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.

what does muffins have to do with anything?

step-by-step explanation:

answer: -2+√x,-2-√x

i hope this , i don't know if you're looking for a pair or not.