Cone Edges in Cylindrical Coordinates

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Homework Help Overview

The discussion revolves around understanding the representation of a cone in cylindrical coordinates, specifically focusing on the equation for the edge of the cone and the calculation of its volume. The cone is defined with a height of 2 and a circular base of radius 1 centered at the origin.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to clarify the meaning of "edge of the cone" and seeks help in deriving the equation for it. Some participants suggest visualizing the cone from the side to find the equation of the slant side. Others raise questions about the setup for calculating the volume, particularly regarding the limits of integration.

Discussion Status

Participants are exploring different interpretations of the problem and discussing the appropriate equations and limits for integration. There is an ongoing exchange of ideas about how to approach the volume calculation, with some guidance provided on setting up the integrals.

Contextual Notes

There is a mention of confusion regarding the phrasing of the problem and the need to clarify the bounds for integration in the context of cylindrical coordinates. The discussion reflects a collaborative effort to understand the geometric representation and mathematical formulation of the cone.

brad sue
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Hi ,
I don't know how to get the edge of the cone in cylindrical coordinates.

For example, we have a cone starting at the origin, of heigth 2 and the top is a circle of radius 1 (center at the origin).

the edge of the cone is z=2r. but I don't know how they find it.
Please can someone help me?

Thank you

B
 
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The phrasing "edge of the cone" is a little confusing! I would have interpreted it to mean the circle where the slant side meets the base but it seems clear that you are talking about the slant (curved) side itself.

Look at the cone "from the side". That is, imagine that you are looking along the y-axis so that positive z is up and positive x is to the right.
The line forming the slant side of the cone, from that perspective, passes through (0,0,0) and (1, 0, 2). Ignoring the y-coordinate, that is (0,0) to (1, 2). What is the equation of a line passing through those coordinates? (Remember to use z instead of y.)
Now, remember that the cone is formed by that line rotating around the z-axis. "x" is really the straight line distance from the z-axis to the point: that's r.
 
HallsofIvy said:
The phrasing "edge of the cone" is a little confusing! I would have interpreted it to mean the circle where the slant side meets the base but it seems clear that you are talking about the slant (curved) side itself.
Look at the cone "from the side". That is, imagine that you are looking along the y-axis so that positive z is up and positive x is to the right.
The line forming the slant side of the cone, from that perspective, passes through (0,0,0) and (1, 0, 2). Ignoring the y-coordinate, that is (0,0) to (1, 2). What is the equation of a line passing through those coordinates? (Remember to use z instead of y.)
Now, remember that the cone is formed by that line rotating around the z-axis. "x" is really the straight line distance from the z-axis to the point: that's r.
Thank you
In fact, I took this explample for a book.
for the same cone, we want to calculate the volume.
I set the integral
(|(a,b) means integral from a to b)
for theta : |(0,2*pi)
for r: |(0,1)
for z: |(X,2) I don't know how to find the lower bound X
 
brad sue said:
Thank you
In fact, I took this explample for a book.
for the same cone, we want to calculate the volume.
I set the integral
(|(a,b) means integral from a to b)
for theta : |(0,2*pi)
for r: |(0,1)
for z: |(X,2) I don't know how to find the lower bound X

IF you are doing it in that order: the [itex]\theta[/itex] integration is the outer integration from [itex]\theta= 0[/itex] to [itex]2\pi[/itex] and then next from r= 0 to r=1, then inner most integral is from the line z= 2r up to 2:
[tex]\int_{\theta = 0}^{2\pi}\int_{r= 0}^1\int_{z=2r}^2 rdzdrd\theta[/tex]

I, personally would be inclined to change the order:
[tex]\int_{\theta= 0}^{2\pi}\int_{z= 0}^2\int_{r= 0}^{\frac{r}{2}} rdrdzd\theta[/tex].

They should give exactly the same answer.
 
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