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Hence the volume would be ∫∫∫rdzdrdθ where Rz/H ≤ r ≤ R, 0 ≤ z ≤ H, and 0 ≤ θ ≤ 2π.

But I’m wondering why I can’t express the volume as ∫∫∫rdzdrdθ where 0 ≤ r ≤ Rz/H, and 0 ≤ z ≤ H and 0 ≤ θ ≤ 2π? Thanks!!

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- Thread starter bodensee9
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- #1

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Hence the volume would be ∫∫∫rdzdrdθ where Rz/H ≤ r ≤ R, 0 ≤ z ≤ H, and 0 ≤ θ ≤ 2π.

But I’m wondering why I can’t express the volume as ∫∫∫rdzdrdθ where 0 ≤ r ≤ Rz/H, and 0 ≤ z ≤ H and 0 ≤ θ ≤ 2π? Thanks!!

- #2

Tom Mattson

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But I’m wondering why I can’t express the volume as ∫∫∫rdzdrdθ where 0 ≤ r ≤ Rz/H, and 0 ≤ z ≤ H and 0 ≤ θ ≤ 2π? Thanks!!

Because you've got the bounds on r in terms of z. That means that the r integration has to come before the z integration.

- #3

HallsofIvy

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No, it would not. The way you have set up this cone, it has vertex at the origin and base at z= H. TheI am wondering if someone can help clarify the following? Suppose that I’m asked to find the volume of a cone. So, the volume would be ∫∫∫dxdydz or using polar coordinates ∫∫∫rdzdrdθ. Therefore the volume would be if I have a cone with base radius R and height H, I can express the radius r at any height z using similar triangles. I would get that r = Rz/H, where z is the height at any point in the cone.

Hence the volume would be ∫∫∫rdzdrdθ where Rz/H ≤ r ≤ R, 0 ≤ z ≤ H, and 0 ≤ θ ≤ 2π.

YouBut I’m wondering why I can’t express the volume as ∫∫∫rdzdrdθ where 0 ≤ r ≤ Rz/H, and 0 ≤ z ≤ H and 0 ≤ θ ≤ 2π? Thanks!!

[tex]\int_{z=0}^H\int_{\theta= 0}^{2\pi}\int_{r= 0}^{Rz/H} rdrd\theta dz[/tex]

[tex]= 2\pi \int_{z=0}^H (1/2)(R^2 z^2)/H^2 dz= \pi R^2/H^2 \int_0^H z^2[/tex]

[tex]= \pi R^2/H^2 (1/3)H^3= (1/3)\pi R^2 H[/tex]

which is the standard formula for area of a cone.

Your original formula,

[tex]\int_{z=0}^H\int_{\theta= 0}^{2\pi}\int_{r= Rz/H}^R rdrd\theta dz[/tex]

[tex]= 2\pi \int_{z=0}^H (1/2)(R^2- R^2z^2/H^2)dz= \pi R^2\int_0^H(1- z^2/H^2)dz[/tex]

[tex]= (2/3)\pi R^2 H[/tex]

Notice that those two volumes add to [itex]\pi R^2 h[/tex], the volume of the cylinder.

(Tom, I think you misinterpreted his question. Well,

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I am wondering how integrating from Rz/H ≤ r ≤ R gives you the volume outside of the cone? I understand that if you integrate from bottom up that would give you 0 ≤ r ≤ Rz/H. But, I thought that since you know that the maximum value for r is R, you can also write it as Rz/H ≤ r ≤ R?

Sorry.

- #5

HallsofIvy

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I am wondering how integrating from Rz/H ≤ r ≤ R gives you the volume outside of the cone? I understand that if you integrate from bottom up that would give you 0 ≤ r ≤ Rz/H. But, I thought that since you know that the maximum value for r is R, you can also write it as Rz/H ≤ r ≤ R?

Sorry.

The axis of the cone is the positive z-axis, isn't it? So that r= 0 is always inside the cone, no matter what z is?

Draw a graph. Take the horizontal axis to be "r" and the vertical axis to be z. Draw the straight lines r= Rz/H or z= Hr/R and z= -Hr/R representing the sides of the cone and the vertical line r= R. Where is the inside of the cone and where is the outside? Where is Rz/H ≤ r ≤ R?

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Okay, I see it now. Thanks so much!!

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Tom Mattson

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(Tom, I think you misinterpreted his question. Well,oneof us did!)

I confess that I didn't really read above the line that said, "why can't I express the volume as...?" in the OP. But my point is that he had the variables of integration out of order. dr has to be to the left of dz in the integrand to use the bounds that he stated.

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