What is the Flux of a Vector Field Over a Box and How Does Divergence Affect It?

[Phymath]

find the flux of the vector
$$\vec{A} = \frac{6ka^2y}{\sqrt{x^2+y^2+a^2}} \hat{e_x}$$
$$\frac{3ka^2z}{\sqrt{y^2+z^2+4a^2}}\hat{e_y}$$

$$\frac{2ka^2x}{\sqrt{x^2+z^2+9a^2}}\hat{e_z}$$

a) intergrating over the surface of the box, of
0 <= x <= 2a
0 <= y <= 3a
0 <= z <= a
b) divergence thrm over the volume of the box


a)...I setup 6 flux intergrals and sum them for the total flux is the first step..
each of these intergrals is setup like (going directly to the dot product result)...
$$\oint \vec{A} \bullet d\vec{a}$$

$$\int_0^a \int_0^{3a} A_x dydz$$ however when i setup the opposite side of the box...
$$\int_0^a \int_0^{3a} -A_x dy dz$$ because of the $$d\vec{a} = dydz(-\vec{\hat{e_x}})$$

which means they just cancel which i do for all of them...and get 0 flux which is not what it should be nothing loops back on itself, and the divergence gives me something totally diffrent so what's up?

 

[StatusX]

The function has a different value at opposite sides of the box.

 

[Phymath]

soooo..? $$\int_0^a \int_0^{3a} A_x dydz$$
$$6ka^3\int_0^{3a} \frac{y}{\sqrt{x^2+y^2+a^2}} dy$$
$$6ka^3(\sqrt{10a^2+x^2}-\sqrt{a^2+x^2})$$


$$\int_0^a \int_0^{3a} -A_x dydz$$
$$-\int_0^a \int_0^{3a} A_x dydz$$ how is this not opposite what I just did?

do i have to setup the first intergral like...
$$x=2a$$ at this side of the surface
$$6ka^3\int_0^{3a} \frac{y}{\sqrt{(2a)^2+y^2+a^2}} dy$$

 

[StatusX]

You're at two different values of x. The final answer should only be in terms of a.

 

[Phymath]

check my edit is that what u mean?

 

[StatusX]

yes, when your integrating at x=2a, you need to plug in 2a for x.