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Evaluating a surface integral

  1. Aug 10, 2016 #1
    1. The problem statement, all variables and given/known data
    1_zpslccngogo.jpg
    It is evaluating a surface integral.
    2. Relevant equations
    s∫ f(x,y,z) dS = ∫R∫ f[x,y,g(x,y)]√(1+[gx(x,y)]2+[gy(x,y)]2) dA

    3. The attempt at a solution
    I set z=g(x) and found my partial derivatives to be gx=√x, and gy=0. I then inserted them back into the radical and came up with √(1+x). After integrating with respect to y (dydx) I had the final integral of 2/3 ∫ x5/2(1+x)1/2 dx. Instructor said to do integration by parts twice, which I've done and it still is a non integrable function. U and V keep increasing/decreasing their exponents without simplifying. I can't get wolfram alpha or symbolab to give me an answer either.
     
  2. jcsd
  3. Aug 10, 2016 #2

    haruspex

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    Have you tried a trig substitution?
     
  4. Aug 10, 2016 #3
    Yes sir, I have. But I couldn't get the √(1+x) to change to a single trig function since its 1+x.
     
  5. Aug 10, 2016 #4

    haruspex

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    What substitution for x will make 1+x the square of a trig function?
     
  6. Aug 10, 2016 #5
    2 cos2(u) + 1. I've substituted x for 2 cos2(u) + 1. It gets rid of the (x+1)1/2, but I also get that identity within the 2/3x5/2. And I can't get that expression to simplify out now.
     
  7. Aug 10, 2016 #6

    haruspex

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    Clearly, it will help if x is also the square of a trig function. I can think of two substitutions that will achieve both. One uses hyperbolic trig functions, but the other is very widely used in integrals.
     
  8. Aug 11, 2016 #7
    Thank you for this. I did come up with a reasonable answer using 1+tan2u=sec2u.
    It still was not the correct answer, but at least I know it how to integrate such a function now. I've attached my work so you all can let me know where else I am miscalculating. Thank you!
    https://lh3.googleusercontent.com/NFVj3nYZrfy8V3XCCiLjnFeN_FbRbxxStVPkryH_dVVHzTPy1Iv3wNmu5HYBFDFVLhFEPxp2q9TcEog9f54qiAWBTAUJtcIpfCO8Hfm185IBQoY_hqqc2sO1Vm05_-m5-ddXtDlH6E7sPeXX6bOZfDsOwZhDqWOv-IoJn5ciGE5sDiNta86vrP9eQbA2zDrb6al1Zf5Kdu5Ne_9cDOe4l913_HGws144IT0fQwoAUb5EWdz1Slm0ARsllpj-Abtqmr3oVxxLBohohZEv3F-KcaApOjdkRpsbLRali_VEfaQDdgl1D2VDInb46D7NTCitR2ZwXOyrTOCh4-oo2OhCEMRpD8QaG_3if7U7yEYFg8MaPVzJ_Z0goZZ3laO2G70Gk_2fL5vl6z3EtkuS8sZMiBLE4-L74LlRausEZYHRZTMesm25FdGcTE43YH9FFJeRjdXI1lNzhRXyFVhBM05NT8LD7U66eHFwrdCzL9rrowQrqb91b68fjAj7MiK1o-yI3Tngw0J3j22DLcl0HjL63Rgqmk2LBbI-cysEaLhmQN54mYvnq4T41o8kl3VF2EoGuWGHImVwIN3GslgX-9i0Q7rFAklYiTo=w924-h1576-no
     
    Last edited: Aug 11, 2016
  9. Aug 11, 2016 #8

    haruspex

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    You failed to take the square root of sec2.
     
  10. Aug 11, 2016 #9
    You're absolutely right. That leaves me with tan6u * sec3u. Wondering how that can be integrated...
     
  11. Aug 11, 2016 #10

    haruspex

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    With an expression like that, I look for a way of getting rid of the trig by using the standard equations sin(x).dx=-d cos(x), etc.
    If you pick the right one, you should get something that you can split up into standard integrals using partial fractions, but I warn you, it looks messy.

    Edit: just realised there are web pages in plenty where you can just plug in your rational function and it will decompose into partial fractions for you. That makes it quite easy.
     
    Last edited: Aug 15, 2016
  12. Aug 19, 2016 #11
    Thank you for your help. I ended up using a web site to decompose it to partial fractions. It was a very messy solution.

    ##u=\frac 1 {288} (61\sqrt 2 - 15 sinh^{-1}(1))##
     
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