Ok then, suppose restrict to just one slice of volume, of thickness dz at position z, where z lies between 0 and 3. It is a circular slice, of radius sqrt(z/2). Imagine just the quadrant where x>0 and y>0. Integrate (2/9) * (x + y + z) over that quadrant. The z value is constant. The area is a double integral, integrated dx and dy, or you can do some polar coordinates version. But suppose just dx and dy. The integral breaks down into three integrals, one int(x dx dy), one int(y dx dy) and one int(z dx dy). The last one is just z (constant) times the area of quadrant. The first two are identical with x and y playing interchangeable roles. The only tricky part is getting the limits of integration right. But you can do that. Try a trig substitution.
That's the way I see the problem. I don't actually know what will be the result, and am going to leave it to you to do the details of the integrations. Perhaps I've mistaken some aspect of this, but you (or some other reader of these posts) will likely correct any error I've made.
The important point, going back to the original formulation of the problem, is that you've been able to evaluate integral of W dot d(area), which could be a nuisance, by using Gauss's theorem to instead evaluate "div W) d(volume).
You can also do the earlier surface integral, W dot d(area), as a check, if you like.
I'm sure there is more insight available by studying this problem. But the above is prettly much a good start, I believe.
Best wishes,
Ken