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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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# Homework Help: Surface of revolution

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