# Thread: Theorem of divergence and Stoke

1. ## Theorem of divergence and Stoke

Where can I use each theorem ?
Could you show me an example?

2. One major example of the Divergence Theorem is Gauss's Law, which is a flux integral:

$\displaystyle \oint_S \bold{E}.\bold{dA} = \frac{Q_{enclosed}}{\epsilon_0}$

More information on Divergence Theorem here: Divergence theorem - Wikipedia, the free encyclopedia

3. Originally Posted by Aryth
One major example of the Divergence Theorem is Gauss's Law, which is a flux integral:

$\displaystyle \oint_S \bold{E}.\bold{dA} = \frac{Q_{enclosed}}{\epsilon_0}$

More information on Divergence Theorem here: Divergence theorem - Wikipedia, the free encyclopedia
Ok. Thank you

4. Stokes I always use to compute flux ? And the divergence theorem ?

example

Determine the flux vector field: $\displaystyle F(x,y,z) = xi +yj + zk$ through the paraboloid $\displaystyle z = x^2+y^2$ below the plane $\displaystyle z = 1$ and vector normal to the surface points to the external side of the paraboloid

I can use Stokes or divergence theorem ?

5. the flux equal

$\displaystyle flux=\int\int F.n ds$

so how you can use Stokes's theorem for flux is there anti curl because Stoke's

$\displaystyle \oint F.dr=\int\int \nabla\times F .n dS$

since $\displaystyle \nabla\times F$ is the curl of F so if there is an anti for curl you can use Stoke's for flux
but the divergent theorem is

$\displaystyle \int\int F.n dS = \int\int\int div F dv$

$\displaystyle divF =\nabla F$ is the gradient

Stoke's theorem change from surface integral to linear integral but divergence theorem change from surface integral to volume

Green's is a special case from Stoke's

that what I think is correct is there any mistake about I said ...

6. in the example that I showed, if I use curl (equal 0) the flux will be equal to 0?
If I use the divergence theorem ?

7. Originally Posted by Apprentice123
in the example that I showed, if I use curl (equal 0) the flux will be equal to 0?
If I use the divergence theorem ?
I think you misunderstand me the flux equal

$\displaystyle \int\int F.n dS \ne \int\int \nabla\times F .n dS$

I need the function that

$\displaystyle F(x,y,z) = \nabla\times G(x,y,z)$ if I can find G I can apply stoke's

$\displaystyle \int\int F.n dS = \int\int \nabla\times G.n.dS = \oint G.dr$

8. Originally Posted by Amer
I think you misunderstand me the flux equal

$\displaystyle \int\int F.n dS \ne \int\int \nabla\times F .n dS$

I need the function that

$\displaystyle F(x,y,z) = \nabla\times G(x,y,z)$ if I can find G I can apply stoke's

$\displaystyle \int\int F.n dS = \int\int \nabla\times G.n.dS = \oint G.dr$
It is possible use Stokes in the previous example ?

9. ok show me you should find G such that

$\displaystyle F = \nabla\times G$

you can use Sotke's for flux if they give you a vector like this

$\displaystyle \nabla\times V(x,y,z)$

so for the flux

$\displaystyle \int\int \nabla\times V(x,y,z).n dS = \oint V .dr$

you can't use divergence if the surface is not closed if it is closed the integral above equal zero because

$\displaystyle \int\int \nabla\times V .n .dS = \int\int\int div(\nabla\times V) dv = 0$

because

$\displaystyle div(\nabla\times V )=0$

div for the curl equal zero

10. Originally Posted by Amer
ok show me you should find G such that

$\displaystyle F = \nabla\times G$

you can use Sotke's for flux if they give you a vector like this

$\displaystyle \nabla\times V(x,y,z)$

so for the flux

$\displaystyle \int\int \nabla\times V(x,y,z).n dS = \oint V .dr$

you can't use divergence if the surface is not closed if it is closed the integral above equal zero because

$\displaystyle \int\int \nabla\times V .n .dS = \int\int\int div(\nabla\times V) dv = 0$

because

$\displaystyle div(\nabla\times V )=0$

div for the curl equal zero

$\displaystyle F(x,y,z) = xi+yj+zk$

$\displaystyle xi+yj+zk = \nabla X G(x,y,z)$

How do I find G?

11. ok

curl G = F do you know the curl l determinate l

l....... i...... j...... k............ l

l..... d/dx.. d/dy...... d/dz... l = xi + yj + zk

l ...... f...... h........... g..... l

$\displaystyle i(\frac{\partial g}{\partial y} -\frac{\partial h}{\partial z} ) -j(\frac{\partial g}{\partial x} -\frac{\partial f }{\partial z})+k(\frac{\partial h}{\partial x} -\frac{\partial f }{\partial y})=xi+yj+zk$

$\displaystyle (\frac{\partial g}{\partial y} -\frac{\partial h}{\partial z} ) = x$

$\displaystyle (\frac{\partial g}{\partial x} -\frac{\partial f }{\partial z}) = y$

$\displaystyle (\frac{\partial h}{\partial x} -\frac{\partial f }{\partial y}) = z$

first you should find the derivative with respect to x,y,z then you should find the f ,h, g after you find them but them like this

$\displaystyle F(x,y,z)=f(x,y,z)i + g(x,y,z)j + h(x,y,z)k$

I will try to find them

12. I think the flux is 0

Paraboloid below the plane z = 1 is a closed surface

$\displaystyle F(x,y,z) = Mi + Nj + Pk$

[ d(N)/dx - d(M)/dy ]dxdy = 0

[ d(P)/dx - d(M)/dz ]dxdz = 0

[ d(P)/dy - d(N)/dz ]dydz = 0

13. Originally Posted by Apprentice123
Stokes I always use to compute flux ? And the divergence theorem ?

example

Determine the flux vector field: $\displaystyle F(x,y,z) = xi +yj + zk$ through the paraboloid $\displaystyle z = x^2+y^2$ below the plane $\displaystyle z = 1$ and vector normal to the surface points to the external side of the paraboloid

I can use Stokes or divergence theorem ?
no

the flux
I will use divergence

$\displaystyle \int\int F.n dS=\int\int\int div(F) dv$

$\displaystyle \int\int\int (\frac{\partial}{\partial x}i+\frac{\partial}{\partial x}j+\frac{\partial}{\partial x}k)(xi+yj+zk) dv$

$\displaystyle \int\int\int \left(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\right) dv=\int\int\int (1+1+1) dv=3\int\int\int dv=3V$ V volume of paraboloid

14. Originally Posted by Amer
no

the flux
I will use divergence

$\displaystyle \int\int F.n dS=\int\int\int div(F) dv$

$\displaystyle \int\int\int (\frac{\partial}{\partial x}i+\frac{\partial}{\partial x}j+\frac{\partial}{\partial x}k)(xi+yj+zk) dv$

$\displaystyle \int\int\int \left(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\right) dv=\int\int\int (1+1+1) dv=3\int\int\int dv=3V$ V volume of paraboloid
Yes I thought about it too. But no important that the paraboloid is below the plane z = 1?

You know the volume the any paraboloid?

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