## modphysnoob Group Title Vector calculs question , please help one year ago one year ago

1. modphysnoob Group Title

2. goformit100 Group Title

@saifoo.khan HELP

3. modphysnoob Group Title

@phi

4. modphysnoob Group Title

@jamesJ

5. modphysnoob Group Title

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 Group Title

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 Group Title

what about phi component?

8. Jemurray3 Group Title

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 Group Title

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

10. Jemurray3 Group Title

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 Group Title

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

12. modphysnoob Group Title
13. Jemurray3 Group Title

oh yeah, sorry. Yet another typo.

14. modphysnoob Group Title

no problem, your help is greatly appreciated

15. Jemurray3 Group Title

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 Group Title

haha, it better be

17. Jemurray3 Group Title

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

18. modphysnoob Group Title

ok, I will try working it out

19. modphysnoob Group Title

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 Group Title

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

21. modphysnoob Group Title

where does this come from?

22. Jemurray3 Group Title

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

23. modphysnoob Group Title

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

24. Jemurray3 Group Title

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 Group Title

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

26. Jemurray3 Group Title

yep

27. modphysnoob Group Title

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

28. modphysnoob Group Title

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

29. modphysnoob Group Title

which is what I got for surface integral above

30. Jemurray3 Group Title

Yep, good job.

31. modphysnoob Group Title

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

32. Jemurray3 Group Title

Thank you!