find surface area of a donut

I have a problem where i dont even know where to begin. Usually, when i post problems here i have something to show but this time i dont. Im sorry for that.


here is the problem:

FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the
y-axis.


would i do something like the Area of a surface of revolution theorem?:

$\displaystyle S = 2 \pi \int_a^b f(t) \sqrt {(dx/dt)^2 + (dy/dt)^2} dt$

if so i am not even sure what to do to get it into the form to put it into that formula?

Any help is appreciated since i am lost!!

Thank you

Quote:

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Originally Posted by icelated

FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the y-axis.

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Use the First Guldin Theorem: the surface area $\displaystyle S $ of a surface of revolution generated by the revolution of a curve about an external axis is equal to the product of the arc length $\displaystyle s$ of the generating curve and the distance $\displaystyle d$ traveled by the curve's geometric centroid $\displaystyle \bar{x}$ .

Quote:

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Originally Posted by FernandoRevilla

Use the First Guldin Theorem:

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Im not suppose to use pappus's theorem. When i googled First Guldin Theorem everything came back to pappus. What is First Guldin Theorem? =(