Vector surface integral

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Homework Statement


Let er be the unit radial vector and r=sqrt(x2+y2+z2). Calculate the integral of F=e-rer over:
a. The upper-hemisphere of x2+y2+z2=9, outward pointing normal
b. The octant x,y,z>=0 of the unit sphere centered at the origin


The Attempt at a Solution


S=<rcos(theta),rsin(theta),sqrt(9-r2)>
int(int(F dot dS))=int(int(F dot erdS))
=int(int(e-r<cos(theta),sin(theta),r> dot <rcos(theta),rsin(theta),sqrt(9-r2)>dr dtheta

The bounds of r are 0 to 3 and theta: 0 to 2pi.
I know the answer from the back of the book (18pi*(e^-3)), but I'm not getting this. Once I have the integral set up correctly, I don't have a problem evaluating it. I apologize for the annoying notation.
 

Answers and Replies

  • #2
tiny-tim
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hi musicmar! :smile:

(have a theta: θ and a phi: φ and a pi: π and an integral: ∫ :wink:)
Let er be the unit radial vector and r=sqrt(x2+y2+z2). Calculate the integral of F=e-rer over:
a. The upper-hemisphere of x2+y2+z2=9, outward pointing normal
b. The octant x,y,z>=0 of the unit sphere centered at the origin

S=<rcos(theta),rsin(theta),sqrt(9-r2)>
int(int(F dot dS))=int(int(F dot erdS))
=int(int(e-r<cos(theta),sin(theta),r> dot <rcos(theta),rsin(theta),sqrt(9-r2)>dr dtheta

The bounds of r are 0 to 3 and theta: 0 to 2pi.
I know the answer from the back of the book (18pi*(e^-3)), but I'm not getting this. Once I have the integral set up correctly, I don't have a problem evaluating it. I apologize for the annoying notation.

erm :redface:why are you using coordinates? :confused:

F.dS = e-r er.er = e-r

just integrate e-r over the area! :wink:
 
  • #3
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F.dS = e-r er.er = e-r …
just integrate e-r over the area!

If you are saying to do

∫e-rdrdθ, 0<=r<=3, 0<=θ<=2π , that doesn't work. You are trying to integrate F, which is e-r times the unit radial vector. Unless I am misinterpreting your suggestion, I don't think it accounts for the radial vector.
 
  • #4
tiny-tim
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hmm …

i] e-r is a constant, so you can just multiply it by the area, can't you? :smile:

ii] anyway, the area element is rdrdθ, not drdθ :wink:
 
  • #5
LCKurtz
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hmm …

i] e-r is a constant, so you can just multiply it by the area, can't you? :smile:

ii] anyway, the area element is rdrdθ, not drdθ :wink:

No, the surface area element with cylindrical parameterization is

[tex]dS = \frac {3r}{\sqrt{9-r^2}}\ drd\theta[/tex]

which will give his answer.
 
  • #6
tiny-tim
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Hi LCKurtz! :smile:

But the surface is a hemisphere, so I was using spherical coordinates …

why would one use cylindrical?
 
  • #7
LCKurtz
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Hi LCKurtz! :smile:

But the surface is a hemisphere, so I was using spherical coordinates …

why would one use cylindrical?

Normally, one wouldn't. But that is how the OP set it up in his question so that's the dS he needs to fix his calculations.
 
  • #8
LCKurtz
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Actually, on looking more closely at what the OP did, there is a more basic mistake. To calculate the dS vector he needs to cross the partials of the parameterization of the surface, not just use the parameterization of the surface itself in his setup.

Instead of R = <rcos(theta),rsin(theta),sqrt(9-r2) in his integral he needs to cross Rr and r_theta.
 
  • #9
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Hi. Thanks for helping. I just went to my professor's office hours and now know how to do this problem. It should be done in spherical coordinates (I don't know why I tried to do it in cylindrical). I'm pretty sure I don't need cross partials. Because r=rho in this case, e-r is simply e-rho. Using spherical coordinates, everything should work out more or less nicely.
 

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