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Surface Integrals

  1. Feb 18, 2012 #1
    surface.png



    Ok, so for Q6, I first said that

    z = 3 - 3x - 1.5y

    Using (∂z/∂x)^2 = 9, (∂z/∂y)^2 = 9/4

    I then did a double integral of (x + y + (3 - 3x - 1.5y)) * sqrt(9 + 9/4 + 1) dA

    Letting y and x be bounded below by 0 as stated, and x bounded above by 1 - 0.5y and y bounded above by 2, I went through the integral and got out 28.

    Using pretty much the exact same method for Q7 I got the answer to be 2.

    This is the first time attempting these sort of questions, really not sure if I'm doing it right. They only take a few minutes to do, so was hoping someone who was more confident in what they were doing could just check this for me, and let me know if I'm actually on the right track or not. Thanks a lot.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 18, 2012 #2

    vela

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    It sounds like you set it up correctly, but your final answer doesn't match what I got, which was 7.

    This is a different type of integral, so you can't use the exact same method to evaluate it. Show us your work in more detail.
     
    Last edited: Feb 18, 2012
  4. Feb 19, 2012 #3
    Thanks a lot for the help.

    Tried Q6 again, got the same answer, 28.

    Q7 I did wrong, realised it's a different type of question and used a different method.

    Put the coordinates into spherical polar coordinates, x=sinθcos∅ y=sinθsin∅ z=cosθ

    dS=sinθdθd∅, and do the integral of F dot producted with the unit normal vector with respect to dS,

    So got an integral of cosθsinθ, limits between 0 and pi/2 and 0 and pi, so got an answer of pi/2 out. That sound better?
     
  5. Feb 19, 2012 #4

    HallsofIvy

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    Here is a standard method for both scalar and vector surface integrals:
    Write the surface in terms of two parameters, x= f(u,v), y= g(u,v), z= h(u,v).
    (This is always possible because a surface is two dimensional.)

    Hence write the "position vector" of any point on the surface
    [tex]\vec{r}(u,v)= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/tex]

    The two derivatives of that position vector
    [tex]\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}[/tex]
    [tex]\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}[/tex]
    lie in the tangent plane to the surface at each point so their cross product
    [itex]\vec{r}_u\times\vec{r}_v[/itex] is perpendicular to the surface and its length gives the "differential of surface area"

    That is [itex]d\vec{S}= \vec{r}_u\times\vec{r}_v dudv[/itex] and

    [itex]dS= \left|\vec{r}_u\times\vec{r}_v\right|dudv[/itex]
     
  6. Feb 19, 2012 #5
    Thanks, that's one of the methods I tried though, and I still got 28 for the first one.
     
  7. Feb 19, 2012 #6

    vela

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    It doesn't really help to simply tell us the answer you got. All I can say is to try again.

    It's close, but not correct. What's the unit normal vector for ##d\vec{S}##?
     
  8. Feb 19, 2012 #7

    HallsofIvy

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    the Well, I tried (6) myself and got 3.

    As for (7), as I said above, I would not calculate dS or the unit normal. To find dS, you calculate the "fundamental vector product", and multiply dudv by the length of that vector. To find the unit normal, you do the same thing and divide by its length. If you take that seriously, you divide and multiply by the same thing!

    Instead, take
    [tex]\vec{r}(\theta, \phi)= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}[/tex]
    find its "fundamental vector product" and multiply that by [itex]d\theta d\phi[/itex].

    But it isn't really necessary to integrate at all! Note that for every point in the upper hemisphere, there exist a corresponding point on the lower hemisphere with the opposite sign on "z".
     
  9. Feb 19, 2012 #8

    vela

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    Hmm, I got 7.

    I just noticed the surface is only for y≥0 and z≥0, so the surface S is only a quarter of the sphere.
     
  10. Feb 20, 2012 #9
    Just realized I made a really dumb mistake, and multiplied everything by 4, so for Q6 I did get the answer to be 7.



    Not quite sure what you mean vela by that.
     
  11. Feb 20, 2012 #10
    Had another go, think I realized my mistake, got 2/3 this time for Q7. Pretty sure that's correct, thanks for the help.
     
  12. Feb 20, 2012 #11

    vela

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    That's not correct.
     
  13. Feb 20, 2012 #12
    working.jpg

    Hope you can read that ok.

    Only thing I can think of that I might not be doing is if I have to square z, as it's squared on the surface..?

    Only recently started vector calculus, and finding it pretty hard to get comfortable with. Really appreciating the advice, thanks a lot.
     
  14. Feb 20, 2012 #13

    vela

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    I'm so sorry. I misread the vector field and thought it was F = zk instead of F = zj. Your answer is correct.
     
  15. Feb 20, 2012 #14
    Oh right, was starting to wonder if you'd done that, I did it too earlier. No worries, very easy mistake to make. Cheers for the help anyway - it's just great having it confirmed, so easy to get an answer, so hard to know if it's correct.
     
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