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Math Help - Application of surface integral

  1. #1
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    Jan 2010
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    9

    Application of surface integral

    Let C be the solid cylinder in \mathbb{R}^3 defined by the equations

    x_1^2 + x_2^2 \leq 1, \text{ } 0 \leq x_2 \leq 1<br />

    and let \partial C be its boundary with the usual orientation; thus \partial C consists of a cylindrical surface segment C_0 together with the top disc D_1 and the bottom disc D_2, each with the appropriate orientation. In addition consider the vector field

    \mathbf{f}(\mathbf{x}) = \left(<br />
\begin{array}{c}<br />
x_1 \\<br />
x_2^2 \\<br />
x_1^2<br />
\end{array} \right)<br />

    and the differential form

    \omega(\mathbf{x}) = x_3 dx_1 \wedge dx_2 + x_1 x_3 dx_2 \wedge dx_3 + x_3 dx_1 \wedge dx_3.


    1. Calculate \int_{\partial C} \mathbf{f} \cdot \mathbf{\nu} dS, where \nu is the outward unit normal vector to the boundary \partial C and dS is the element of surface area, by using Stokes's Theorem.

    2. Calculate \int_{D_2} \mathbf{f} \cdot \mathbf{\nu} dS.

    3. Calculate \int_{C_0} \omega.

    Note especially that the normal in part (2) is not (0,0,1), but (0,0,-1).
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  2. #2
    Member Abu-Khalil's Avatar
    Joined
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    Santiago
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    Maybe -1\leq x_2\leq 1.

    Your normal vectors are \hat k,-\hat k and (\cos\theta,\sin\theta),\theta\in[0,2\pi).

    1) Using divergence:

    \int_{\partial C}f\cdot d\vec S=\int_C \nabla \cdot f dV, \nabla \cdot f = 1+2x_2.

    With cilindrical coordinates: \int_0^{2\pi}\int_0^1 r+2r^2\sin \theta drd\theta=\pi.

    2)

    \int_{D_2}f\cdot d\vec S = \int_{D_2} -x_1^2 dA=\int_0^{2\pi}\int_0^1 -r^3\cos^2\theta drd\theta=-\frac{\pi}{4}.
    Last edited by Abu-Khalil; January 7th 2010 at 12:23 PM.
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