Question

Vector calculs question , please help

Answer 1

@saifoo.khan HELP

Answer 2

@phi

Answer 3

@jamesJ

Answer 4

I actually reworked it using
http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

and got this
\[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]

Answer 5

Oops, typo. Sorry. The radial component is

\[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]

Answer 6

what about phi component?

Answer 7

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
\[ \hat{n} = \hat{r} \]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png

Answer 12

oh yeah, sorry. Yet another typo.

Answer 13

no problem, your help is greatly appreciated

Answer 14

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

Answer 15

haha, it better be

Answer 16

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

Answer 17

ok, I will try working it out

Answer 18

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

Answer 19

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

Answer 20

where does this come from?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

yep

Answer 26

well,

A_phi= sin (theta)^2

so integrating it would result in

sin(theta)^2*(2*pi)

Answer 27

since it is in x-y plane ,evaluate at pi/2

sin(pi/2)^2 *(2*Pi)=2pi

Answer 28

which is what I got for surface integral above

Answer 29

Yep, good job.

Answer 30

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

Answer 31

Thank you!