Calculating Surface of Revolution for a Torus | Math Problem Solution

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To calculate the surface area of a torus using the surface of revolution method, the equation for a circle is adapted to represent the torus with a distance to the center of 12 and a radius of 2. The equation simplifies to \(y = 12 \pm \sqrt{4 - x^2}\), representing both the upper and lower halves of the circle. The user expresses confusion about how to proceed since y is not a function of x, prompting a discussion on whether to treat the top and bottom halves as separate surfaces. A suggested resource provides a method for calculating the volume by considering the volumes of the upper and lower semi-circles rotated about the x-axis. The discussion emphasizes the importance of correctly interpreting the equations for accurate surface area calculation.
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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,

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

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

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

Expanding,

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

Solving for y,

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

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?
 
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Yes, use just what you have.
But it is not
\[y = \pm \left( {\sqrt {4 - x^2 } - 12} \right)\]
it is
y = 12 \pm\sqrt {4 - x^2 }
One function is
y = 12+ \sqrt {4 - x^2 }
and the other is
y = 12-\sqrt {4 - x^2 }
 
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Fifty Ways To Find The Volume Of A Torus

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

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: .
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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