Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

HeLp: Surface Integral

  1. Jun 20, 2005 #1
    Here is the question:

    Evaluate the surface integral ∫∫s (X^4 + Y^4 + Z^4) dσ, where dσ is the surface element and S = { (X,Y,Z) : X^2 + y^2 + Z^2 = 1}

    I know you have to take the square root of 1 + (dz/dx)^2 + (dz/dy)^2 dxdy. And I got -2X/2Z and -2Y/2Z, respectively. Then, I must incorporate this into Stoke's. Can anyone tell me what to do next? The answer to the quesion -2pi/5. THANKS!
     
  2. jcsd
  3. Jun 20, 2005 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    How do you get a negative answer to an integral where the integrand is positive?
     
  4. Jun 20, 2005 #3

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Hello Defang. Welcome to PF. This is what I have:

    [tex]g(x,y,z)=x^4+y^4+z^4[/tex]

    [tex]S=\{(x,y,z):x^2+y^2+z^2=1\}[/tex]

    So, we're going to integrate g over a sphere at the origin with radius 1. Since everything is spherically symmetrical, let's just cut it in half and just do the top part:

    [tex]I=2\int\int_K (x^4+y^4+z^4)ds[/tex]

    where K is the surface above the x-y plane described by the function:

    [tex]K=z(x,y)=\sqrt{1-x^2-y^2}[/tex]

    Now, the projection of this surface onto the x-y plane is just the circle:

    [tex]R=x^2+y^2+1[/tex]

    It is over this area that we'll integrate using the standard formula:

    [tex]I=2\int\int_R g(x,y,z(x,y)) \sqrt{(z_x)^2+(z_y)^2+1} dA[/tex]
     
  5. Jun 21, 2005 #4

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Defang, I now suspect the function:

    [tex]g(x,y,z)=x^4+y^4+z^4[/tex]

    is not spherically symmetrical. Can someone confirm this? If so then my analysis above is not correct. Sorry.
     
    Last edited: Jun 21, 2005
  6. Jun 21, 2005 #5

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    You know, sometimes its good just to keep your mouth shut. Unfortunately for me that doesn't work in math: better to get in trouble if that's what it takes to get the problem straight in my head. I now believe the top hemisphere of the function is symmetrical with the bottom part in which case the analysis above WOULD work.

    Can someone please provide another opinion about this? I'd like to see the problem completed not to mention attempt to solve it numerically and compare the results.
     
  7. Jun 21, 2005 #6

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Alright, go ahead, say it:

    [tex]g(x,y,a)=g(x,y,-a)[/tex]

    I'm outta' here . . .got some french toast to make . . .
     
  8. Jun 21, 2005 #7
    Thanks saltydog. Then, I have 0 ∫ 2pi 0 ∫ 1 (r^4 + √ 1-r^2) (√1 + 4r^2) r dr dθ. Is this correct? If it is correct, any tips on how to solve this integration?
     
  9. Jun 21, 2005 #8
    "[ tex ] \int_0^{2 \pi} \int_0^1 \left( r^4 + \sqrt{1 - r^2} \right) \left( \sqrt{1 + 4r^2} \right) r \; dr \; d \theta [ /tex ]"

    [tex] \int_0^{2 \pi} \int_0^1 \left( r^4 + \sqrt{1 - r^2} \right) \left( \sqrt{1+4r^2} \right) r \; dr \; d \theta [/tex]

    edit to add link: https://www.physicsforums.com/showthread.php?t=8997&page=1
     
    Last edited by a moderator: Jun 21, 2005
  10. Jun 21, 2005 #9
    [tex]r^4 = \left( x^2+y^2+z^2 \right) ^2 \neq x^4+y^4+z^4[/tex]
     
  11. Jun 21, 2005 #10
    Here's my attempt, which is different from saltydog's:

    [tex] \left( x^2 + y^2 + z^2 \right) ^2 = x^4 + y^4 + z^4 + x^2 \left( y^2 + z^2 \right) + y^2 \left( x^2 + z^2 \right) + z^2 \left( x^2 + y^2 \right)[/tex]

