Evaluating Surface Integral with Divergence Theorem

In summary, the problem involves using the divergence theorem to evaluate the surface integral of a given function over a solid bounded by a hyperboloid and two planes. The limits of integration are determined by the volume of the solid. To solve the problem, one can integrate over a ring of radius r, from r=0 to the largest possible r value for a specific z, and from z=-2 to z=2. Then, the function should be converted to polar coordinates and integrated accordingly.
  • #1
jcertel
2
0
1.The problem asks " use the divergence theorem to evaluate the surface integral [tex]\int\int[/tex] F.ds
for F(x,y,z) = <x3y,x2y2,−x2yz>


where S is the solid bounded by the hyperboloid x^2 + y^2 - z^2 =1 and the planes z = -2 and z=2.

i know that the
[tex]\int\int[/tex] F.ds = [tex]\int\int\int[/tex] divFdv
but in not sure on the limits of integration.

when i found the divF i got 4x^2y where do i gor from here i turned it into polar with x=[rcos]\theta[/tex]
y=rsin[tex]\theta[/tex]
but I am not sure where to go from here?
 
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  • #2
i know that the
[tex]\int\int[/tex] F.ds = [tex]\int\int\int[/tex] divFdv
but in not sure on the limits of integration.

You want to integrate over all of the volume bound by the surface S, so the limits are whatever you need to include all of the volume.
when i found the divF i got 4x^2y where do i gor from here i turned it into polar with x=[rcos]\theta[/tex]
y=rsin[tex]\theta[/tex]
but I am not sure where to go from here?

OK, that's good. Now you integrate 4x^2y over the entire volume of the hyperboloid that's bound by z=2 and z=-2. I would personally do this:

1. For a specific z value and specific r value, integrate around a ring of radius r. This should get rid of theta.
2. Integrate from r=0 to the largest possible r value for this specific z.
3. Integrate from z=-2 to z=2.

But that's just my personal preference.
 
  • #3
ok i that with the limits

-2<Z<2
0<[tex]\theta[/tex]<[tex]\pi[/tex]
0<r<4

now do i need to change 4(x^2)y to 4(r^2)(cos^2[tex]\theta[/tex])sin[tex]\theta[/tex]? when i integrate that get 0? I am not sure if that's correct.
 

FAQ: Evaluating Surface Integral with Divergence Theorem

What is the Divergence Theorem?

The Divergence Theorem is a mathematical theorem that relates a surface integral over a closed surface to a volume integral over the region enclosed by that surface. It states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of that vector field over the region enclosed by the surface.

How is the Divergence Theorem used to evaluate surface integrals?

The Divergence Theorem allows us to evaluate surface integrals by converting them into volume integrals, which are often easier to calculate. By using the Divergence Theorem, we can reduce the complexity of the integral and solve it in a more efficient way.

What are the necessary conditions for using the Divergence Theorem?

In order to use the Divergence Theorem, the vector field must be continuous and have a continuous derivative over the region enclosed by the closed surface. The surface itself must also be smooth and closed, meaning that it has no boundary.

How does the Divergence Theorem relate to Gauss's Law?

Gauss's Law is a special case of the Divergence Theorem, where the vector field is the electric field and the surface is a closed surface surrounding a charge distribution. In this case, the Divergence Theorem states that the total electric flux through the closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space.

Are there any limitations to using the Divergence Theorem?

The Divergence Theorem can only be applied to closed surfaces and vector fields that meet the necessary conditions. It also cannot be used to evaluate surface integrals over surfaces with boundaries, as it only applies to closed surfaces. Additionally, the Divergence Theorem is only valid in three-dimensional space and cannot be extended to higher dimensions.

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