1. modphysnoob

2. goformit100

@saifoo.khan HELP

3. modphysnoob

@phi

4. modphysnoob

@jamesJ

5. modphysnoob

I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this $\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }$

6. Jemurray3

Oops, typo. Sorry. The radial component is $\frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r}$

7. modphysnoob

8. Jemurray3

Notice that it doesn't actually matter, because you'll be taking the dot product with the normal vector to the surface of the hemisphere, which is just $\hat{n} = \hat{r}$

9. modphysnoob

oh, wow, didn't see that; that does make it a lot easier

10. Jemurray3

Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating $\int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta$

11. modphysnoob

shouldn't there be sin(theta ) from spherical form of ds?

12. modphysnoob
13. Jemurray3

oh yeah, sorry. Yet another typo.

14. modphysnoob

no problem, your help is greatly appreciated

15. Jemurray3

Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.

16. modphysnoob

haha, it better be

17. Jemurray3

The line integral is trivially simple, don't worry :)

18. modphysnoob

ok, I will try working it out

19. modphysnoob

http://upload.wikimedia.org/math/0/a/c/0ac2a22b40a49a17025d8ad27f49cf6f.png we only use middle term,their phi is our theta since we actually have don't have any theta vector it is 0

20. Jemurray3

No, the line integral should be $\int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi$

21. modphysnoob

where does this come from?

22. Jemurray3

$d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi$ $\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi$

23. modphysnoob

we don't use r vector becuase it is set to constant?

24. Jemurray3

What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.

25. modphysnoob

just to make sure we are on same page , phi is angle from x axis on x-y plane right

26. Jemurray3

yep

27. modphysnoob

well, A_phi= sin (theta)^2 so integrating it would result in sin(theta)^2*(2*pi)

28. modphysnoob

since it is in x-y plane ,evaluate at pi/2 sin(pi/2)^2 *(2*Pi)=2pi

29. modphysnoob

which is what I got for surface integral above

30. Jemurray3

Yep, good job.

31. modphysnoob

thanks you , you are not only very smart also very patient

32. Jemurray3

Thank you!