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$\newcommand{\Div}{\operatorname{Div}}$I'm going to use an example to explain what I'm trying to ask.

Let $T =\{(x,y,z): x^2+y^2=z^2, 0\leq z\leq3\}$, I'm asked to calculate $\iint_T x\,dy\,dz+y^2\,dz\,dx-2z\,dx\,dy = \iint_TF$ with the outer pointing normal vector, and I need to use the Divergence Theorem. So, I need to close the surface $T$ with another surface:

$S=\{x^2+y^2=9, z=3\}$, so I can use the Divergence Theorem. I receive:

$$ \iint_T F+\iint_SF = \iiint_V \Div(F) $$

Where $V=\{x^2+y^2\leq z^2, 0\leq z \leq 3\}$ So I need to find: $$ \iiint_V \Div(F)-\iint_SF $$ using the outer pointing normal vector.

Now, calculating $\iint_SF$ with an outer pointing normal vector is straightforward: the cross product of the derivatives of the parametrization of $S$ is the normal vector, and you change the sign of the vector based on whether you want the vector to be outer pointing or inner pointing.

However, the integral $\iiint_V \Div(F)$ is calculated without finding a normal vector for the volume. It is calculated by plugging in the parametrization in $F$ and multiplying by the Jacobian.

Based on whether we are calculating $\iint_TF$ with inner or outer pointing normal vector, the sign of $\iiint_V \Div(F)$ should change. I'm asking how do I decide the sign of the volume integral.

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  • $\begingroup$ Can you explain a little better what $\iint_T F$ means? $\endgroup$ – Ian Aug 27 '14 at 13:23
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    $\begingroup$ I'm a bit confused. Are you integrating over $T$ as defined initially, or are you treating that as the definition of $V$ and taking $T$ to be its surface? Note as well that the statement of the divergence theorem for the vector field $\mathbf{F}$ over the volume $R$ is properly $$\int\!\!\int\!\!\int_R \text{div}(\mathbf{F})\,dV=\int\!\!\int_{\partial R} \mathbf{F}\cdot d\mathbf{S}$$ $\endgroup$ – Semiclassical Aug 27 '14 at 13:23
  • $\begingroup$ T is initially defined. To use the integral over V I need to close the surface, so I add the surface S, and the volume inside T and S is V. Also, edited the OP, hopefully I'm more clear. $\endgroup$ – sagooz Aug 27 '14 at 13:45
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    $\begingroup$ The Divergence Theorem requires that the normal vector to the boundary be outward-pointing. So if instead you choose inward-pointing normals for both $S$ and $T$, then indeed you would end up reversing the sign of the volume integral. $\endgroup$ – rogerl Aug 27 '14 at 14:07
  • $\begingroup$ @guyuzB : In the expression $T =\{(x,y,z): x^2+y^2=z^2, 0\leq z\leq3\}$, you put the {curly braces} outside of MathJax. I edited so now they're inside. That assures proper spacing and alignment and matching fonts, and is standard usage. And of course they need backslashes: T =\{(x,y,z): x^2+y^2=z^2, 0\leq z\leq3\} (I also created a new command making Div and operator name.) ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 27 '14 at 14:30

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