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Surface integral

  1. Oct 8, 2011 #1
    How do i calculate the surface integral

    [itex]∫∫xdS[/itex] where [itex]z=x^{2}[/itex] is the parabolic cylinder



    over the area [itex]x^{2}+y^{2}=1[/itex]


    I do not know how to solve this task because i cant express the surface parametrization in r(x,y).

    But when i express it as g(x,z)=0 i get the double integral dependet on dxdz. But i cant use this when the circular transformation is rdrdθ=dxdy


    Help!
     
  2. jcsd
  3. Oct 8, 2011 #2
    Why not just muscle-through it in cartesian coordinates first. So you have a surface integral of a function of three variables f(x,y,z)=x over the surface z=g(x,y)=x^2 that is above the region x^2+y^2=1 in the x-y plane. How would that surface integral look in terms of dydx?
     
  4. Oct 8, 2011 #3
    I do not know how to do it.

    r=xi + yj +zk

    z=x^2

    or

    g(x,z)=z-(x^2/2)

    is the only thing popping out of my head right now....
     
  5. Oct 8, 2011 #4
    You gotta' look in the book dude:

    [tex]\mathop\iint\limits_{\begin{array}{cc}z=f(x,y)=x^2\\x^2+y^2=1\end{array}} x ds=\mathop\iint\limits_{x^2+y^2=1} x \sqrt{(f_x)^2+(f_y)^2+1}dA=\mathop\iint\limits_{x^2+y^2=1} x \sqrt{(f_x)^2+(f_y)^2+1}dydx[/tex]

    Now I bet you can then convert that last double integral into an explicit one in terms of just x and y for now with upper and lower limits as well as compute those simple partials right?
     
  6. Oct 9, 2011 #5
    df/dx = 2x

    df/dy = 0

    sqrt(4x^2 + 1)


    The primitive is ((4x^2+1)^1.5)/12

    Integration limits over y is then -1 and 1. And -sqrt(1-y^2) and sqrt(1+y^2) over x.

    If im right, then i do not know how to solve the integral involving dx. It looks like it results into zero because of symmetry. (x^2, positive for both limits)



    im doing my best, help me out if you want.
     
  7. Oct 9, 2011 #6
    You got the limits conceptually right but reversed:

    [tex]\mathop\iint\limits_{x^2+y^2=1} x\sqrt{4x^2+1}dydx=\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} x \sqrt{4x^2+1} dydx[/tex]

    and I agree with you: because of symmetry, that's zero.
     
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