Surface Area of Revolution

Calculus 2 · Axiom Academy

LESSON Surface Area of Revolution Spin a curve around an axis and you sweep out a surface — each arc-length sliver traces a frustum band whose area builds the integral. Consider a smooth curve y = f(x) on the interval [ a , b ]. When we rotate this curve around the x-axis, it sweeps out a surface of revolution — the profile is the outline, the rotation fills in the solid skin. The profile curve, ready to spin around the x-axis 2. Approximating with Frustums Divide [ a , b ] into n subintervals. When we rotate each small piece of the curve, we get approximately a frustum — a cone with the tip sliced off. Read its two radii straight off the curve: the band's bottom and top edges sit at the curve's height at each cut. For our curve, r_1 = f(x_i) , r_2 = f(x_ i+1 ) , and the slant height ℓ is approximately the arc length of that curve segment. The slant height of each frustum is exactly the arc length of the curve segment. Zoom in on one piece from x to x + Δx: the slanted chord is the hypotenuse of a tiny right triangle with legs Δx and Δy. This makes sense: the steeper the curve, the longer the slant height, and the more surface area each band contributes. Now multiply each band's circumference 2πf(x) by its slant ds and add them up. Letting the number of frustums go to infinity turns the sum into a definite integral. where f(x) is the curve being rotated and f′(x) is its derivative. Same slant factor — surface area just weights every sliver by its circumference 2πf(x).

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