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Arc Circle Volume

by badtwistoffate
Tags: arc, area, circle, geometry, volume
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badtwistoffate
#1
Feb9-10, 10:29 PM
P: 85
1. The problem statement, all variables and given/known data
Trying to either derive or find a formula for the volume given by the space carved out from
the area between an arc and the center of a circle. Essentially, the area between an arc and the center of a circle rotated 360 degrees, almost a cone shape with the upper boundary being that of a circular arc.

2. Relevant equations

area between an arc and the center of a circle = .5 r2θ

volume of a circular cone = (1/3) pi r2 h

3. The attempt at a solution

Not really sure where to start. A web search reveals many formulas for volumes but none for this situation - surprising to me. I imagine a circle first and then a sphere carving out that volume from the area between an arc and the center of the circle. Attached is a pic to help explain. Any guidance?

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Mentallic
#2
Feb10-10, 01:04 AM
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I'm not exactly understanding your question. You've mentioned volume multiple times and even have the volume of a cone formula, but when you describe what you want it seems like you just want to find the area of a segment of a circle.

Can you please give a little more clarification?
badtwistoffate
#3
Feb10-10, 08:55 AM
P: 85
Sorry. What I am trying to find is the volume of a segment of a sphere. I can find the area in 2-D space but cannot figure out a formula for volume for 3-D space.

Mentallic
#4
Feb10-10, 08:44 PM
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Arc Circle Volume

Ahh well that makes a lot more sense now (I hope)

This is actually a very similar problem to the 2-d diagram you showed. If we take a circle and try to find the area of a segment of that circle that has an arc of some distance, then the area of the segment is the total area of the circle multiplied by the fraction that the arc length is compared to the entire circumference.

Mathematically, [tex]A=\pi r^2 \frac{\theta}{2\pi}= \frac{1}{2}r^2\theta[/tex] where [itex]\theta[/itex] is the angle in the centre of the circle, or equivalently the length of the arc in a circle radius 1.

As for your 3-d sphere problem, it is essentially the same thing except that now you'll have the volume of a sphere multiplied by the fraction that the surface area of the "base" of the cone is compared to the entire surface area of the sphere.

The tricky part would be to find the surface area occupied on the sphere. I couldn't see a way to do it without the use of calculus but we might be able to work with the angle subtended just like in the 2-d problem....

actually while writing this I've realized the surface area subtended by the angle at the centre isn't a linear relationship (or any other intuitive relationship for that fact).

e.g. For an angle of [itex]\pi[/itex] radians, the area subtended is half the entire surface area of the sphere. For an angle of [itex]\pi /2[/itex] the area is not a quarter, but much less, and I'm unsure what the actual answer is. For [itex]3\pi /2[/itex] we have an area much more than 3/4 of the sphere.

If anyone else can shed a bit of light on this problem...?
LCKurtz
#5
Feb10-10, 09:58 PM
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It is a simple integral to set up in spherical coordinates. I will call the angle you denote y by [itex]\alpha[/itex] and the radius a:

[tex] V = \int_0^{2\pi}\int_0^\alpha\int_0^a \rho^2\sin{(\phi)}d\rho\ d\phi\ d\theta =

\frac{2\pi a^3}{3}(1 - cos{(\alpha)})[/tex]
Mentallic
#6
Feb11-10, 01:13 AM
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But isn't it possible without the use of calculus? FYI we are in the precalc. forum.
And it doesn't seem correct.

Let's take the radius as 1:

[itex]\alpha=\pi[/itex] then we should get half a sphere, which is [itex]2/3\pi[/itex] but your formula suggests it's [itex]4/3\pi[/itex].
LCKurtz
#7
Feb11-10, 12:14 PM
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Quote Quote by Mentallic View Post
But isn't it possible without the use of calculus? FYI we are in the precalc. forum.
Yes, I'm aware this is a pre-calc forum. Yet the OP asked to "derive or find" the formula, so I provided the answer. I realize he may or may not understand the integrals.


And it doesn't seem correct.

