Surface area of a cone-inconsistency?

[betelguese05]

Surface area of a cone--inconsistency?

Geometry tells us that the surface area of a cone with a circular base is

$$SA = \pi rs$$

where s is the slant height of the cone, or

$$SA = \pi r \sqrt{r^2 + h^2}$$

Take a cone with a circular base of radius 1 and a height of 4. This formula tells us that

$$SA = \pi\sqrt{17}$$

However, trying to approach this problem from a calculus standpoint:

$$SA = \int 2 \pi r dh$$

Considering the cone as a triangle for a moment, we have the point (0,4) on the h-axis and (1,0) and the r-axis (where h is the analogue of y and r the analogue of x), and a line connecting them.

This leads to a linear relationship between r and h, such that:

$$h = -4r+4$$

or

$$r = (0.25)(4-h)$$

Integrating:

$$2\pi\int^{4}_{0}(0.25)(4-h)dh$$

=$$0.5\pi(4h - 0.5h^2)$$

evaluated from h=0 to 4. However, this yields $$4\pi$$, not $$\pi\sqrt{17}$$.

 

[g_edgar]

However, trying to approach this problem from a calculus standpoint:

$$SA = \int 2 \pi r dh$$

This is incorrect. Check your text for the integral to do a surface area.

 

[betelguese05]

I disagree. This would be analogous to a volume of revolution--rather, we are taking a surface area. Each disk has radius r and therefore circumference 2*pi*r; since r varies as a function of h, we can evaluate the integral of 2*pi*r*dh from 0 to 4.

 

[Mute]

I disagree. This would be analogous to a volume of revolution--rather, we are taking a surface area. Each disk has radius r and therefore circumference 2*pi*r; since r varies as a function of h, we can evaluate the integral of 2*pi*r*dh from 0 to 4.

The circle has circumference r but the height of the small disk is not dh, it's

$$ds = \sqrt{dr^2 + dh^2} = \sqrt{\left(\frac{dr}{dh}\right)^2 + 1}~dh$$.

http://en.wikipedia.org/wiki/Surface_of_revolution

 

[betelguese05]

Ah, well done. Thanks.

 

[HallsofIvy]

Here's another way to find the surface area. If the cone has base radius r and height h, then the "slant height" is $R= \sqrt(r^2+ h^2)$. Imagine cutting the cone along a line from the base to the tip. Flatten it out to form part of a circle. The circle it fits into has radius R, circumference $2\pi R$, and area $\pi R^2$. The base circle of the cone has radius r and so circumference $2\pi r$. That means that the "flattened out" cone takes up $(2\pi r)/(2\pi R)= r/R$ of the entire circle and so its area is that fraction of the area of the entire disk: $(r/R)(\pi R^2)= \pi Rr= \pi\sqrt{r^2+ h^2}r$.

 

[LCKurtz]

One thing that has always bothered me about the "cut and flatten it out" method is the assertion that it will, in fact, flatten out. I agree it is "obvious" that a cone will and a sphere won't, but I don't recollect ever seeing a convincing argument for it.

 

[Prologue]

One thing that has always bothered me about the "cut and flatten it out" method is the assertion that it will, in fact, flatten out. I agree it is "obvious" that a cone will and a sphere won't, but I don't recollect ever seeing a convincing argument for it.

Maybe that every where you need to 'bend' the cone is along a straight line (tip to base)? If it is not a straight line it will not bend properly.

 

[HallsofIvy]

One thing that has always bothered me about the "cut and flatten it out" method is the assertion that it will, in fact, flatten out. I agree it is "obvious" that a cone will and a sphere won't, but I don't recollect ever seeing a convincing argument for it.

A cone has the property that, at any point, there exist a line through the point that is contained in the cone (the line through the vertex of the cone). Such a surface is called a "developable surface" (also "ruled" surface) and it can be shown, in differential geometry, that any "developable surface" can be flattened.

See http://en.wikipedia.org/wiki/Developable_surface

 

[IPhO' 2008]

In the area formula.What is the t?

 

[HallsofIvy]

In the area formula.What is the t?

? What area formula are you talking about? I can find no "t" in any formula anyone has mentioned here.

 

[Identity]

In a physics assignment earlier this year I was asked to prove that for a shell of a sphere with charge q and charge density function ρ(x),

$$dq = \rho (r)4\pi r^2 dr$$

What I did was I cut the spherical shell in half, then I flattened out each half of the shell. Of course, stretching would have to be involved in this flattening process, so you get two frustums (like a 3D trapezium thingy). The inner radius, 'r' of the shell contributes to the smaller flat side of the frustum, and the outer radius, 'R', of the shell contributes to the larger flat side of the frustum. Due to circular symmetry the two flat sides of the frustum are circular.

The advantage of flattening it out is that the height of the frustum becomes $$dr$$.

The smaller side has area $$4\pi r^2$$

The larger side has area $$4\pi R^2$$

As $$dr \to 0$$, $$R \to r$$, so by the squeeze theorem the volume of the frustum approaches the volume of a cylinder with radius r = R and height dr.

Therefore the volume of the shell is $$4\pi r^2 dr$$

And from an application of $$\rho (r) = \frac{dq}{dv}$$ the result follows.

What do you think of this?

 

[Borek]

A cone has the property that, at any point, there exist a line through the point that is contained in the cone (the line through the vertex of the cone). Such a surface is called a "developable surface" (also "ruled" surface) and it can be shown, in differential geometry, that any "developable surface" can be flattened.

It makes me wonder... Seems to me like for helicoid there exist at any point line through that point that is contained in the helicoid - is helicoid a developable surface?

Edit: I hope I am correctly using name of helicoid, I am thinking about surface that can be made by rotating and shifting line segment perpendicular to the line that is axis of its rotation.

Edit 2: OK, I think I can see what is my problem, I was trying to make it rectangular after developing, while it makes nice circle, just 'multilayered'.