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Area of a sphere |
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| May25-05, 07:18 PM | #1 |
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Area of a sphere
How do you find the area of a sphere based on it's volume?
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| May25-05, 07:20 PM | #2 |
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surface area of sphere: 4pi r^2
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| May25-05, 07:22 PM | #3 |
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Well, usually you find both of the above based on the circle.
Integrating half a circle's arc length gives surface area, and integrating half a circle's area gives volume. However, the sphere's volume [itex] \frac{4}{3}\pi r^3 [/tex], and surface area [itex] 4\pi r^2 [/tex] are related as integrals/derivatives of the radius 'r'. |
| May25-05, 07:26 PM | #4 |
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Area of a sphere
Can you put that in an equation form like Area= volume x _ ?
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| May25-05, 07:29 PM | #5 |
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You mean turn the volume into the SA?
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| May25-05, 07:32 PM | #6 |
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| May25-05, 07:34 PM | #7 |
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Well I guess a really ugly way of doing it is just saying
[tex] SA = \frac{3V}{r} [/tex] Do you know calculus? |
| May25-05, 07:38 PM | #8 |
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No, i just finished algebra II. Is there another way to say it? Without having to first calculate radius?
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| May25-05, 07:39 PM | #9 |
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You have all you need to figure that out yourself...
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| May25-05, 08:10 PM | #10 |
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Serj:
Do you agree that the radius r in terms of volum V is: [tex]r=(\frac{3V}{4\pi})^{\frac{1}{3}}[/tex]?? Use this to express the surface area in terms of the volume. |
| May25-05, 08:20 PM | #11 |
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Serj,the sphere's volume ("it's (sic!) volume") is zero...So the area is simply
[tex]\mbox{Area}_{\mbox{Sphere}} =\mbox{Area}_{\mbox{Sphere}} \ + 0 [/tex] [tex] \Longrightarrow \mbox{Area}_{\mbox{Sphere}} =\mbox{Area}_{\mbox{Sphere}} \ + \ \mbox{Volume}_{\mbox{Sphere}} [/tex] Daniel. P.S.Apart from the notation,there's no joke in this post...
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| May25-05, 08:21 PM | #12 |
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Daniel is of course right.
You are to express the sphere's area in terms of the enclosed ball's volume. |
| May25-05, 08:38 PM | #13 |
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Recognitions:
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I could see that one coming a mile away. LOL !! :D
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| May25-05, 08:55 PM | #14 |
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[tex]A = (36 \pi V^2)^{1/3}[/tex]
- Warren |
| May25-05, 09:34 PM | #15 |
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Recognitions:
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to find the volume of a sphere one computes how the volume form the portion of the sphere at or below height y, changes as y changes. it turns out that the rate of change of the volume at height y, equals the area of the circular slice at height y.
from this one can write a formula for the dertivative of the rising voilume formula, and from that one can write a formula for the volume of a sphere of radius R as (4/3)pi R^3). then one writes the volume of the sphere in terms of the radius, and sees that the rate of change in that case, at the point where the radius is r, equals the area of the sphere of radius r. Since we know the volume, and that the area is the derivative of the volume wrt the radius, we get that the area is 4 pi R^2. so first you find the volume of the sphere, knowing the area of a circle. Then knowing the volume of a sphere you find the area of the sphere. I am not sure, since I have not seen his works, but i suspect this is the way Archimedes did it. |
| May26-05, 04:14 AM | #16 |
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Recognitions:
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I was waiting for the same thing to happen too. |
| May26-05, 05:53 PM | #17 |
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