1. ## Flux integral

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

R centred at the origin and located in the x, y-plane.

2. Originally Posted by Croc
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 :P

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 $\displaystyle y = r sin \theta$

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

3. I'm afraid AllanCuz has completely misunderstood the problem. The problem does not ask about the flux through the sides of a cylinder, only about the flux through a disk.

Croc, the unit normal to the xy-plane is, of course, $\displaystyle \vec{k}$. The "vector differential of surface area" is just $\displaystyle \vec{n}dS= \vec{k}dxdy$. Since $\displaystyle \vec{F}(x,y,z)= 2xy\vec{i}+ yz\vec{j}+ \vec{k}$, $\displaystyle \vec{F}\cdot \vec{n}dS= dxdy$.

That is, since the disk is in the xy-plane, only the z-component cause a flux through it. And since that component is 1, the flux is just the area of the circle.

4. Originally Posted by HallsofIvy
I'm afraid AllanCuz has completely misunderstood the problem. The problem does not ask about the flux through the sides of a cylinder, only about the flux through a disk.

Croc, the unit normal to the xy-plane is, of course, $\displaystyle \vec{k}$. The "vector differential of surface area" is just $\displaystyle \vec{n}dS= \vec{k}dxdy$. Since $\displaystyle \vec{F}(x,y,z)= 2xy\vec{i}+ yz\vec{j}+ \vec{k}$, $\displaystyle \vec{F}\cdot \vec{n}dS= dxdy$.

That is, since the disk is in the xy-plane, only the z-component cause a flux through it. And since that component is 1, the flux is just the area of the circle.
I assumed the question meant a cylinder because I've been given problems such as the one above and we were meant to interpret/bound it as a cylinder.