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Revolutions of integrals - torus

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The circle x=acost, y=asint, 0≦t≦2pi is revolved about the line x=b, 0<a<b, thus generating a torus (doughnut). Find its surface area.

Area if the torus:_____________.

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Solution Summary

The area of the torus is found by revolutions of integrals.

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A surface of revolution is formed by the rotation of a planar curve C about an axis in the plane of the curve and not cutting the curve. The Pappus--Guldinus theorem says that:
• The area ...

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