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