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Surface area of a function

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the area of the surface with parametric equations x=uv, y=u+v, z=u-v.
    u^2 + v^2 <= 1

    2. Relevant equations
    A(S)= double integral over the domain D of the norm of the cross product (r_u X r_v) DA

    3. The attempt at a solution
    I started off with finding the norm of the cross product and got sqrt [4+2(v^2) + 2(u^2)]. I found it almost impossible to solve that integral. So I'm trying to use polar coordinates, but I don't know how to set the equations in terms of theta and radius.
  2. jcsd
  3. Jul 27, 2008 #2


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    What you are thinking is exactly right. Since the region of the surface in which you are interested is given by [itex]u^2+ v^2\le 1[/itex] in the uv-plane, you use [itex]u= r cos(\theta)[/itex], [itex]v= rsin(\theta)[/itex]. Then [itex]dudv= r dr d\theta[/itex] and the integral is just
    [tex]\int_{r= 0}^1\int_{\theta= 0}^{2\pi}\sqrt{4+ 2r^2}rdrd\theta[/itex]
  4. Jul 27, 2008 #3
    When I compute the cross product, does r_theta X r_radius = r_u X r_v? In which case I could just compute r_u X r_v since it's easier right?
  5. Jul 27, 2008 #4


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    Yes! Notice that I specifically said "the uv-plane". The reason you want to change to "polar coordinates" is because you have "[itex]u^2+ v^2\le 1[/itex]" and [itex]u^2+ v^2[/itex] in your integral. This have nothing to do with the xy-plane.
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