Arc Length & Surface Area

Arc Length & Surface Area

Part 1 of 7 — Arc Length of Cartesian Curves

The length of a smooth curve $y = f(x)$ from $x = a$ to $x = b$ is:

$\boxed{L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx}$

Derivation Sketch

Approximate the curve with short line segments of length $\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}$. Factor out $\Delta x$:

$\Delta s \approx \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}\,\Delta x \xrightarrow{\Delta x \to 0} \sqrt{1 + (f'(x))^2}\,dx$

Key Fact: The integrand $\sqrt{1 + (f')^2}$ rarely simplifies to an elementary antiderivative. Many arc length problems require a calculator.

Classic Examples

Example 1. $y = x^{3/2}$ from $x = 0$ to $x = 4$.

$y' = \frac{3}{2}x^{1/2}$, so $(y')^2 = \frac{9}{4}x$.

$L = \int_0^4 \sqrt{1 + \tfrac{9}{4}x}\,dx$

Let $u = 1 + \frac{9}{4}x$, $du = \frac{9}{4}dx$:

$L = \frac{4}{9}\cdot\frac{2}{3}\left[u^{3/2}\right]_1^{10} = \frac{8}{27}\left(10\sqrt{10} - 1\right)$

Example 2. $y = \frac{x^2}{4} - \frac{\ln x}{2}$ from $x = 1$ to $x = e$.

$y' = \frac{x}{2} - \frac{1}{2x}$. Then $1 + (y')^2 = 1 + \frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2} = \left(\frac{x}{2} + \frac{1}{2x}\right)^2$.

$L = \int_1^e \left(\frac{x}{2} + \frac{1}{2x}\right)dx = \left[\frac{x^2}{4} + \frac{\ln x}{2}\right]_1^e = \frac{e^2 - 1}{4} + \frac{1}{2}$

AP Tip: Example 2 is the "perfect square" type — designed so the square root simplifies. AP problems often feature this pattern.

Practice Problems

Concept Checks

Computation

Summary

• Arc length: $L = \int_a^b \sqrt{1 + (f'(x))^2}\,dx$
• Most arc length integrals need a calculator
• "Perfect square" problems are designed for hand computation
• $ds = \sqrt{1 + (y')^2}\,dx = \sqrt{(dx)^2 + (dy)^2}$

Next: Part 2 — Arc length in parametric form.

Arc Length & Surface Area — Parametric & Polar Arc Length

Part 2 of 7 — Arc Length in Parametric and Polar Forms

Parametric Arc Length

For $x = f(t)$, $y = g(t)$, $a \le t \le b$:

$\boxed{L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt}$

Polar Arc Length

For $r = f(\theta)$, $\alpha \le \theta \le \beta$:

$\boxed{L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta}$

Key Fact: The polar formula comes from substituting $x = r\cos\theta$, $y = r\sin\theta$ into the parametric formula.

Examples

Parametric: $x = \cos t$, $y = \sin t$, $0 \le t \le 2\pi$.

$dx/dt = -\sin t$, $dy/dt = \cos t$.

$L = \int_0^{2\pi}\sqrt{\sin^2 t + \cos^2 t}\,dt = \int_0^{2\pi} 1\,dt = 2\pi$ ✓

Polar: $r = 1$ (unit circle), $0 \le \theta \le 2\pi$.

$dr/d\theta = 0$.

$L = \int_0^{2\pi}\sqrt{1 + 0}\,d\theta = 2\pi$ ✓

Polar arc length of $r = e^\theta$ from $\theta = 0$ to $\theta = \ln 2$:

$dr/d\theta = e^\theta$. $L = \int_0^{\ln 2}\sqrt{e^{2\theta} + e^{2\theta}}\,d\theta = \int_0^{\ln 2}e^\theta\sqrt{2}\,d\theta = \sqrt{2}[e^\theta]_0^{\ln 2} = \sqrt{2}(2-1) = \sqrt{2}$

Practice Problems

Form Selection

Computation

Summary

• Parametric: $L = \int_a^b \sqrt{(x')^2 + (y')^2}\,dt$
• Polar: $L = \int_\alpha^\beta \sqrt{r^2 + (r')^2}\,d\theta$
• All arc length formulas come from $ds = \sqrt{dx^2 + dy^2}$

Next: Part 3 — Surface area of revolution.

Arc Length & Surface Area — Surface Area of Revolution

Part 3 of 7 — Surfaces of Revolution

When a curve is rotated about an axis, it sweeps out a surface. The surface area is:

About the $x$-axis ($y \ge 0$):

$\boxed{S = 2\pi \int_a^b y\,ds = 2\pi \int_a^b f(x)\sqrt{1 + (f'(x))^2}\,dx}$

About the $y$-axis ($x \ge 0$):

$\boxed{S = 2\pi \int_a^b x\,ds = 2\pi \int_a^b x\sqrt{1 + (f'(x))^2}\,dx}$

Key Fact: The formula is $S = 2\pi\int(\text{radius})(ds)$. The "radius" is the distance from the curve to the axis of rotation.

