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Surface Integral From Div, Grad, Curl and all that

  1. Aug 31, 2013 #1
    First, this is my first time actually posting anything so hi PF!


    Second, I have been working out of Div, Grad, Curl and all that. This problem has me stumped for some reason. My answer never comes out to be the same as the books. If you could help me figure out where I am going wrong I would greatly appreciate it.

    1. The problem statement, all variables and given/known data
    Evaluate the surface integral using the following equation.
    2. Relevant equations
    [itex]\int[/itex][itex]\int[/itex][itex]\vec{F}[/itex](x,y,z)[itex]\hat{n}[/itex]=[itex]\int[/itex][itex]\int[/itex]-F[itex]_{x}[/itex]∂S[itex]/[/itex]∂x-F[itex]_{y}[/itex]∂S[itex]/[/itex]∂y+F[itex]_{z}[/itex]dS

    Where,
    S(x,y,f(x,y))=z=1-x[itex]^{2}[/itex]-y[itex]^{2}[/itex]

    above the xy plane

    and

    [itex]\vec{F}[/itex](x,y,z)=[itex]\hat{j}[/itex]y+[itex]\hat{k}[/itex]


    3. The attempt at a solution
    After setting up the integral I attempt to switch to polar:

    [itex]\int[/itex][itex]\int[/itex]-F[itex]_{x}[/itex]∂S[itex]/[/itex]∂x-F[itex]_{y}[/itex]∂S[itex]/[/itex]∂y+F[itex]_{z}[/itex]dS=[itex]\int[/itex][itex]\int[/itex]2y[itex]^{2}[/itex]+1 dxdy

    =[itex]\int[/itex][itex]\int[/itex]2((rSin[θ])[itex]^{2}[/itex]+1)rdrdθ
    from r=0 to r=1 and θ=0 to θ=2[itex]\pi[/itex]
    =[itex]\int[/itex]Sin[itex]^{2}[/itex][θ][itex]/2[/itex]dθ+[itex]\pi[/itex]

    [itex]\int[/itex]Sin[itex]^{2}[/itex][θ]dθ=[itex]\int[/itex](1-Cos[2θ])[itex]/[/itex]2dθ
    =θ[itex]/[/itex]2-Sin[2θ][itex]/[/itex]4

    [itex]\int[/itex]Sin[itex]^{2}[/itex][θ][itex]/2[/itex]dθ+[itex]\pi[/itex]=[itex]\pi[/itex][itex]/[/itex]4+[itex]\pi[/itex]

    The book says the answer is [itex]\pi[/itex][itex]/[/itex]2. Looking at what I have I I could say 1=r[itex]^{2}[/itex]
    Then [itex]\int[/itex][itex]\int[/itex]2((rSin[θ])[itex]^{2}[/itex]+1)rdrdθ=[itex]\int[/itex][itex]\int[/itex]2((rSin[θ])[itex]^{2}[/itex]+r[itex]^{2}[/itex])rdrdθ and thus equal to [itex]\pi[/itex][itex]/[/itex]2.

    Where am I going wrong? Thanks in advance.
     
  2. jcsd
  3. Aug 31, 2013 #2

    vela

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    You should get ##3\pi/2##. The first term works out to ##\pi/2##, not ##\pi/4##.

    To get an answer of ##\pi/2##, you need to include the contribution from the bottom surface. What exactly did the problem ask you to calculate?
     
  4. Sep 1, 2013 #3
    Actually, after looking, it appears I may have looked at the wrong answer. 3(pi)/2 is the answer. Sorry for my carelessness.
     
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