Question on triple integrals w/ domain x^2+y^2<=z<=4

[Superyoshiom]

I have to find the triple integral of xy dV, where the domain D is the region x^2 + y^2 <= z <= 4.

. . .\(\displaystyle \mbox{(b) Evaluate}\)

. . . . .\(\displaystyle \displaystyle \int\, \int\, \int_D\, x\, dV\)

. . .\(\displaystyle \mbox{where }\, D\, \mbox{ is the region }\, x^2\, +\, y^2\, \leq \, z\, \leq \, 4\)

I tried setting my initial integral for dz with bounds from x^2 + y^2 to 4, but I don't know how to do this with the other two integrals. I've uploaded the question for clarity. Any help would be greatly appreciated!

 

[tkhunny]

How would you do the double integral - without any mention of z?

 

[Superyoshiom]

How would you do the double integral - without any mention of z?

I'm not too sure, at first I tried setting up an inequality in terms of x and y but I didn't know how to proceed from there.

 

[Superyoshiom]

How would you do the double integral - without any mention of z?

I'm pretty sure I need to set up two more inequalities but I don't know how to do that in terms of x and y.

 

[Deleted member 4993]

I have to find the triple integral of xy dV, where the domain D is the region x^2 + y^2 <= z <= 4.

. . .\(\displaystyle \mbox{(b) Evaluate}\)

. . . . .\(\displaystyle \displaystyle \int\, \int\, \int_D\, x\, dV\)

. . .\(\displaystyle \mbox{where }\, D\, \mbox{ is the region }\, x^2\, +\, y^2\, \leq \, z\, \leq \, 4\)

I tried setting my initial integral for dz with bounds from x^2 + y^2 to 4, but I don't know how to do this with the other two integrals. I've uploaded the question for clarity. Any help would be greatly appreciated!

Do you see that the inequality describes an inverted right circular cone?

 

[tkhunny]

Do you see that the inequality describes an inverted right circular cone?

Yes, please take advantage of symmetries.