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A surface integral

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Compute the integral
    [itex]\iint_S \sin y dS[/itex]
    where S is part of the surface [itex]x^2 +z^2 = \cos^2(y)[/itex] lying between the planes [itex]y=0[/itex] and [itex]y=\pi/2[/itex].

    2. Relevant equations
    [itex]\iint_S f(x,y,z) dS = \iint_D f(x,y, g(x,y)) \sqrt{g_x^2 +g_y^2 +1}dA[/itex]
    [itex]\iint_S f(x,y,z) dS = \iint_D f(\vec{r}(u,v)) \|\vec{r}_v\times\vec{r}_v\| dA[/itex]


    3. The attempt at a solution
    So my attempt was to parametrize to surface as [itex] x=x , y=y, z=\sqrt{\cos^2y - x^2}[/itex], then put everything in the equation, and ended up with this integral [itex]\int_0^{\pi/2}\int_0^{cosy} siny \sqrt{\frac{x^2+sin^2ycos^2y}{cos^2y-x^2} +1} dxdy [/itex]

    This is exactly what I would get if I took the partial derivatives and used the other equation. I don't think I've made a mistake in my calculations, but that's more than likely to happen.
    I'm not sure how to proceed from here. Is there some kind of factorization I can do? Some identity to apply? Some substitution to use?
     
  2. jcsd
  3. Dec 15, 2012 #2

    haruspex

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    I've never worked in this area, so I may have this completely wrong, but I tried r = (x, sin(y), z), taking x and z as the parameters. This simplified nicely and I ended up with pi/4. Is that valid, or must I use r = (x, y, z)?
     
  4. Dec 15, 2012 #3
    How did you end up with r = (x, sin(y), z)? If you take x and z to be the parameters, wouldn't y = cos-1√(x^2+z^2) ?

    When parameterizing a surface, you use two variables, right?
    With what I did, my surface ended up being parametrized as [itex]\vec{r}(x,y) = <x, y, \sqrt{cos^2y-x^2}> [/itex]
     
  5. Dec 15, 2012 #4

    haruspex

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    I was not happy with my r = (x, sin(y), z), so I tried again with the more convincing r = (x, y, z). Taking x and z as parameters, r = (x, √(1-x2-z2), z). ||rx x rz|| = 1/sin(y), so the integral reduces to ∫∫dxdz. All that's left is to be careful with the ranges in the double integral. (This is the same result as I got with r = (x, sin(y), z), so maybe it was valid.)
     
  6. Dec 15, 2012 #5
    Hm... ok. I think I see what's happening now.
    Thank a lot for the help.
     
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