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Surface Integrals of Vector Fields question

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data
    F=<0,3,x^2> computer the surface integral over the hemisphere x^2 + y^2 + z^2 = 9
    z greater than or equal to 0, outward pointing normal.

    2. Relevant equations

    3. The attempt at a solution

    I don't know why I keep getting this problem wrong. The general formula for surface integrals of vector fields is [tex]\int[/tex][tex]\int[/tex]F([tex]\Phi[/tex]([tex]\theta[/tex],[tex]\phi[/tex])*(dot product) n([tex]\theta[/tex],[tex]\phi[/tex]) For a sphere the normal is defined in the chapter of my text as R^2 sin([tex]\phi[/tex])<cos([tex]\theta[/tex])sin([tex]\phi[/tex]),sin([tex]\theta[/tex])sin([tex]\phi[/tex]),cos([tex]\phi[/tex])> However in the solutions manual when they are setting up the Integral they just have the sin([tex]\phi[/tex])<cos([tex]\theta[/tex])sin([tex]\phi[/tex]),sin([tex]\theta[/tex])sin([tex]\phi[/tex]),cos([tex]\phi[/tex])> but no R^2 ? I was looking over a sample problem on Pauls online notes and he did include the R^2 when setting up the integral so I am a little confused They end up getting 9pi/4. Can anyone help this isn't a homework problem I am just studying for my final which is in a few days thanks a lot.
  2. jcsd
  3. Dec 19, 2008 #2
    I will try to help, but I am by no means a pro at surface integrals. I a having a little difficulty understanding where you are stuck, so let's go through it quickly.

    I assume you have
    1.) parametrized your surface S as [itex]\vec{r}(\phi,\thete)=<x(\phi,\theta),y(\phi,\theta),z(\phi,\theta)>[/itex]

    2.) Computed the cross product [tex]\frac{\partial{r}}{\partial{\phi}}\times\frac{\partial{r}}{\partial{\theta}}[/tex]


    3.) Written the vector Field restricted to the parametrization of S=[itex]\vec{F}(\vec{r}(\phi,\theta))[/itex]

  4. Dec 19, 2008 #3
    hey thanks, my surface I have parameterized as (3cos(theta)sin(phi),3sin(theta)sin(phi),3cos(phi)) 0<theta<2pi 0<phi<pi/2

    the cross product came out to 9sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> I didn't actually do this cross product but in the chapter of my text the normal vector for a spherical surface is defined as R^2*sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)>

    and I am given that the R=3. And the vector field restricted to the parameterization is 27sin(theta)(sin(phi))^2 + 81(cos(theta))^2 * (sin(phi))^3 * cos(phi) However my solution manual has this as 3sin(theta)(sin(phi))^2 + 9(cos(theta))^2 * (sin(phi))^3 * cos(phi) this is because what they are using for their normal is just sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> I don't see why they wouldn't have the R^2 in it it really doesn't make sense to me so that is where I am stuck! Any help would be greatly appreciated.
  5. Dec 19, 2008 #4
    Your parameterization is correct.

    Your cross-product is also correct.

    You are saying though that value that YOU GET for [itex]\vec{F}(\vec{r}(\phi,\theta))[/itex] is different from that of the text because of what they use for the cross-product?

    Either you have made a typo, or you should see why that statement does not make sense. The value that they use for n has nothing to do with F(r) does it?
  6. Dec 19, 2008 #5
    Sorry if I caused confusion I typed that mess up fast cause I didn't want you to think I wasn't going to respond, I am saying mine and the text's answer vary because of what we got for the cross products. I have the same exact thing as the text for the Parametrization and the Vector field restricted to it, where my answer and the text's stray are when it does the dot product because their cross product answer has no 9 in front of it but mine does.
  7. Dec 19, 2008 #6
    Well in that case I would need to see what the book got for a cross-product, but I have a feeling that they either made a typo or what they have written is equivalent and you may just not be seeing how they wrote it. Solutions manuals are very prone to errors and they like to write things in ways that are tricky.

    But you and I both got the same thing for the cross-product though.
  8. Dec 19, 2008 #7
    ah ok well that makes me feel better than because my prof won't care about simplification, I had the same feeling as you about me not seeing it so I just went ahead and calculated the integral out but I got a different answer. And I got the other problems in the set correct, I just wanted to be sure I wasn't doing the cross product wrong because my finals coming up and this is my first time learning this chapter...heh..skipped a few classes towards the end ...thanks a lot !
  9. Dec 19, 2008 #8
    oh by the way the book got sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> for the cross product.
  10. Dec 19, 2008 #9
    Ah yes, of course. It is a vector, so they simply removed the factor of 9. You can do that since the information that a vector provides is not lost by doing so.
  11. Dec 19, 2008 #10
    ooo, sigh you are correct thanks a lot for the help I feel really stupid now hahah
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