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12 years ago
Hi- I need help with a calculus homework problem.  It's an even-numbered problem (not in back of book or solution manual) and is also about a torus, which I have never heard or seen before.  There isn't any other question in the section that resembles this one at all.  

A torus is formed by revolving the graph of (x-1)^2 plus y-squared equals 1.  Find the surface area of the torus.  

How would I do this?  The example in the book talks about the volume of the torus, but not the surface area.  

Thanks for any help!
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12 years ago
A Torus has the form of a donut or innertube. Its cross section is a circle. The circle they gave you is a translation of the unit circle x² + y² = 1 with center (0, 0). This one has the center at (1, 0). I assume they want you to generate a torus by revolving the circle on the x-y plane around the origin. The center of the generating circle would go around in a unit circle itself but totally contained in the z-plane.

The surface element will be the circumference of the generating circle times the surface of a differential radial element (a slice of the donut). By integrating around the unit circle path on the z-plane you will get the total surface.

It seems to me that using a polar coordinate representation of the unit circle on the z plane (i.e. r = 1, &theta &isin [0, 2&pi) you can construct the differential element of the surface in simpler manner.

All the calculus aside, if you look at the problem intuitively, the circumference of the generating circle is 2&pi (it as radius 1). So you are moving a line of length 2&pi around a unit circle in the z plane, which is a length of 2&pi, so the surface should be the product of these two, i.e. 4&pi².
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