Surface of revolution

  • #1
Hello,

I am trying to find the surface area of a torus. I know there are equations
for finding this but in this problem I must use a surface of revolution.

Given in the problem is the distance to the center of the torus is 12 and
the radius of the torus is 2.

By using the equation for a circle I come up with the following
equation for my surface of revolution,

[tex] \[(x - h)^2 + (y - k)^2 = r^2\] [/tex]

I replace x with 0 since I suppose it doesn't matter where on the
x axis I revolve this surface. The equation becomes,

[tex] \[x^2 + (y - 12)^2 = 2^2\] [/tex]

Expanding,

[tex] \[x^2 + y^2 - 24y + 140\] [/tex]

Solving for y,

[tex] \[y = \pm \left( {\sqrt {4 - x^2 } - 12} \right)\] [/tex]

A quick plot in Maple to show my surface of revolution,

http://img256.imageshack.us/img256/5137/surfacecopy0df.jpg [Broken]

I am confused on where I go next because y is not a function of x,
nor is x a function of y. Do I need have two different surfaces of
revolution? The top half of the circle and the bottom half of the
circle?
 
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Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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964
Yes, use just what you have.
But it is not
[tex] \[y = \pm \left( {\sqrt {4 - x^2 } - 12} \right)\] [/tex]
it is
[tex] y = 12 \pm\sqrt {4 - x^2 } [/tex]
One function is
[tex] y = 12+ \sqrt {4 - x^2 } [/tex]
and the other is
[tex] y = 12-\sqrt {4 - x^2 }[/tex]
 
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  • #3
benorin
Homework Helper
Insights Author
Gold Member
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108
Fifty Ways To Find The Volume Of A Torus

opticaltempest said:
...[tex] \[x^2 + (y - 12)^2 = 2^2\] [/tex]...
Solving for y,
[tex] \[y = \pm \left( {\sqrt {4 - x^2 } - 12} \right)\] [/tex]

A quick plot in Maple to show my surface of revolution,

http://img256.imageshack.us/img256/5137/surfacecopy0df.jpg [Broken]

I am confused on where I go next because y is not a function of x,
nor is x a function of y. Do I need have two different surfaces of
revolution? The top half of the circle and the bottom half of the
circle?

There is a perfect answer to your exact question: check this out,http://www.math.umn.edu/~drake/pdfs/fifty-ways.pdf [Broken], it starts on the bottom of pg 2, under the heading (3) Slip out the back, Jack.

Basically:

Vol(torus) = Vol(upper semi-circle rotated about the x-axis) - Vol(lower semi-circle rotated about the x-axis).

Have fun :rofl: .
 
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