Without calculation, decide if each of the integrals below are positive, negative, or zero. Let W be the solid bounded by z=sqrt (x2+y2) and z=2.1. ∭ (z-2) dV2. ∭e-xyzdV3. ∭ (z-sqrt (x2+y2)) dV

1) ∭ (z-2) dV negative.

2) ∭e^{-xyz} dV positive.

3) ∭ (z-/sqrt{x²+y²}) positive.

Step-by-step explanation:

From Exercise we have:

z=/sqrt{x²+y²}

z=2

⇒2=/sqrt{x²+y²}

4=x²+y²

Therefore, we get that the solid bounded by:

/sqrt{x²+y²}≤z≤2

4=x²+y²

1) From initial condition we have that

/sqrt{x²+y²}≤z≤2

⇒ 2-z≤0

Therefore, we get that the triple integral is

∭ (z-2) dV negative.

2) We know that e^{-xyz} is always positive number.

Therefore, we get that the triple integral is

∭e^{-xyz} dV positive.

3) From initial condition we have that

/sqrt{x²+y²}≤z≤2

⇒ z-/sqrt{x²+y²}>0

Therefore, we get that the triple integral is

∭ (z-/sqrt{x²+y²}) positive.