Surface Area of Sphere

Calculus 2 · Axiom Academy

EXAMPLE Surface Area of a Sphere Derive by revolving a semicircle about the x -axis and integrating the surface of revolution. Rotate the upper semicircle of radius r about the x -axis to sweep out a sphere, then use the surface-of-revolution formula to prove its surface area is . The upper semicircle (solid arc) sweeps out the full sphere (dashed) when revolved a full turn about the x -axis. You derived the surface area of a sphere from scratch with the surface-of-revolution integral. Here's what carried the proof: Surface of revolution: revolving a curve about an axis builds a surface whose area is . Implicit differentiation: from x^2+y^2=r^2 we got dy/dx = -x/y without solving for y first. The key cancellation: y^2 + x^2 = r^2 collapsed the radical to r/y , and the y then cancelled against the out front. A constant integrand: everything reduced to . The result: — exactly four times the area of the sphere's great circle. The same machinery handles any solid of revolution — swap in a different generating curve and the integral does the rest.

This is the written version of the interactive lesson above. See the full Calculus 2 course.