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Math Help - Surface integral.

  1. #1
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    Surface integral.

    Calculate $\int {x^2 zds}$ Being S the external surface of $x^2  + y^2  = a^2$ understood between $z = -2$ and $z = 2$
    Last edited by mr fantastic; July 19th 2010 at 12:03 AM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by rikelme23 View Post
    Calculate $\int {x^2 zds}$ Being S the external surface of $x^2  + y^2  = a^2$ understood between $z = -2$ and $z = 2$
    x^2+ y^2 is a cylinder, with radius a, in three dimensions. It can be written in parametric equations using cylindrical coordinates but taking r= a: x= a cos(\theta), y= a sin(\theta), z= z.

    The "position vector" of any point on the cylinder can be written as \vec{p}= a cos(\theta)\vec{i}+ a sin(\theta)\vec{j}+ z\vec{k}. The two derivatives:
    \vec{p}_\theta= -a sin(\theta)\vec{i}+ a cos(\theta)\vec{j} and
    \vec{p}_z= \vec{k}
    are tangent vectors to the cylinder. Their cross product is
    -a sin(\theta)\vec{i}+ a cos(\theta)\vec{j} and the "scalar differential of surface area" is given by the length of that: a d\theta dz.

    (For this simple geometry, you could have just noted that the differential of length around the cylinder is a d\theta so the differential of area is a d\theta dz.)

    Since x^2= a^2 cos^2(\theta) this integral is
    \int_{z= -2}^2\int_{\theta= 0}^{2\pi} (a^2 cos^2(\theta)) z (a d\theta dz).

    (In fact, just thinking about the "symmetry" (or, more correctly "asymmetry") makes this problem trivial.)
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