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Math Help - Area of the Surface

  1. #1
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    Area of the Surface

    Hello -

    I am to find the part of the plane 2x+5y+z=10 that lies inside the cylinder x^2+y^2=9

    I'm not sure how to start, and this is all i need help with. Do i parametrize the cylinder?
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  2. #2
    Jen
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    Using cylindrical coordinates, we know that
    x=rcos\theta and
    y=rsin\theta

    0\le r \le 3
    and 0\le\theta\le\ 2 \pi

    Substituting all this for x and y after isolating z in the equation of the plane we get,

    z=10-2(3)cos\theta -5(3)sin\theta

    which simplifies to

    z=10-6cos\theta-15sin\theta
    Hope that helps!
    Last edited by Jen; April 17th 2008 at 06:07 PM. Reason: ooops :)
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  3. #3
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    i think its z=10-6cos\theta-15sin\theta, but i'm confused as to how to use this, since we only have an equation for dS in cartesian coordinates => \sqrt{(1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2)}
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  4. #4
    Jen
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    I should have read the title to the post

    Right, now that I have figured out what we are doing.

    \int \int 1 dS is the surface area.

    so we get

    \int \int \sqrt{1+(-2)^2+(-5)^2}dA

    since we want only the area inside the circle we convert the integral to polar coordinates

    \sqrt{30} \int \int dA=\sqrt{30} \int_{0}^{2 \pi} \int_{0}^{3}rdrd\theta

    Hope that helps!!
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