1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Identifying a Solid By Its Bounds

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    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


    3. 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?
     
  2. jcsd
  3. Nov 18, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can't have a 'dummy' variable of integration like 'r' in your problem in the integration dz to also be the upper bound for the z variable. There must be two different r's there. I'd fix that first. That may be what's making this confusing.
     
  4. Nov 19, 2011 #3

    I like Serena

    User Avatar
    Homework Helper

    Why not? :confused:

    After the first 2 integrations we have a cylinder surface with radius r and height r.


    You have it right. The solid is bounded by a couple of surfaces of which one is a cone.
    However, the resulting solid is not a solid cone.
    Consider from where to where the bounds run.

    Since the max z-coordinate varies with the radius, it's not a solid cylinder, although a cylinder is another bounding surface.
     
  5. Nov 19, 2011 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You're right, of course. Too late at night, I think.
     
  6. Nov 19, 2011 #5
    So you say this object is bounded both a cone and a cylinder. Then I can only imagine a solid that has a cylindrical surface (with a height from 0 to 2), yet (and my descriptive ability is failing me) also has a depression in the shape of a cone on its top (radius from 0 to 2). To say it another way, suppose a round pencil is tightly enclosed enclosed by a cylinder, then the region I thinking of will be the empty space between the tip of the pencil and where the pencil achieves its maximum diameter.

    Yes? No? Incomprehensible?
     
  7. Nov 19, 2011 #6

    I like Serena

    User Avatar
    Homework Helper

    Yes, very comprehensible!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook