Hello, everyone. So here is the problem in question with their respective answers:

Let W1W1 be the solid half-cone bounded by z=x2+y2−−−−−−√z=x2+y2, z=3z=3 and the yzyz-plane with x≥0x≥0, and let Let W2W2 be the solid half-cone bounded by z=x2+y2−−−−−−√z=x2+y2, z=5z=5 and the xzxz-plane with y≤0y≤0.

For each of the following, decide (without calculating its value) whether the integral is positive, negative, or zero.

(a)∫W2x2+y2−−−−−−√dV is ? positive negative zero

(b)∫W1yzdV is ? positive negative zero

(c)∫W2xdV is ? positive negative zero

So in an attempt to solve this problem, I drew what the boundaries of W1 and W2 looked like and ended up with a triple integral for both cases. For W1, I got ∫(Θ=-pi/2 to pi/2)∫(r=0 to 3)∫(z=r to 3) dzdrdΘ and for W2, I got ∫(Θ=0 to pi)∫(r=0 to 5)∫(z=r to 5) dzdrdΘ. I'm not actually sure if those boundaries are even correct, but for (b) and (c) I'm confused about why the answer would be zero. Also, is there some intuitive method to figuring out whether the solution to a triple integral is postive, zero or negative without actually solving it? Moreover, since triple integrals find the volume of solids, how is it possible to obtain a negative volume at all? Thanks for your time and help in advance.

∫

∫

∫