Calculating Surface of Revolution for a Torus | Math Problem Solution

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SUMMARY

The discussion focuses on calculating the surface area of a torus using the method of surfaces of revolution. The parameters provided are a distance to the center of the torus of 12 and a radius of 2. The equation derived is \(x^2 + (y - 12)^2 = 2^2\), leading to two functions for y: \(y = 12 + \sqrt{4 - x^2}\) and \(y = 12 - \sqrt{4 - x^2}\). The solution involves calculating the volume of the torus by subtracting the volume of the lower semi-circle from that of the upper semi-circle, as detailed in the referenced document.

PREREQUISITES
  • Understanding of surfaces of revolution in calculus
  • Familiarity with the equation of a circle
  • Basic knowledge of volume calculations for solids of revolution
  • Proficiency in using mathematical software like Maple for plotting
NEXT STEPS
  • Study the method of calculating volumes using the disk and washer methods
  • Learn about the application of the shell method for solids of revolution
  • Explore the concept of parametric equations in surface area calculations
  • Review the document "Fifty Ways to Find the Volume of a Torus" for advanced techniques
USEFUL FOR

Students and educators in mathematics, particularly those focused on calculus and geometry, as well as professionals involved in mathematical modeling and engineering applications related to surfaces of revolution.

opticaltempest
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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

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?
 
Last edited by a moderator:
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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]
 
Last edited by a moderator:
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

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 , 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 :smile: .
 
Last edited by a moderator:

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