Another flux problem

1. Apr 28, 2008

kasse

1. The problem statement, all variables and given/known data

Calculate the outward flux of the vector field F = <2x, -3y, 3z> across the boundary of the first-octant unit cube with opposite vertices (0, 0, 0) and (1, 1, 1).

2. The attempt at a solution

The top and the bottom of the box is easy. The flux here is 3 and 0 respectively. But how do I calculate the other sides?

Last edited: Apr 28, 2008
2. Apr 28, 2008

HallsofIvy

I hate to be the one to break it to you but the flux over the bottom is NOT -3. The unit, outward, normal to the bottom, z= 0, is <0, 0, -1> so you are integrating $<2x,-3y,2z>\cdot<0, 0, ->= <2x, -3y, 0>\cdot<0, 0, -1>= 0$, since z= 0 there, over the xy-plane, 0< x< 1, 0< y, 1. That flux is 0. It is correct that the flux over the top, z= 1 is
[tex]\int_{x=0}^1\int_{y= 0}^1 <x, y, 3>\cdot<0, 0, 1> dxdy= 3[/itex].

The "left" face is x= 0 and an outward unit normal is <-1, 0, 0>. On that face, x= 0 so the integrand is $<0, -3y, 3z>\cdot<-1, 0, 0>= 0$ so the flux there is also 0.
The "right" face is x= 1 and an outward unit normal is <1, 0, 0>. On that face, x= 1 so the integrand is $<2, -3y, 3z>\cdot<1, 0, 0>$ integrate that for 0<y<1, 0< z< 1.

The "front" face is y= 0 and an outward unit normal is <0, -1, 0>. On that face, y= 0 so the integrand is $<2x, 0, 3z>\cdot<0, -1, 0>$.
The "back" face is y= 1 and an outward unit normal is <0, 1, 0>. On that face, y= 1 so the integrand is $<2x, -3, 3z>\cdot<0, 1, 0>$.

Of course, it would be easy to very easy to use the divergence theorem to get the answer.

3. Apr 28, 2008

kasse

Yes, the divergenze theorem is part of the next chapter, but it would definitely be easier. How about this one:

Calculate the outward flux of the vector field F = <0, 0, z^2> out of the boundary of te solid bounded by the paraboloids z= z^2 and z = 18 - z^2 - y^2

I change to polar coordinates: z = 18 - r^2 and z = r^2. These functions intersect in the circle r=3.

Then I use the divergence theorem, integrating 2z dz dr d(tetha) within the limits [r^2, 18-r^2], [0, 3] og [0, 2*pi] respectively. The answer I get is 1296*pi. The correct answer is 1458*pi. Where's my mistake?

4. Apr 28, 2008

HallsofIvy

The differential of volume in cylindrical coordinates is $r dzdrd\theta$. You forgot the "r".