I've been playing around with integrals lately to derive the surface area as well as the volume a sphere with a radius $R$.
The thought process for the volume of the sphere is super intuitive:
Focus on the top half of a sphere and then sum all the elementary volumes of the discs with area $\pi r^2(y)$ and height dy, where r(y) is the function of the radius of each circle depending on y.
$\int_0^R \pi r^2(y) dy$
Now, if we define an angle $\theta$ that traverses from $0$ to $\pi/2%$, y can be defined as $y= R \sin(\theta)$, where $R$ is the radius of the sphere and the differential $dy$ becomes $dy = R \cos(\theta)$. Also the funtion r(y) can be expressed as $r(\theta) = R \cos(\theta)$. Then the integral becomes
$V = \int_0^{\pi/2} \pi R^3 \cos^3(\theta) d\theta = \pi R^3 \int_0^{\pi/2} \cos^3(\theta) d\theta = \frac{2}{3} \pi R^3$.
That's for the hemisphere, so the duplicate calculates the volue of the sphere as expected.
Now, I'm trying to follow the same process of thought for the surface area. Again, we can visualize slicing the sphere, but this time we seek for the area of this elementary volume which is essentially a rectangle with a width of $2\pi r(y)$ and height $dy$.
The problem is that once I follow this process I come up with the following integral:
$A = \int_0^{\pi/2} 2 \pi R^2 \cos^2(\theta) d\theta = \frac{1}{2} \pi^2 R^2$.
It turns out that the correct differential for $dy$, for the calculation of the surface area of a sphere, should be derived from the definition of the arc length of a circle, i.e., $y = R \theta$, and therefore, $dy = R d\theta$.
Using this differential the above integral turns out perfectly fine for the calculation of the area of the hemisphere
$A = \int_0^{\pi/2} 2 \pi R^2 \cos(\theta) d\theta = 2 \pi R^2$.
I'm trying to understand why the differential of y in both cases should be different. Intuitively it makes perfect sense that the y axis is traversed the same way for both the area as well as the volume of the sphere. But it seems that this is not the case.
