This problem is related to question i asked before. Link


Formula for area was proven in the other post but i still have some problem with integration (with the arithmetic itself).

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

This is formula for area of cap of a sphere if revovling function is known. $f(x)$ is the revolving function in this formula.

Our revolving function is:

$$ f(x)=\sqrt{R^2-x^2} $$

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

We wanted to know area when $$ (R-h) \le x \le R $$

Now simply putting all these together we get.

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

Now when i try to compute indefinite integral i don't get so far. I have no clue how to integrate something like this. $$ A=2\pi \int |\sqrt{R^2-x^2}| \sqrt{1+(-\frac{x}{\sqrt{R^2-x^2}})^2}dx $$ $$ A=2\pi \int |({R^2-x^2})^\frac{1}{2}| ({1+(-\frac{x}{R^2-x^2})^2})^{\frac{1}{2}}dx $$

Well i tried this to integrate this with wolframalpha and it returns following:

Indefinite integral:

No result found in terms of standard mathematical functions.

However when you put in the integration limits in wolframalpha it gives correct answer. Problem is i don't understand why ?

  • $\begingroup$ Is this not a duplicate? math.stackexchange.com/questions/2539853/… $\endgroup$ – Guy Fsone Nov 28 '17 at 9:41
  • $\begingroup$ I am not sure if you are plugging in the correct f'; f' is given in the answer to the previous question (that you already accepted); I am surprised wolfram gives the right answer with the integration limits; maybe you are typing the right thing in your expression with limits? $\endgroup$ – E-A Nov 28 '17 at 9:42
  • $\begingroup$ There is solution provided but it doesn't explain it $\endgroup$ – Tuki Nov 28 '17 at 9:42
  • $\begingroup$ @E-A I believe i have correct f'(x) inserted into formula. wolframalpha.com/input/?i=d%2Fdx+(sqrt(R%5E2-x%5E2)) wolfram alpha approves this $\endgroup$ – Tuki Nov 28 '17 at 9:46
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    $\begingroup$ @Guy Fsone It's not a duplicate, both OPs introduced different errors/typos. Their problem is with elementary algebra, not with calculus. Here, it should be $$A=2\pi \int |\sqrt{R^2-x^2}| \sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}dx$$ $\endgroup$ – Professor Vector Nov 28 '17 at 9:47

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