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Thread: Stuck on Surface integral problem

  1. #1
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    Stuck on Surface integral problem

    Stuck on Surface integral problem-capture.png

    I have 2 tries left, and I am having trouble figuring out how to set up the integral to find surface area.
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  2. #2
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    Re: Stuck on Surface integral problem

    $z = 1 - x - y$

    $z_x = -1$

    $z_y = -1$

    $D = \sqrt{1 + z_x + z_y} = \sqrt{3}$

    $I = \displaystyle \int_S f(1-x-y)\cdot D = \int_S 4(1-x-y)^2 \cdot \sqrt{3}$

    The integral over the surface is

    $I = 4\sqrt{3} \displaystyle \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}~(1-x-y)^2~dy~dx$

    In polar coordinates

    $I = \displaystyle 4\sqrt{3}\int_0^{2\pi}\int_0^1 (1-r \cos(\theta)-\sin(\theta))^2\cdot r ~dr ~d\theta$

    I'll let you decide which way you want to actually integrate it.

    The answer is $I = 6\sqrt{3}\pi$
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  3. #3
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    Re: Stuck on Surface integral problem

    Is it z=-x-y you mean? I think the 1 you have in the z function is not suppose to be there. Would that change the answer?

    Update: I got the answer with your help. I just changed up the function inside the integral . Thanks!
    Last edited by lc99; Nov 24th 2018 at 01:38 PM.
    Thanks from romsek
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