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modphysnoob
 3 years ago
Vector calculs question , please help
modphysnoob
 3 years ago
Vector calculs question , please help

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modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this \[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1what about phi component?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Notice 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 \[ \hat{n} = \hat{r} \]

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1oh, wow, didn't see that; that does make it a lot easier

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating \[ \int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta\]

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1shouldn't there be sin(theta ) from spherical form of ds?

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1oh yeah, sorry. Yet another typo.

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1no problem, your help is greatly appreciated

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1The line integral is trivially simple, don't worry :)

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1ok, I will try working it out

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1http://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

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1where does this come from?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1\[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1we don't use r vector becuase it is set to constant?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1just to make sure we are on same page , phi is angle from x axis on xy plane right

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1well, A_phi= sin (theta)^2 so integrating it would result in sin(theta)^2*(2*pi)

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1since it is in xy plane ,evaluate at pi/2 sin(pi/2)^2 *(2*Pi)=2pi

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1which is what I got for surface integral above

modphysnoob
 3 years ago
Best ResponseYou've already chosen the best response.1thanks you , you are not only very smart also very patient
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