Surface integral of half sphere

JamesMay
Messages
5
Reaction score
0

Homework Statement



I am trying to sort out surface integrals in my head, and have become more confused when attempting to calculate the surface integral of a hemisphere. I am getting confused about which values to use as boundaries.



Homework Equations



da=R^2 sinθdθdφ

The Attempt at a Solution





To get half a sphere I integrated dφ between 0 and 2pi, and integrated sinθdθ between 0 and pi/2.

This gives R^2*0*2pi=0

Whereas if I integrate dφ between 0 and pi, and integrate sinθdθ between 0 and pi i get

R^2*1*pi

which is the answer I am looking for but shouldn't I get the same using both sets of boundary values? Confused...
 
Physics news on Phys.org
? Not if you integrate correctly! d\phi, integrated from 0 to 2\pi is, of course, 2\pi and sin\theta d\theta integrated from 0 to \pi/2 is -cos\theta evaluated between 0 and 2\pi: -(0- (1))= 1, not 0. The half sphere area is 2\pi R^2.

If, instead, you integrate d\phi from 0 to \pi and sin\theta d\theta from 0 to \pi you get \pi for the first integral and -cos(\theta) evaluated from 0 to \pi, which is -(-1- (1))= 2 for the second, again getting 2\pi R^2.
 
Oh right, I always forget that cos(0) isn't just 0. You would think I'd have learned that by now. Thanks!
 
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...
Back
Top