1. anonymous

2. goformit100

@saifoo.khan HELP

3. anonymous

@phi

4. anonymous

@jamesJ

5. anonymous

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. anonymous

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. anonymous

8. anonymous

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. anonymous

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

10. anonymous

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. anonymous

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

12. anonymous
13. anonymous

oh yeah, sorry. Yet another typo.

14. anonymous

no problem, your help is greatly appreciated

15. anonymous

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

16. anonymous

haha, it better be

17. anonymous

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

18. anonymous

ok, I will try working it out

19. anonymous

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. anonymous

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

21. anonymous

where does this come from?

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

yep

27. anonymous

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

28. anonymous

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

29. anonymous

which is what I got for surface integral above

30. anonymous

Yep, good job.

31. anonymous

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

32. anonymous

Thank you!