When is integration good enough

  • Thread starter Thread starter stewman_phil
  • Start date Start date
  • Tags Tags
    Integration
stewman_phil
Messages
2
Reaction score
0
When is integration "good enough"

1. I was attempting to find the surface area of a hemisphere by summing up the surface areas of infinitesimal cylinders (or ribbons) of increasing radii (excluding the top and bottom areas). I wanted to solve this integration problem this way as opposed to using spherical coordinates (or any of the other numerous methods).

2. The attempt at a Solution.

My method for solving this problem is exactly the same IBY describes in the following post: https://www.physicsforums.com/showthread.php?t=427695

Dick explains why this attempted solution is incorrect. By claiming: "It gives you the area of the ribbon if it's vertical (i.e. parallel to the direction dx). If not it ignores the horizontal component. That's not a good enough approximation."

The value that my attempted solution outputs shows that the solution I tried is incorrect; however, this brought up a question for which I have no understanding. What makes an integration a "good enough approximation"? I had thought that since the ribbons' heights are infinitesimal there would be no problem, but there is. Why, and how would one predict that there would be a problem?

If dick's explanation is correct, then why does integration ever converge to the area under a curve? For example if one was to find the area of under y=x (a 45 degree line) for any discrete change in x, the rectangles whose sum results in the area under the curve is not quite accurate. Since the change in x ceases to be discrete in integration (it becomes infinitely small), that problem is solved. Why is it not so with the above problem?

Thank you very much! I appreciate any help I can get.​
 
Physics news on Phys.org


Your integral is wrong, you made a change of variables. So the limits a and 0 isn't the same anymore when you made dx = a\cos\theta d\theta

EDIT: Also, I would use from - a to a. Because when you simplfy the integrand, the limits of integration isn't even aproblem
 
Last edited:


flyingpig said:
Your integral is wrong, you made a change of variables. So the limits a and 0 isn't the same anymore when you made dx = a\cos\theta d\theta

You are correct, and I should have noted that my limits of integration are correct in what I did as opposed to IBY. I did not make that mistake. Even with correct limits of integration, the value it results in is incorrect. My question is what makes that method of integrating not a "good enough" approximation.
 


I am not 100% sure myself, but here is the intuition I think I can give you.

Instead of summing up the horizontal dx, ds sums the curvy part.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top