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> Setting R=r should give you the area for two spheres, thus:
'fraid not.
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Kevin Wampler wrote:
> I'm pretty sure that these are correct.
Well, except for a really whopping error I made they were.
I redid my calculations and now I think I have the correct formulas:
V = 4*pi*(1/6*r1^2*sqrt(r2^2-r1^2)+1/4*pi*r1*r2^2
-1/2*r1*r2^2*asin(-r1/r2)+1/3*r2^2*sqrt(r2^2-r1^2))
A = 4*pi*(r1*r2*acos(-r1/r2)+r2*sqrt(r2^2-r1^2))
Now, if I didn't make a typo, those should work much better.
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Wow, how'd you get these? (No, don't answer, my poor little brain can't
handle it :-) )
Well they certainly do seem to work although they took forever to store
in my calc...
--
Homepage: http://www.faricy.net/~davidf/
___ ______________________________
| \ |_ <dav### [at] faricy net>
|_/avid |ontaine <ICQ 55354965>
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David Fontaine wrote:
> Wow, how'd you get these? (No, don't answer, my poor little brain can't
> handle it :-) )
Actually it wasn't too bad, even if the answers look complex. And now
that I've had a good nights sleep I'm ready to type it all up again.
First of all, volume:
The formula for a circle of radius r2, centered at the origin is f(x) =
sqrt(r2^2-x^2)
If you imagine breaking up this circle into rectangles (like a sum
approximation of its area), and then revolving it about the y-axis, each
of these rectangles sweeps out a cylindrical shell. As the number of
these shells approaches infinity, the volume of each of them becomes
2*pi*x*f(x)*dx
This circle, however, when it is swept out to create the tours, is not
centered at the origin, it is moved r1 units along the x-axis, to
compensate for this, you will need to perform an integral from -r1 to r2.
This also means that you need to change the x in the formula for the
infinitesimal volume of a shell to reflect this by replacing the x with
(x+r1), notice, however, that f(x) remains unchanged. Also notice that
the volume of the tours is actually twice what would normally be given by
the integral, because f(x) gives the length of a vertical line segment
connecting 0 and the circle (not the two sides of the circle).
The integral to solve for the volume of the tours then is given by:
/r2
|
4*pi* | (x+r1)*sqrt(r2^2-x^2) dx
|
/-r1
/r2 /r2
| |
= 4*pi*( | x*sqrt(r2^2-x^2) dx + | r1*sqrt(r2^2-x^2) dx )
| |
/-r1 /-r1
The first integral is fairly easy to solve with u-substitution, and
evaluates to
-1/3*(r2^2-x^2)^(3/2)
The second integral is a bit more tricky, but fortunately there is a
formula for it. It evaluates to:
r1/2*(x*sqrt(r2^2-x^2)+r2^2*asin(x/r2))
adding these together, multiplying by the 4*pi and subtracting the value
at x=-r1 from the value at x=r2 gives the answer.
Now for surface area:
The area of the infinitesimal side of the truncated cone formed by
sweeping f(x) around the y axis is 2*pi*f(x)*sqrt(1+f'(x)^2)
Set up the integral in pretty much the same way as was done for finding
the volume, giving:
/r2
|
4*pi* | (x+r1)*sqrt(1+x^2/(r2^2-x^2)) dx
|
/-r1
/r2
|
= 4*pi* | (x+r1)*sqrt(r2^2/(r2^2-x^2)) dx
|
/-r1
/r2
|
= 4*pi* | (x+r1)*r2/sqrt(r2^2-x^2)) dx
|
/-r1
/r2 /r2
| |
= 4*pi*( | x*r2/sqrt(r2^2-x^2) dx + | r1*r2/sqrt(r2^2-x^2) dx )
| |
/-r1 /-r1
Once again, the first integral is easy to solve with u-substitution, and
evaluates to:
-r2*sqrt(r2^2-x^2)
The other side is again, trickier, but in this case has a simple solution:
-r1*r2*acos(x/r2)
If you wanted you could replace the -acos(x/r2) with asin(x/r2) and it
would still be correct, I did it this way so the signs would match up.
Evaluate them at x=r2 and x=-r1, subtract, multiply by 4*pi and again you
should have the answer.
I probably made that a bit more confusing than it really is, but hopefully
it will help you to solve similar problems in the future.
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Sigmund Kyrre Aas wrote:
> Darn. Both are wrong. Setting R=r should be equal to a sphere, but it's
> not:
> A!=-4*pi*r^2( (sin(1)-pi )
> V!=2*pi^2*r^3
Setting r=R should actually dive you the volume and area of a horn thorus:
A=4*pi^2*r^2
V=2*pi^2*r^3
So you were actually closer than you thought. Setting R=0 should give you
a sphere.
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Me is 15. Me in Algebra 2 Trig.
--
Homepage: http://www.faricy.net/~davidf/
___ ______________________________
| \ |_ <dav### [at] faricynet>
|_/avid |ontaine <ICQ 55354965>
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> So you were actually closer than you thought. Setting R=0 should give you
> a sphere.
