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Thread: Surface integral.

  1. #1
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    Surface integral.

    Calculate $\displaystyle $\int {x^2 zds}$$ Being S the external surface of $\displaystyle $x^2 + y^2 = a^2$$ understood between $\displaystyle $z = -2$$ and $\displaystyle $z = 2$$
    Last edited by mr fantastic; Jul 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 $\displaystyle $\int {x^2 zds}$$ Being S the external surface of $\displaystyle $x^2 + y^2 = a^2$$ understood between $\displaystyle $z = -2$$ and $\displaystyle $z = 2$$
    $\displaystyle x^2+ y^2$ is a cylinder, with radius a, in three dimensions. It can be written in parametric equations using cylindrical coordinates but taking r= a: $\displaystyle x= a cos(\theta)$, $\displaystyle y= a sin(\theta)$, $\displaystyle z= z$.

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

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

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

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