Answer:
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.