Not with those equations!
--
Homepage: http://www.faricy.net/~davidf/
___ ______________________________
| \ |_ <dav### [at] faricynet>
|_/avid |ontaine <ICQ 55354965>
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David Fontaine wrote:
> Me is 15. Me in Algebra 2 Trig.
Bah, if your school's not challenging you it's the least I can do to
give you something a bit more difficult to learn. If you actually are
interested in learning some calculus (which you very understandably
might not be), try http://mathworld.wolfram.com/ Take a look under
headings like Calculus, Derivative, Integral, etc. I suggest you print
out the information you want to learn and study it when your not paying
attention in your school's math class. If you have any questions about
anything feel free to email me.
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In article <388### [at] tapestrytucsonazus>, Kevin Wampler
<kev### [at] tapestrytucsonazus> wrote:
> If you actually are
> interested in learning some calculus (which you very understandably
> might not be), try http://mathworld.wolfram.com/
Well, I don't know about David, but I have been looking for something
like this for a long time now. :-)
--
Chris Huff
e-mail: chr### [at] yahoocom
Web page: http://chrishuff.dhs.org/
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ahhh. It starts coming back.
I used x=r1=R as the area center, but it's not the same for a sliced
sircle, of course. Finding the new area center and using that expression
in the revolution integral would have left me with just as much work as
you had there, perhaps even more. (except that I'm cheating; ti89)
sig.
Kevin Wampler wrote:
>
> David Fontaine wrote:
>
> > Wow, how'd you get these? (No, don't answer, my poor little brain can't
> > handle it :-) )
>
> Actually it wasn't too bad, even if the answers look complex. And now
> that I've had a good nights sleep I'm ready to type it all up again.
>
> First of all, volume:
> The formula for a circle of radius r2, centered at the origin is f(x) =
> sqrt(r2^2-x^2)
>
> If you imagine breaking up this circle into rectangles (like a sum
> approximation of its area), and then revolving it about the y-axis, each
> of these rectangles sweeps out a cylindrical shell. As the number of
> these shells approaches infinity, the volume of each of them becomes
> 2*pi*x*f(x)*dx
>
> This circle, however, when it is swept out to create the tours, is not
> centered at the origin, it is moved r1 units along the x-axis, to
> compensate for this, you will need to perform an integral from -r1 to r2.
> This also means that you need to change the x in the formula for the
> infinitesimal volume of a shell to reflect this by replacing the x with
> (x+r1), notice, however, that f(x) remains unchanged. Also notice that
> the volume of the tours is actually twice what would normally be given by
> the integral, because f(x) gives the length of a vertical line segment
> connecting 0 and the circle (not the two sides of the circle).
>
> The integral to solve for the volume of the tours then is given by:
>
> /r2
> |
> 4*pi* | (x+r1)*sqrt(r2^2-x^2) dx
> |
> /-r1
>
> /r2 /r2
> | |
> = 4*pi*( | x*sqrt(r2^2-x^2) dx + | r1*sqrt(r2^2-x^2) dx )
> | |
> /-r1 /-r1
>
> The first integral is fairly easy to solve with u-substitution, and
> evaluates to
>
> -1/3*(r2^2-x^2)^(3/2)
>
> The second integral is a bit more tricky, but fortunately there is a
> formula for it. It evaluates to:
>
> r1/2*(x*sqrt(r2^2-x^2)+r2^2*asin(x/r2))
>
> adding these together, multiplying by the 4*pi and subtracting the value
> at x=-r1 from the value at x=r2 gives the answer.
>
> Now for surface area:
> The area of the infinitesimal side of the truncated cone formed by
> sweeping f(x) around the y axis is 2*pi*f(x)*sqrt(1+f'(x)^2)
>
> Set up the integral in pretty much the same way as was done for finding
> the volume, giving:
>
> /r2
> |
> 4*pi* | (x+r1)*sqrt(1+x^2/(r2^2-x^2)) dx
> |
> /-r1
>
> /r2
> |
> = 4*pi* | (x+r1)*sqrt(r2^2/(r2^2-x^2)) dx
> |
> /-r1
>
> /r2
> |
> = 4*pi* | (x+r1)*r2/sqrt(r2^2-x^2)) dx
> |
> /-r1
>
> /r2 /r2
> | |
> = 4*pi*( | x*r2/sqrt(r2^2-x^2) dx + | r1*r2/sqrt(r2^2-x^2) dx )
> | |
> /-r1 /-r1
>
> Once again, the first integral is easy to solve with u-substitution, and
> evaluates to:
>
> -r2*sqrt(r2^2-x^2)
>
> The other side is again, trickier, but in this case has a simple solution:
>
> -r1*r2*acos(x/r2)
>
> If you wanted you could replace the -acos(x/r2) with asin(x/r2) and it
> would still be correct, I did it this way so the signs would match up.
> Evaluate them at x=r2 and x=-r1, subtract, multiply by 4*pi and again you
> should have the answer.
>
> I probably made that a bit more confusing than it really is, but hopefully
> it will help you to solve similar problems in the future.
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