \int_0^r 2\pi x dy$$
$$=> \displaystyle A = \int_0^r 2\pi \sqrt{r^2-y^2} dy$$
But this is wrong. Why?

Orodruin
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But this is wrong. Why?
You are failing to account for the fact that the surface element is not orthogonal to the y-direction.

Also, if I'm integrating from 0 to r, wouldn't this give me the area of only half the sphere?
Yes, but you have multiplied your integral by 2, so it's okay. Or you can omit the 2 and integrate from -r to +r. Same thing

You are failing to account for the fact that the surface element is not orthogonal to the y-direction.

I'm sorry, I don't understand. How is this any different from the volume case?

nrqed