Calculating Outward Flux of a Vector Field Across a Unit Cube Boundary

[kasse]

Homework Statement

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?

 

[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].<br /> <br /> 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.<br /> 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.<br /> <br /> The "front" face is y= 0 and an outward unit normal is <0, -1, 0>. On that face, y= 0 so the integrand is $<2x, 0, 3z>\cdot<0, -1, 0>$.<br /> The "back" face is y= 1 and an outward unit normal is <0, 1, 0>. On that face, y= 1 so the integrand is $<2x, -3, 3z>\cdot<0, 1, 0>$.<br /> <br /> Of course, it would be easy to very easy to use the divergence theorem to get the answer.[/tex]

 

[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?

 

[HallsofIvy]

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