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Homework Help: 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


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    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

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    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


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    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

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    Yes, very comprehensible!
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