Calculus II Area of Torus

Question

The circle of radius a centered at (0,b) such that a < b can be parametrized by x = acost and y = b + asint for 0 <= t <= 2pi.

This circle is revolved around the x-axis to generate a torus (donut-shaped surface). Find the surface area of this torus.

Answer 1

Here is the general Calc II method for solving a surface of revolution with a parametrization.

For an x-axis rotation, the formula for surface area is given by the integral S = ∫_t1^t2 y(t) √(x'(t)^2+y'(t)^2)dt where t1 and t2 are the start and end times for the parametrization, x(t) and y(t) are the parametrized equations for the circle, and y(t) is never negative. Here are the specific values for the problem.

t1 = 0, t2 = 2pi

x(t) = acos(t)

y(t) = asin(t) + b

The derivatives of the functions are

x'(t) = -asin(t)

y'(t) = acos(t)

So the integral becomes

S = ∫₀^2pi (asin(t) + b) √(a²sin²(t)+a²cos²(t))dt

= ∫₀^2pi (asin(t) + b) √(a²(sin²(t)+cos²(t))dt

= ∫₀^2pi(asin(t) + b) a√(1)

= a ∫₀^2pi(asin(t) + b)dt

= a (-acos(t) + bt)|₀^2pi

= a (-acos(2pi) + 2pi b + acos(0) - 0 b)

= a(-a + 2pi b + a - 0)

= 2pi ab

For more information on why the surface integral is that particular formula, feel free to message me!