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Surface integral of a cylinder

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface integral ∫A[itex]\bullet[/itex][itex]\hat{n}[/itex]dS where A = z[itex]\hat{x}[/itex]+x[itex]\hat{y}[/itex]+3y^2z[itex]\hat{z}[/itex] and S is the cylinder x^2+y^2=16 for the range of x[itex]\geq[/itex]0,y[itex]\geq[/itex]0, 0[itex]\leq[/itex]z[itex]\leq[/itex]5

    2. Relevant equations
    I used this page as an example way to do this. I'm not good with surface integrals.

    3. The attempt at a solution
    I know I did this wrong because my answer seems wrong but... following that page linked to above

    [itex]\Phi[/itex](θ,t) = 4cos(θ)[itex]\hat{x}[/itex]+4sin(θ)[itex]\hat{y}[/itex]+t[itex]\hat{z}[/itex]

    where 0[itex]\leq[/itex]θ[itex]\leq[/itex]2∏,0[itex]\leq[/itex]t[itex]\leq[/itex]5

    [itex]\partial[/itex][itex]\Phi[/itex]/[itex]\partial[/itex]θ = -4sin(θ)[itex]\hat{x}[/itex]+4cos(θ)[itex]\hat{y}[/itex]

    [itex]\partial[/itex][itex]\Phi[/itex]/[itex]\partial[/itex]t = [itex]\hat{z}[/itex]

    normal vector is the cross product between those two derivatives which I found it equal to


    setting up the integral
    (the integral on the left is from 0 to 5 and the second integral is from 0 to 2∏)

    ∫∫A([itex]\Phi[/itex](θ,t))[itex]\bullet[/itex](that cross product I got)dθdt

    = ∫∫(t[itex]\hat{x}[/itex]+4cos(θ)[itex]\hat{y}[/itex]+3*16*(sin(θ))^2*t)[itex]\bullet[/itex](4cos(θ)[itex]\hat{x}[/itex]+4sin(θ)[itex]\hat{y}[/itex])dθdt

    = ∫∫(4t*cos(θ)+16sin(θ)cos(θ))dθdt

    =∫ (4t*sin(θ)-4cos(2θ))dt evaluated for 0 to 2π

    and this is where I get zero.... where did I go wrong?

  2. jcsd
  3. Feb 2, 2012 #2
    I guess the easiest way is to convert into a volume integral, [itex]\int A\cdot\hat{n}\,dS=\int \nabla\cdot A\,dV[/itex], then you realize the divergence of A is really simple...
  4. Feb 2, 2012 #3
    Well if that equation you have there is equal then it would be very simple. The divergence of A is 3y^2. This is a triple integral and the answer would be 1280... even if I screwed up this integral... this method seems way too easy. What is the logic behind that equation you gave me?
  5. Feb 2, 2012 #4
    I guess you haven't learned Gauss's theorem, which states exactly that identity I wrote:
  6. Feb 2, 2012 #5
    Well I can't argue with that. Just one last question. Is my new answer right? I'll show my work if you want me too.
  7. Feb 2, 2012 #6
    Since you haven't known Gauss's theorem, the problem you have in hand is obviously not intended for you to use Gauss's theorem, but to practice your skill of straightforward calculation. So stick to your original method and make sure every step is right, you can use the other method to check your answer, which I can't do for you. Also note that in Gauss's theorem the surface integral also includes the two end surfaces of the cylinder.
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