Question: When the circle x 2 + (y - a) 2 = r 2 on the interval [-r, r] is revolved about the x-axis, the result

When the circle x² + (y - a)² = r² on the interval [-r, r] is revolved about the x-axis, the result is the surface of a torus, where 0 < r < a. Show that the surface area of the torus is S = 4π²ar.

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