Calculating Surface of Revolution for a Torus | Math Problem Solution

  • Thread starter Thread starter opticaltempest
  • Start date Start date
  • Tags Tags
    Revolution Surface
opticaltempest
Messages
135
Reaction score
0
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?
 
Last edited by a moderator:
Physics news on Phys.org
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 }
 
Last edited by a moderator:
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: .
 
Last edited by a moderator:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top