The Timpson Lyceum · Calculus 1 (2024)

Arc Length & Surface Area - Lecture 18 (Calculus 1)

Overview

This lecture extends the "integral as an infinite sum of arbitrarily good approximations" idea from volumes to two new geometric quantities: the arc length of a curve $y = f(x)$ and the surface area of the solid generated when that curve is revolved about an axis. The arc-length formula is built by approximating the curve with many tiny straight segments and applying the Pythagorean theorem to each. The surface-area formula then reuses that same arc-length element $ds$, multiplying it by the circumference $2\pi r$ swept out as the curve rotates. Together these complete the geometric applications of the integral begun with volumes.

Key Concepts

• The arc-length element $ds$. Approximate a curve by small chords. Each chord over a subinterval has horizontal change $\Delta x_k$ and vertical change $\Delta y_k$, so by the Pythagorean theorem its length is $\Delta s_k = \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2}$. Summing these and taking the limit gives $ds = \sqrt{(dx)^2 + (dy)^2}.$

• Arc length, integrating in $x$. Factor $dx$ out of $ds$ (multiply inside the radical by $dx^2/dx^2$) to get the form used when the curve is a function of $x$ on $[a,b]$: $L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.$

• Arc length, integrating in $y$. When the curve is better described as $x = g(y)$ on $[c,d]$ — for example where the slope $dy/dx$ is undefined (a vertical tangent) — factor out $dy$ instead: $L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy.$ The two formulas describe the same length; choose the variable that keeps the derivative defined and the algebra clean.

• Perfect-square radicands. Textbook arc-length integrands are constructed so that $1 + [f'(x)]^2$ simplifies to a perfect square, letting the square root cancel. If you cannot integrate $\sqrt{1+[f'(x)]^2}$ directly, suspect (and look for) a perfect square — often found by completing the square.

• Surface of revolution — the frustum band. Revolving the curve about an axis and slicing gives thin band-like pieces shaped like the lateral surface of a frustum. The lateral surface of a cylinder of radius $r$ is (circumference)$\times$(height) $= 2\pi r\,h$. For the slanted band, the slant length is exactly the arc element $ds$, so each band contributes $2\pi r\,ds$.

• Surface-area formula. Summing the bands gives $S = \int 2\pi r\,ds.$ The radius $r$ is set by the axis of revolution:

  • Revolving $y = f(x)$ about the $x$-axis: $r = y$, so $S = \displaystyle\int_a^b 2\pi y\,\sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}\,dx$.
  • Revolving about the $y$-axis: $r = x$, so $S = \displaystyle\int 2\pi x\,ds$.

• Same $ds$ throughout. The surface-area integrand differs from arc length only by the factor $2\pi r$; the $ds = \sqrt{1+(y')^2}\,dx$ (or $\sqrt{1+(x')^2}\,dy$) factor is identical.

Worked Example

Find the surface area generated by revolving $y = 2\sqrt{x}$ on $1 \le x \le 2$ about the $x$-axis.

Since we revolve about the $x$-axis, $r = y$ and we integrate in $x$ with $S = \int_a^b 2\pi y\,\sqrt{1 + (y')^2}\,dx$. Here $y' = x^{-1/2}$, so $(y')^2 = \tfrac{1}{x}$ and $1+(y')^2 = \tfrac{x+1}{x}$.

Summary

• The arc-length element comes from the Pythagorean theorem on tiny chords: $ds = \sqrt{(dx)^2+(dy)^2}$.
• Arc length of $y=f(x)$ on $[a,b]$ is $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$; switch to $L = \int_c^d \sqrt{1 + (dx/dy)^2}\,dy$ when integrating in $y$ is easier or the $x$-slope is undefined.
• These integrands are engineered to be perfect squares so the radical cancels — complete the square if it isn't obvious.
• Revolving a curve produces frustum-like bands; each contributes $2\pi r\,ds$, giving $S = \int 2\pi r\,ds$.
• The radius $r$ is dictated by the axis: $r = y$ for revolution about the $x$-axis, $r = x$ for revolution about the $y$-axis.
• Surface area reuses the exact same $ds$ as arc length, just weighted by the circumference factor $2\pi r$.