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Thread: Surface Integral Setup

  1. #1
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    Surface Integral Setup

    Problem 15g of section 17.6 of Calculus with Analytic Geometry by Purcell and Varberg reads:

    Calculate $\displaystyle \iint_{\partial S}\mathbf{F\cdot n}dS$ for $\displaystyle \mathbf{F} = (x\mathbf{i} + y\mathbf{j})\ln(x^2 + y^2)$; S is the solid cylinder $\displaystyle x^2 + y^2 \le 4, 0 \le z \le 2$.

    This book uses $\displaystyle \partial S$ to refer to the boundary of the set $\displaystyle S$.

    This section of the book deals with the divergence theorem but the divergence of the function (and the function itself) is undefined on the z-axis, making a volume integral meaningless. I set up the surface integral thus:

    $\displaystyle \iint_{\partial S}(x\mathbf{i} + y\mathbf{j})\ln(x^2 + y^2)\cdot\frac{x\mathbf{i} + y\mathbf{j}}{\sqrt{x^2+y^2}}dS = \iint_{\partial S}\frac{x^2+y^2}{\sqrt{x^2+y^2}}\ln(x^2+y^2)dS = \int_0^2\int_0^{2\pi}2\ln4d\theta dz$

    The last equality represents a conversion to cylindrical coordinates, $\displaystyle r = \sqrt{x^2+y^2}$ having the constant value $\displaystyle 2$ on the surface. This results in $\displaystyle 8\pi\ln4 = 16\pi\ln2$.

    WolframAlpha says I evaluated the integral properly (https://www.wolframalpha.com/input/?...2%CF%80%7D%5D). The answer key has $\displaystyle 32\pi\ln2$. This would suggest that I set up the integral incorrectly.
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  2. #2
    Member Walagaster's Avatar
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    Re: Surface Integral Setup

    In your last integral, $dS = rd\theta dz = 2 d\theta dz$. To be complete, you should observe what happens on the top and bottom surfaces too.
    Thanks from Zexuo
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    Re: Surface Integral Setup

    Thanks for the catch! $\displaystyle \mathbf{F}\perp\mathbf{n}\longrightarrow\mathbf{F} \cdot\mathbf{n}=0$ on the top and bottom so those surfaces do not contribute to the integral.
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