# Volume Question

bodensee9
I 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.

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!

Staff Emeritus
Gold Member
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.

Homework Helper
I 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.
No, it would not. The way you have set up this cone, it has vertex at the origin and base at z= H. The inside of the cone has 0≤ r≤ Rz/H. Your limits of integration gives the volume of the part of the cylinder of radius R and height H outside the cone.

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!
You can. This is the correct integral!
$$\int_{z=0}^H\int_{\theta= 0}^{2\pi}\int_{r= 0}^{Rz/H} rdrd\theta dz$$
$$= 2\pi \int_{z=0}^H (1/2)(R^2 z^2)/H^2 dz= \pi R^2/H^2 \int_0^H z^2$$
$$= \pi R^2/H^2 (1/3)H^3= (1/3)\pi R^2 H$$
which is the standard formula for area of a cone.

$$\int_{z=0}^H\int_{\theta= 0}^{2\pi}\int_{r= Rz/H}^R rdrd\theta dz$$
$$= 2\pi \int_{z=0}^H (1/2)(R^2- R^2z^2/H^2)dz= \pi R^2\int_0^H(1- z^2/H^2)dz$$
$$= (2/3)\pi R^2 H$$
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, one of us did!)

bodensee9
Thanks, that was helpful. But I'm still a little bit confused.
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.

Homework Helper
Thanks, that was helpful. But I'm still a little bit confused.
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?

bodensee9
Okay, I see it now. Thanks so much!

Staff Emeritus