Hi

Having a bit of trouble with this question. Any ideas? Im thinking its going to be the integral of F.N? With N being the normal...using jacobian. Sorry, just throwing out ideas! Cheers.

Calculate the flux of the magnetic field B = (2xy, yz, 1) through the surface of

the planar circle of radius R centred at the origin and located in the x, y-plane.

Before we continue, we should note a few things. The first of which is our "circle" will actually become a cylinder in 3 space. When we extend this upward and downward along the z plain we will get a cylinder bounded by heights -h and h. We label these as such because the flux out of an infinately large shape is going to be infinity, so we bound our z such that our result is actually useful.

\(\displaystyle -h \le z \le h \)

We can do this 1 of 2 ways. The first of which is to compute the flux integrals for the 3 distinct sides of the cylinder (top, bottem and side) or use the divergence theorem.

I will use the divergence theorem,

\(\displaystyle \iiint_D div F dV= \iint F \cdot \hat N dS \)

There should be a closed symbol on the right side but I dont know the latex code for this

Our Function, \(\displaystyle F = 2xy \hat i + yz \hat j + 1 \hat k \) has the divergence such that,

\(\displaystyle div F = 2y + z + 1 \)

Our flux is therefore,

\(\displaystyle Flux = \iiint (2y + z + 1) dxdydz \)

Noting our dz bounds are the bounds we defined for z at the start of this problem, and our x and y bounds can be most easily expressed in cylindrical co-ordinates (because it is a cylinder!).

\(\displaystyle Flux = \int_0^{2 \pi } d \theta \int_0^r r dr \int_{-h}^{h} (2y + z + 1) dz \)

Where \(\displaystyle y = r sin \theta \)

\(\displaystyle Flux = \int_0^{2 \pi } d \theta \int_0^r r dr \int_{-h}^{h} (2rsin \theta + z + 1) dz \)