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## Homework Statement

Sketch the solid whose volume is given by the following integral and then evaluate this integral:

∫(0≤r≤2)∫(0≤θ≤2π)∫(0≤z≤r) (r) dzdθdr

## The Attempt at a Solution

So, I kind of cheated and evaluated the integral first. So I have

∫(0≤z≤r) (r) dz = rz, from z=0 to z=r, which I equate to r

^{2}.

Observing no θ term in r

^{2}, I multiply r

^{2}by 2π to get: 2π*r

^{2}. I integrate this with respect to r.

2π ∫(0≤r≤2) r

^{2}= 16π/3.

This seems to resemble hr

^{2}π/3, which is the volume of a cone. So I am tempted to believe this solid is a cone. My sketch of the cone has its "nose" at the origin, and it extends upward about the z-axis until z=2. The radius of this cone steadily increases from 0 at the nose to 2 at z=2.

But how would I be able to tell this solid from a cylinder? I think both would have the same bounds, right?