calculus - Area of north cap of a sphere

Problem

Derive formula for area of cap of sphere where $h$ is height of the cap. Derive formula $A=2\pi Rh$.

Hint: (by rotating function $f(x)=\sqrt{R^2-x^2}$ in between $R-h\le x\le R$)

Attempt to solve

Now area of revolving function should be:

$$A=2\pi \int_{a}^{b}|f(x)|\sqrt{1+f'(x)^2}dx$$

I've tried to find where this formula comes from but i wasn't able to find proof for this. Now if someone knows how to prove this is valid formula / or if it isn't that would be great. (Possibly the cause of confusion on this problem).

We have the function given by the hint:

$$f(x)=\sqrt{R^2-x^2}$$
$$f'(x)=\frac{d}{dx}(\sqrt{R^2-x^2})=-\frac{-x}{\sqrt{R^2-x^2}}$$

Now to find out area this would be simply inserting function into the formula ?

$$A=2\pi \int_{R-h}^{R}\sqrt{R^2-x^2}\sqrt{1+(-\frac{x}{\sqrt{R^2-x^2}})^2}dx=2\pi Rh$$

I have tried to calculate the indefinite integral

$$A=2\pi \int\sqrt{R^2-x^2}\sqrt{1+(-\frac{x}{\sqrt{R^2-x^2}})^2}dx=2\pi Rh$$
$$A=2\pi(\frac{1}{4}(2x+1)\sqrt{R^2-x^2}\sqrt{\frac{-R^2+x^2+x}{x^2-R}}-\frac{1}{8}(4R^2+1)\tan^{-1}(\frac{(2x+1\sqrt{R^2-x^2}\sqrt{\frac{-R^2+x^2+x}{x^2-R^2}})}{2(-R^2+x^2+x)})) +c$$
I've tried to integrate this but with little success. However this is suppose to be same as $2\pi Rh$ (hence the equality at the end) but i don't have definite proof of it.

You can notice that $2\pi$ is only constant and does not come from the integration itself. So integration of something will most likely produce $Rh$

$$\int_{R-h}^{R}\sqrt{R^2-x^2}\sqrt{1+(-\frac{x}{\sqrt{R^2-x^2}})^2}dx=Rh$$

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To sum up the (possible) problems

(a) is formula for area correct in this. If it is where does it come from ?

(b) If we assume formula is correct then there has to be problem with my integration.

Answer

First, the formula is correct. You can find the derivation of the formula here.

Second, you made some mistakes in simplifying the expressions. In particular,

$$f'(x) = -\frac{x}{\sqrt{R^2-x^2}} \implies \sqrt{1+f'(x)^2} = \frac{R}{\sqrt{R^2-x^2}}.$$
Consequently,
$$A = 2\pi \int_{R-h}^{R}|f(x)|\sqrt{1+f'(x)^2}dx = 2\pi\int_{R-h}^{R}R\,dx=2\pi Rh.$$