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2 Surface integral problems

  1. Feb 17, 2006 #1
    I have two problems on surface integrals.

    1] I have a constant vector [itex]\vec v = v_0\hat k[/itex]. I have to evaluate the flux of this vector field through a curved hemispherical surface defined by [itex]x^2 + y^2 + z^2 = r^2[/itex], for z>0. The question says use Stoke's theorem.

    Stoke's theorem suggests:
    [tex]\int_s \left(\vec \nabla \times \vec v\right) \cdot d\vec a = \int_p \vec v \cdot d\vec l[/tex]

    But the curl of this vector comes out to be zero :yuck:. Am I going right? How is the surface integral evaluated?

    2] I have a vector field [itex]\vec A = y\hat i + z\hat j + x\hat k[/itex]. I have to find the value of the surface integral:
    [tex]\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a[/tex]

    The surface S here is a paraboloid defined by:
    [tex]z = 1 - x^2 - y^2[/tex]

    I evaluated the curl and it comes out to be:
    [tex]\vec \nabla \times \vec A = -1\left(\hat i + \hat j + \hat k\right)[/tex]

    I need help here on the procedure to evaluate the surface integral.
    Last edited: Feb 17, 2006
  2. jcsd
  3. Feb 17, 2006 #2


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    Not :yuck:, :biggrin: ! The integral of 0 over any volume is 0 so you have the answer right in front of you! Since you are using Stoke's theorem, you don't need to evaluate the surface integral.

    You probably have a formula for [itex]d\vec a[/itex] but here is how I like to think about it: The surface is given by z= 1- x2- y2 which is the same as x2+ y2+ z= 1. We can think of that as a "level surface" of the function f(x,y,z)= x2+y2+ z. The gradient of that: 2xi+ 2yj+ k is normal to the surface and, since it is "normalized" to the xy-plane in the sense that the k component is 1, [itex]d\vec a= (2xi+ 2yj+ k)dxdy[/itex].
    Take the dot product of that with -(i+ j+ k) to get the integrand. You don't say over what region of the paraboloid that is to be integrated. If it is over the region above z= 0, then projected into the xy-plane, you have the circle x2+ y2= 1. Integrate over that circle. It will probably be simplest to do it in polar coordinates.
    Last edited: Feb 17, 2006
  4. Feb 17, 2006 #3
    Thank you very, very much!!! o:)

    [tex]\left(\vec \nabla \times \vec A\right) \cdot d\vec a = -\left(2x + 2y + 1\right)dxdy[/tex]

    The region of integration is over z>0. Just a little doubt...will the projected region on the xy-plane be a circle even if z>0?

    [tex]\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a = -\int_0^1 \int_0^1 \left(2x + 2y + 1\right)dxdy[/tex]

    [tex]\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a = -\int_0^1 \int_0^1 \left(2xdx + 2ydx + 1dx\right)dy[/tex]

    [tex]\int_s \left(\vec \nabla \times \vec A\right) \cdot d\vec a = -2\int_0^1 (1+y)dy[/tex]

    Evaluation of this gives me: -3
    Last edited: Feb 17, 2006
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