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 r2.
Observing no θ term in r2, I multiply r2 by 2π to get: 2π*r2. I integrate this with respect to r.
2π ∫(0≤r≤2) r2 = 16π/3.
This seems to resemble hr2π/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?