Issue converting dV to dx dy dz

[Addez123]

Homework Statement

Integrate the vector field
$$A = (2xy + x^2, 2 +yz, 2z^4)$$
over the sphere surface:
$$x^2+(y-1)^2+z^2 = 4$$
where y >= 0

Relevant Equations

Divergence theorem

I can calculate the divergence
$$div A = 2y + 2x + z + 8z^3$$
Now I have to integrate over this cut-off sphere.

So I decide I'll cut it up into small discs with height dy and integrate over that
$$dV = \pi(4 - (y-1)^2)^2 * dy$$

My issue here is I don't know how to integrate 2x + z + 8z^3.
Not only that but it's suppose to be a triple integral and all I get is dy..

I must have missed a step but idk which :/

 

[ergospherical]

There's not sufficient symmetry to use discs as volume elements, because the vector field $\mathbf{A}$ varies within each disc. Exploit the symmetry. Shift $y' \equiv y-1$ and write the resulting integral in terms of spherical coordinates (with $dV = r^2 \sin{\theta} dr d\theta d\phi$).

 

[Orodruin]

Homework Statement:: Integrate the vector field
$$A = (2xy + x^2, 2 +yz, 2z^4)$$
over the sphere surface:
$$x^2+(y-1)^2+z^2 = 4$$
where y >= 0
Relevant Equations:: Divergence theorem

So I decide I'll cut it up into small discs with height dy and integrate over that
dV=π(4−(y−1)2)2∗dy

This is not dV. It is dV integrated over the x and z coordinates assuming that the integrand does not depend on them. (Assuming you get rid of that additional square you inserted.)

However, there may be a way to get rid of that pesky x and z dependence if you think a few extra times before writing down the integral.