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Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) §8.3 Area of a Surface of Revolution, #11 · Calculus · Applications of integrals

A cone with base radius … and slant height … is laid…

A cone with base radius $R$ and slant height $s$ is laid out flat. Explain why the angle (in radians) is $θ = 2 π R / s$ . Then the surface area is a fraction of a circle:
$area = (\frac{θ}{2 π}) π s^{2} = (\frac{R}{s}) π s^{2} = π R s .$

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Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §8.3 Area of a Surface of Revolution, #11

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