    Spherical substitution:
    [tex]\begin{align*} &x=R \sin \theta \cos \phi \\
    &y = R \sin \theta \sin \phi \\
    &z = R \cos \theta \end{align}[/tex]

    Which gives us:
    [tex] \begin{align*} x^4 + y^4 + z^4 &= \left( x^2 + y^2 + z^2 \right) ^2 - x^2(y^2+z^2) - y^2(x^2+z^2) - z^2(x^2+y^2) \\
    &= R^4 - 2x^2y^2 - 2x^2z^2 - 2y^2z^2 \\
    &= R^4 \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \cos^2 \phi - 2 \sin^2 \theta \cos^2 \theta \sin^2 \phi \right) \\
    &= R^4\left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \left( \cos^2 \phi + \sin^2 \phi \right) \right) \\
    &= R^4\left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) \end{align}[/tex]

    So [tex]g(R, \theta, \phi ) = R^4 \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) [/tex]

    And the integral we want is
    [tex] \int_0^{2 \pi} \int_0^{\pi} g(1, \theta, \phi) dS = \int_0^{2 \pi} \int_0^{\pi} \left( 1 - 2 \sin^4 \theta \cos^2 \phi \sin^2 \phi - 2 \sin^2 \theta \cos^2 \theta \right) \sin \theta \; d \theta \; d \phi [/tex]
     
    Last edited by a moderator: Jun 21, 2005
  12. Jun 21, 2005 #11
    Just to be sure, you did mean

    [tex]\iint_S x^4 + y^4 + z^4 \; dS [/tex]

    and not

    [tex]\iint_S x^4 \hat{i} + y^4 \hat{j} + z^4 \hat{k} \; dS [/tex]

    right?
     
  13. Jun 21, 2005 #12

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Guys . . . I got:

    [tex] \frac{12\pi}{5}[/tex]

    I mean, if I'm wrong I want to be right Ok. What did you get?
     
  14. Jun 21, 2005 #13
    My calculator tells me its

    [tex]\frac{13 \pi^2}{16}[/tex]

    but it's old and the batteries are low, so it could be wrong.
     
  15. Jun 21, 2005 #14
    Oops! I entered it wrong! The answer is more likely to be

    [tex]\frac{17 \pi}{5} [/tex].
     
  16. Jun 21, 2005 #15

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    I don't know Rachmaninoff. Suppose I'll go over my work. How about you Defang? I just converted the integral to polar coordinates and plugged it into Mathematica and multiplied by 2.
     
  17. Jun 21, 2005 #16
    okay... what's your final integral look like? mine was derived above.

    -rachmaninoff
     
  18. Jun 21, 2005 #17

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Here it is:

    [tex]\int_0^{2\pi}\int_0^1(\frac{r^5Cos^4(\theta)}{\sqrt{1-r^2}}+\frac{r^5Sin^4(\theta)}{\sqrt{1-r^2}}+\frac{r(1-r^2)^2}{\sqrt{1-r^2}})drd\theta[/tex]

    That's just the top part. Multiply by 2 to get the answer I got (via Mathematica)
     
  19. Jun 22, 2005 #18

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Well, I hate to be a pain but I request some closure in this matter. Now:

    Rachmaninoff said:

    [tex]\frac{17\pi}{5}[/tex]

    I said:

    [tex]\frac{12\pi}{5}[/tex]

    Now, I'm pretty sure mine is correct. So sure, I'd bet a dollar. But Rachmaninoff feels otherwise and I've followed his other postings; he's very good so I'm worried. Can someone supply a third opinion?

    Thanks,
    Salty
     
  20. Jun 28, 2005 #19
    Indeed, after what looked like endless calculation, I have finally got an answer: 12pi/5 (still learning this LaTex program). 2 0 ∫ 2pi 0∫1 [ r^4 (3/4 + 1/4 cos4θ ) + (1-r^2)^2 ] 1/(√1-r^2) r dr dθ. saltydog & rachmaninoff, thanks for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: HeLp: Surface Integral
  1. Surface integral (Replies: 2)

  2. Surface integrals. (Replies: 8)

  3. Surface integral (Replies: 4)

Loading...