Let's take the radius as 1:

[itex]\alpha=\pi[/itex] then we should get half a sphere, which is [itex]2/3\pi[/itex] but your formula suggests it's [itex]4/3\pi[/itex].
The formula is correct. If you want half the sphere take [itex]\alpha = \pi/2[/itex]. Remember that [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex].
badtwistoffate
#8
Feb11-10, 01:52 PM
P: 85
Thanks for the help guys.

Actually, my math background goes to ODE - so it makes since, I just assumed a simpler method existed. I am going to look in my multivariable calculus text to revitalize my head for this approach...
badtwistoffate
#9
Feb11-10, 02:33 PM
P: 85
Quote Quote by LCKurtz View Post
It is a simple integral to set up in spherical coordinates. I will call the angle you denote y by [itex]\alpha[/itex] and the radius a:

[tex] V = \int_0^{2\pi}\int_0^\alpha\int_0^a \rho^2\sin{(\phi)}d\rho\ d\phi\ d\theta =

\frac{2\pi a^3}{3}(1 - cos{(\alpha)})[/tex]
Did you derive this triple integral?
LCKurtz
#10
Feb11-10, 03:01 PM
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If you mean did I figure out the limits and work it out, yes. It's a typical exercise like you might find in a calculus text in the section of spherical coordinates and triple integrals.
badtwistoffate
#11
Feb11-10, 03:20 PM
P: 85
Quote Quote by LCKurtz View Post
If you mean did I figure out the limits and work it out, yes. It's a typical exercise like you might find in a calculus text in the section of spherical coordinates and triple integrals.
Just to clarify:
[tex] \phi\ [/tex], [tex] \theta\ [/tex], [tex] \rho\ [/tex]
are respectively: elevation angle, azimuth angle, and radius...
LCKurtz
#12
Feb11-10, 03:33 PM
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Quote Quote by badtwistoffate View Post
Just to clarify:
[tex] \phi\[/tex], [tex]\theta\ [/tex], [tex]\rho\ [/tex]
are respectively: elevation angle, azimuth angle, and radius...
I don't use those terms, so to be clear:

[itex]\phi[/itex] is the angle measured from the positive z axis to the radius [itex]\rho[/itex], and [itex]\theta[/itex] is the polar coordinate angle between the x axis and the projection of the radius [itex]\rho[/itex] on the xy plane. Standard spherical coordinates.
badtwistoffate
#13
Feb11-10, 04:53 PM
P: 85
Quote Quote by LCKurtz View Post
I don't use those terms, so to be clear:

[itex]\phi[/itex] is the angle measured from the positive z axis to the radius [itex]\rho[/itex], and [itex]\theta[/itex] is the polar coordinate angle between the x axis and the projection of the radius [itex]\rho[/itex] on the xy plane. Standard spherical coordinates.
[tex]
V = \int_0^{2\pi}\int_0^\alpha\int_0^a \rho^2\sin{(\phi)}d\rho\ d\phi\ d\theta =

\frac{2\pi a^3}{3}(1 - cos{(\alpha)})
[/tex]

Reason why I ask was I am just evaluating your limits...

I see for outermost integral, limits are from 0 to 2[tex] \pi\ [/tex]. makes since, because were rotating around 360 degrees for the sphere.

I see for middle integral, limits are from 0 to [tex] \alpha\ [/tex]. Alpha being the angle "measured from the positive z axis to the radius". This makes since, so will be 35 degree converted into radians.

Then, for the inner integral, limits are from 0 to [tex] \rho\ [/tex]. With rho being the radius.

Thus, its the "last" integral that scopes out the 3d volume obtain from rotating 360 degrees from the angle formed by the positive z axis down to the radius.
LCKurtz
#14
Feb12-10, 12:42 PM
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Quote Quote by badtwistoffate View Post

Then, for the inner integral, limits are from 0 to [tex] \rho\ [/tex]. With rho being the radius.
[tex] \rho [/tex] goes from 0 to a with a being the radius.

Thus, its the "last" integral that scopes out the 3d volume obtain from rotating 360 degrees from the angle formed by the positive z axis down to the radius.
I think you have the idea.


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