Use triple integration to find the volume of the solid enclosed by the cylinder x^2 +y^2 = 9 and the planes y + z = 5 and z = 1.

Question

anonymous · Answer

is that your boyfriend

anonymous · Answer

ok triple integration is dV

anonymous · Answer

SSSdv

anonymous · Answer

so SSS f(x,y,z) dV

anonymous · Answer

How do I find the boundaries of integration though? I was trying to draw it but it's a pain.

anonymous · Answer

it is

anonymous · Answer

its a great question though, a real doozie

anonymous · Answer

It's supposed to give me a really nice answer.

across · Answer

On the xy-plane, you will have a circle centered at the origin with a radius of 3. However, this circle creates a cylinder with a base that starts on the plane z = 1 (that is, 1 unit up) and cut off at the top at an angle described by the line through the yz-plane with equation y = -z + 5.

anonymous · Answer

cut off by a plane ? on the top

anonymous · Answer

I can picture the cylinder and the plane at z = 1, but I'm having trouble seeing where y+z = 5 is cutting this shape

anonymous · Answer

wouldnt this be double integration?

across · Answer

No, this is a legit triple-integral problem.

Have you tried this?$$\iint\left [  \int_{1}^{-y+5}f(x,y,z)dz \right ]dA$$

anonymous · Answer

That makes sense. I hadn't thought of that. I get 36π.

anonymous · Answer

so the integral f(x,y,z) dz from 1..-y+5 , what does that do?

across · Answer

Yes, that's the correct answer, 481991.

fermats, those are the boundaries of the z-axis, so to speak.

across · Answer

Basically, the laid-out integral he had to compute was the following:$$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{1}^{-y+5}dzdydx.$$