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#1
Feb910, 10:29 PM

#2
Feb1010, 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? 


#3
Feb1010, 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 2D space but cannot figure out a formula for volume for 3D space.



#4
Feb1010, 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 2d 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 3d 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 2d 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...? 


#5
Feb1010, 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] 


#6
Feb1110, 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]. 


#7
Feb1110, 12:14 PM

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#8
Feb1110, 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... 


#9
Feb1110, 02:33 PM

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#10
Feb1110, 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.



#11
Feb1110, 03:20 PM

P: 85

[tex] \phi\ [/tex], [tex] \theta\ [/tex], [tex] \rho\ [/tex] are respectively: elevation angle, azimuth angle, and radius... 


#12
Feb1110, 03:33 PM

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[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. 


#13
Feb1110, 04:53 PM

P: 85

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. 


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