Identifying a Solid By Its Bounds

  • #1

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 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?
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
I like Serena
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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.
Why not? :confused:

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


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?
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.
 
  • #4
Dick
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  • #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?
 
  • #6
I like Serena
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Yes, very comprehensible!
 

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