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Thread: Application of surface integral

  1. #1
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    Application of surface integral

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

    $\displaystyle x_1^2 + x_2^2 \leq 1, \text{ } 0 \leq x_2 \leq 1
    $

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

    $\displaystyle \mathbf{f}(\mathbf{x}) = \left(
    \begin{array}{c}
    x_1 \\
    x_2^2 \\
    x_1^2
    \end{array} \right)
    $

    and the differential form

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


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

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

    3. Calculate $\displaystyle \int_{C_0} \omega$.

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

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

    1) Using divergence:

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

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

    2)

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