Verifying the divergence theorem- half done, help needed

[mcc123pa]

Homework Statement

Verify the divergence theorem for F(x,y,z) = (x,y,2z^2) and T is the region bounded by the paraboloid z=x^2+y^2 and the plane z=1.

Homework Equations

F(ds) = div(F)dV

The Attempt at a Solution

I have successfully evaluated the integral and come up with an answer:
div(F) = x/dx + y/dy + 2z^2/dz = 2+4z
Cylindrical limits of integration:
z: r^2 to 1
r: 0 to 1
theta: o to 2(pi)

I found the answer to be 7/3 * pi.
How do I verify this? Do I need to do any more to do so? Please advise ASAP! Thanks!

 

[mighty2000]

Thank you for your post. I am happy to help you verify the divergence theorem for the given vector field and region.

First, let's review the divergence theorem. It states that the flux of a vector field F through a closed surface S is equal to the volume integral of the divergence of F over the region T enclosed by S. Mathematically, it can be written as:

∫∫S F · ds = ∭T div(F) dV

Where ds is the differential element of surface area and dV is the differential element of volume.

Now, let's apply this theorem to the given problem. We have the vector field F(x,y,z) = (x,y,2z^2) and the region T bounded by the paraboloid z=x^2+y^2 and the plane z=1. To verify the divergence theorem, we need to evaluate both sides of the equation and show that they are equal.

On the left hand side, we have the flux of F through the closed surface S, which in this case is the boundary of the region T. Since the region T is bounded by the paraboloid and the plane, we can use the divergence theorem to write the left hand side as:

∫∫S F · ds = ∫∫S (x,y,2z^2) · ds = ∫∫S (x,y,2z^2) · n ds

Where n is the unit normal vector to the surface S. Since the surface S is curved, we can use the fact that the dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. In this case, the angle between the vectors (x,y,2z^2) and n is 90 degrees, so the cosine is 0. Therefore, the left hand side becomes:

∫∫S F · ds = ∫∫S (x,y,2z^2) · n ds = 0

On the right hand side, we have the volume integral of the divergence of F over the region T. We can use your calculated value for the divergence of F, which is 2+4z, and the cylindrical limits of integration you provided to evaluate this integral:

∭T div(F) dV = ∫∫∫T (2+4z) dV =