Integration by Parts in Several Variables

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Main Question or Discussion Point

My professor gave me the following formula for integration by parts in my multivariable calculus class. He said that we wouldn't find it in our book, and he didn't provide a proof. I have tried to work through it, but I am still left with one question: Why is it necessary that the curve is closed (the line integral)?

[tex]\int\int_{D}f(x,y)\frac{\partial g}{\partial x,y}dA=\oint_{\Sigma}f(x,y)g(x,y)\mathbf{n}\cdot d\mathbf{s}-\int\int_{D}g(x,y)\frac{\partial f}{\partial x,y}dA[/tex]

For lack of a better notation, I used [tex]\frac{\partial f}{\partial x,y}[/tex] to represent the fact that the derivative could be with respect to either x or y.

Thanks for your help.
 

Answers and Replies

Dr Transport
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Gauss's theorem states

[tex] \int\int\int_{V} \vec{\bigtriangledown} \cdot \vec{F} d \tau = \oint_{S} \vec{F}\bullet\textbf{n}dS [/tex]

substitute [tex] F = fg [/tex] anad do the algebra
 
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Ok, by Gauss's Theorem do you mean the Divergence Theorem? I haven't heard of it referred to as that before and wanted to make sure they're the same :smile:

[tex]\iint\limits_S\mathbf{F}\cdot d\mathbf{S}=\iiint\limits_V\operatorname{div}\mathbf{F}dV[/tex]

Thanks for the help.
 
Last edited:
TD
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apmcavoy said:
Ok, by Gauss's Theorem do you mean the Divergence Theorem?
That is correct :smile:
 

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