Flux of vector field through box

[jameson2]

Homework Statement

Consider the vector field $$\vec F=\frac{\vec r}{r^3}$$ with $$\vec r =x\hat{i} +y\hat{j} +z\hat{k}$$
Compute the flux of F out of the box $$1\leq x \leq 2, 0\leq y \leq 1, 0\leq z \leq 1$$

Homework Equations

I can't use the Gauss divergence theorem since the divergence of the vector field is zero. So I think that the only thing I can use is the surface integral formula, $$\int_S \vec F \cdot d\vec A=\int_D dudv (\vec F(u,v))\cdot (\frac{dw}{du} \times \frac{dw}{dv} )$$ for a suitable parametrization of the surface $$w(x,y,z)=w(x(u,v),y(u,v),z(u,v))$$

The Attempt at a Solution

I'm not sure how i should treat the box. My attempt was to try and work out the flux through each side of the box and sum the results. I think that maybe there is no outward flux through the sides of the box that are in the xz and xy planes, since they are parallel to the field. Also, I don't think there is any flux going out of the side of the box in the x=1 plane, since there is only a flux inward. If I try the problem this way, I get an awkward integration, for example one of the sides gives the integration of the following for the flux:
$$\int dxdy \frac{2}{(4+y^2 +z^2)^3^/^2}$$
This is where I'm stuck, I can't find a way to integrate this. An help would be great, as I'm not sure if this is the most efficient way, or even if it is the right way to do the problem.

 

[3d51]

If the divergence of a vector field is zero, then the divergence theorem tells you that the outward flux through the box is equal to zero. However, in this example I don't think the divergence is zero...

You are right to consider the flux across each side and add the results.

To calculate integrals like the one you gave, you can use a trigonometric substitution. If you're doing the y integration first, try the substitution
$$y = (4+z^2)^\frac{1}{2} \tan(t)$$
Then, the denominator should become something like
$$(4+z^2)^\frac{3}{2} (1 + \tan^2 (t))^\frac{3}{2} = (4+z^2)^\frac{3}{2} \sec^3 (t)$$
and so the whole integral becomes
$$\int_a^b \frac{\cos(t)}{4+z^2} dt = \left.\frac{\sin(t)}{4+z^2}\right|_a^b$$
where a and b are the new limits of the t integration (i.e. not the y limits!).