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Surface Area Integral

  1. Apr 27, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Find the area of the surface z=7+2x+2y2 that lies above the triangle with vertices (0,0) (0,1) and (2,1)

    3. The attempt at a solution

    It's not a particularly difficult problem to set up, I just can't seem to get a simple integrand.

    ∫∫√(16y2+5) dA is what I come up with, and I'm not quite sure how to find the anti-derivative of that. dA=dy dx, 0<y<(x/2) & 0<x<2.

    I can't do a u-sub because I don't have anything to cancel out dy with. I've looked in some integral tables also, and the closest thing I could think of would give me an arcsin function and I'm pretty sure that's not how I'm supposed to do it.

    Thanks in advance
     
  2. jcsd
  3. Apr 27, 2012 #2

    jambaugh

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    First integrate with respect to x, you'll get a y factor from one of your limits. You can then do the integration with a simple substitution d(a y^2 + b) = 2y dy.

    You'll have another term however with no "extra y". To integrate [itex]\sqrt{a y^2 + b}dy[/itex] you'll need a trig substitution.
    First divide out the coefficient:
    [itex]\sqrt{a(y^2+b/a)}[/itex]
    and then its a matter of integrating [itex] \int \sqrt{y^2 + c^2}dy[/itex].
    Try the trig substitution y = c * tan(theta).
    (Or if you like hyperbolic trig, y = c * sinh(z).)
     
  4. Apr 27, 2012 #3

    ElijahRockers

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    Thanks!

    I integrated with respect to x first, and as it turned out, the y factor from my limit actually enabled me to do a simple u-substitution. The algebra was a little hairy, but the answer was correct!
     
    Last edited: Apr 27, 2012
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