How Does the Divergence Theorem Apply to Vector Fields and Surface Integrals?

[danni7070]

Homework Statement

Let D be an area in R^3 and S be its surface. D fulfills the Divergence theorem. Let N be the unit normal on S and let the volume, V, be known. Let (\overline{x},\overline{y}, \overline{z}) coordinates of the centre of mass of D be known (and the density delta is constant).

Lets define three vector fields:

F = (x)i + (y)j + (x)k
G = (xz)i + (2yz)j + (2z^2)k
H = (y^2)i + (2xy)j + (xz)k

Find a formula for:
\int\int_S F\bullet N dS, \int\int_S F\bullet N dS, \int\int_S F\bullet N dS

where you use V, (\overline{x},\overline{y}, \overline{z}) !

Homework Equations

Divergence theroem states:

\int\int_R div F dA = \oint_C F\bullet\widehat{N}ds

The Attempt at a Solution

divF is easily found but that doesn't tell me anything at the moment so there is nothin special going on really because I'm completely clueless what to do with the volume and the centre of mass. Any hints would be great.

Thanks in advanced and forgive my english.

 

[benorin]

The Divergence Theorem

The version of divergence theorem you stated is known as Green's Theorem and applies to a region in the plane (2-d). You need the 3-d version of http://mathworld.wolfram.com/DivergenceTheorem.html" which states that:

\iint_S \vec{F}\cdot \vec{N}\, dS = \iiint_D \mbox{div } \vec{F}\, dV​

Also, remember that the coordinates of the centroid of D are given by

\left( \bar{x},\bar{y},\bar{z}\right) = \left( \frac{1}{V}\iiint_D x\, dV,\frac{1}{V}\iiint_D y\, dV,\frac{1}{V}\iiint_D z\, dV \right)​

 

[HallsofIvy]

I hope you determined that the divergence of F is the constant, 2. Of course, the integral of 2 over a region is just 2 times the volume, which you know!

The divergence of G is z+ 2z+ 4z= 7z. The "z" in the integrand should remind you of the calculation of the z-coordinate of the centroid (i.e. "center of mass").

 

[danni7070]

Thanks, this is solved. The teacher did not set this problem correctly up and it made me confused. In fact this was easy :P

But here is another problem!

Let S be 0 <= z = 1- \sqrt{x^2+y^2} and

F : R^3\rightarrow R^3, (x,y,z) \rightarrow (x,z^2e^z+y,z) be vector field.

a) Calculate \int\int_S F\cdot NdS where N has positive z-coordinate.
b) calculate \int\int_S curl F\cdot NdS where N has negative z-coordinate.

Attempts:

a) I use Gauss theorem and find \int\int\int_D divDdV = \int\int\int_D 3 dV but now I hit the brick wall because I'm not sure what to do with z ... What is it? A cone or Cylinder I can't see it.

 

[benorin]

A half-cone with it's vertex at (0,0,1) opening downward. http://mathworld.wolfram.com/QuadraticSurface.html" .

 

[danni7070]

Ahh, very nice! thanks. Now I get it.

\int\int\int_D 3dV = 3\cdot volume of cone = 3\cdot \frac{1}{3} \cdot \pi \cdot 1^2 \cdot 1 = \pi

 

[danni7070]

the b) is still unsure because what I get from the curl part is -ze^z(2+z) and that confuses me. There is an example in the book but as almost always, they are pretty "easy" :P

 

[benorin]

It doesn't matter: \mbox{div }\left(\mbox{curl }\vec{F}\right)=0 for any \vec{F} for which the operation is defined.

 

[danni7070]

Ok. The teacher told me to use Stoke's Theorem which I did but the answer was 0 and there is an example in the book with simular results which says that the theorem "fails" in that particular example.

benorin: I'm not sure what div(curlF) = 0 tells me in this example other than I just know this rule from the book.Could you possibly inform me something more? :)

Thanks

 

[benorin]

div(curl F)=0 + divergence theorem

By Divergence theorem

\iint_S \left(\mbox{curl }\vec{F}\right)\cdot \vec{N}\,dS = \iiint_D \mbox{div }\left(\mbox{curl }\vec{F}\right) \, dV​

Consider that if \vec{F}=\left&lt; P,Q,R\right&gt;, then

\mbox{div }\left(\mbox{curl }\vec{F}\right) = \nabla\cdot\left(\nabla \times\vec{F}\right) = \left|\begin{array}{ccc}\frac{\partial}{\partial x}&amp;\frac{\partial}{\partial y}&amp;\frac{\partial}{\partial z}\\ \frac{\partial}{\partial x}&amp;\frac{\partial}{\partial y}&amp;\frac{\partial}{\partial z}\\ P &amp; Q &amp; R\end{arrary}\right| = 0​

since two rows are identical.

Hence the Triple integral above is zero.