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This problem is related to question i asked before. Link

Problem

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 ?

Tuki
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  • Is this not a duplicate? https://math.stackexchange.com/questions/2539853/area-of-north-cap-of-a-sphere – Guy Fsone Nov 28 '17 at 09:41
  • 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? – E-A Nov 28 '17 at 09:42
  • There is solution provided but it doesn't explain it – Tuki Nov 28 '17 at 09:42
  • @E-A I believe i have correct f'(x) inserted into formula. https://www.wolframalpha.com/input/?i=d%2Fdx+(sqrt(R%5E2-x%5E2)) wolfram alpha approves this – Tuki Nov 28 '17 at 09:46
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    @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 $$A=2\pi \int |\sqrt{R^2-x^2}| \sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}dx$$ –  Nov 28 '17 at 09:47
  • @ProfessorVector I realizzed ti that is why I did not flag it yet – Guy Fsone Nov 28 '17 at 09:48
  • yeah they forgot a square root; it is not a huge issue but I wanted to slowly guide them to it; should this question stay or should it be closed once OP sees this? it is not exactly an informative question or should be on this website I think (i dont know the rules well myself since i am very much a newbie to this website) – E-A Nov 28 '17 at 09:49
  • @E-A yes there is typo in first indefinite integral. There is suppose to be $\sqrt{R^2-x^2}$ in the denominator but this is not the cause of this problem. It's just simply typo. Is this what you meant ? – Tuki Nov 28 '17 at 09:52
  • @E-A i think i got the typos fixed now. Let me know if there are more of them. – Tuki Nov 28 '17 at 09:56
  • Yeah, that is what I meant; try going through the calculus without it! You should see that that term cancels rather nicely – E-A Nov 28 '17 at 09:56
  • try to add that 1 to that fraction – E-A Nov 28 '17 at 09:57
  • @E-A I can see that it becomes $$A= 2\pi \int |\sqrt{R^2-x^2}|\cdot \sqrt{\frac{x^2}{R^2-x^2}+1} $$ if this is what you mean by cancelling ? – Tuki Nov 28 '17 at 09:57
  • It's still wrong in the last equation. When correct, simplify the expression in the second square root. –  Nov 28 '17 at 09:58
  • Yeah it looks right right now; now add that 1 (also you can remove the absolute value signs since square roots are always nonnegative) – E-A Nov 28 '17 at 10:01
  • @E-A Well one way to put it would be $$ R\sqrt{\frac{1}{R^2-x^2}} $$ it becomes $$ A=2\pi \int \sqrt{R^2-x^2}\cdot R\sqrt{\frac{1}{R^2-x^2}} $$ right ? – Tuki Nov 28 '17 at 10:06
  • Yup! Now, notice that those two terms cancel. – E-A Nov 28 '17 at 10:08
  • @E-A $$\frac{R}{\sqrt{R^2-x^2}}$$ ? – Tuki Nov 28 '17 at 10:14
  • I meant the terms in the square root signs – E-A Nov 28 '17 at 10:16
  • @E-A I don't think i see it – Tuki Nov 28 '17 at 10:20
  • Let us continue this discussion in chat. – E-A Nov 28 '17 at 10:20

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