Example 1 — Sphere Surface Area

Rotate $y = \sqrt{r^2 - x^2}$ (semicircle) about the $x$-axis, $-r \le x \le r$.

$y' = \frac{-x}{\sqrt{r^2-x^2}}$, $1 + (y')^2 = \frac{r^2}{r^2 - x^2}$

$S = 2\pi\int_{-r}^{r} \sqrt{r^2 - x^2}\cdot\frac{r}{\sqrt{r^2-x^2}}\,dx = 2\pi r\int_{-r}^{r}dx = 2\pi r(2r) = 4\pi r^2$

This confirms the known sphere surface area formula. ✓

Example 2 — Cone Lateral Surface

Rotate $y = 2x$ from $x = 0$ to $x = 3$ about the $x$-axis.

$y' = 2$, $ds = \sqrt{1+4}\,dx = \sqrt{5}\,dx$

$S = 2\pi\int_0^3 2x\cdot\sqrt{5}\,dx = 4\pi\sqrt{5}\cdot\frac{9}{2} = 18\pi\sqrt{5}$

Practice Problems

Concept Checks

Verification

Summary

• Surface area of revolution = $2\pi\int(\text{radius})(ds)$
• About $x$-axis: radius $= |y|$
• About $y$-axis: radius $= |x|$
• Verify with known shapes: sphere ($4\pi r^2$), cylinder ($2\pi rh$), cone ($\pi r\ell$)

Next: Part 4 — Parametric and polar surface area formulas.

Arc Length & Surface Area — Parametric & Polar Surface Area

Part 4 of 7 — Surface Area in Parametric and Polar Forms

Parametric Surface Area (about the $x$-axis)

$\boxed{S = 2\pi\int_a^b y(t)\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt}$

Polar Surface Area (about the polar axis / $x$-axis)

Since $y = r\sin\theta$ and $ds = \sqrt{r^2 + (r')^2}\,d\theta$:

$\boxed{S = 2\pi\int_\alpha^\beta r\sin\theta\,\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta}$

Example — Parametric

Rotate $x = \cos t$, $y = \sin t$ ($0 \le t \le \pi$) about the $x$-axis.

This is the upper semicircle of radius 1 — should give $4\pi$ (sphere).

$ds = \sqrt{\sin^2 t + \cos^2 t}\,dt = dt$

$S = 2\pi\int_0^\pi \sin t\,dt = 2\pi[-\cos t]_0^\pi = 2\pi(1+1) = 4\pi$ ✓

Example — Polar

Rotate $r = 2\cos\theta$ ($0 \le \theta \le \pi/2$) about the polar axis.

$r' = -2\sin\theta$. $r^2 + (r')^2 = 4\cos^2\theta + 4\sin^2\theta = 4$.

$y = r\sin\theta = 2\cos\theta\sin\theta = \sin 2\theta$.

$S = 2\pi\int_0^{\pi/2}\sin(2\theta)\cdot 2\,d\theta = 4\pi\int_0^{\pi/2}\sin 2\theta\,d\theta = 4\pi\left[-\tfrac{1}{2}\cos 2\theta\right]_0^{\pi/2} = 4\pi$

Practice Problems

Concept Checks

Computation

Summary

• Parametric: $S = 2\pi\int y(t)\,ds$ (about $x$-axis) or $2\pi\int x(t)\,ds$ (about $y$-axis)
• Polar: $S = 2\pi\int r\sin\theta\,ds$ (about polar axis) or $2\pi\int r\cos\theta\,ds$ (about $\theta = \pi/2$)
• All formulas follow the pattern: $S = 2\pi\int(\text{radius})\,ds$

Next: Part 5 — Comparison of arc length methods and exam strategies.

Arc Length & Surface Area — Exam Strategies

Part 5 of 7 — Choosing the Right Formula & AP Tips

Decision Tree

Common AP Patterns

1. "Set up but do not evaluate" — Write the complete integral with limits and integrand
2. Calculator-required — Write the integral, then give decimal to 3 places
3. Perfect square — $(y')^2$ is chosen so $1 + (y')^2$ is a perfect square
4. Parametric motion — Arc length = total distance; same integral

Scoring: Setup points and computation points are awarded separately. A correct integral with a computation error still earns most credit.

Tricky Cases

When $x = g(y)$: Integrate with respect to $y$.

$y = \ln x \iff x = e^y$. Arc length from $y = 0$ to $y = 1$:

$L = \int_0^1 \sqrt{1 + e^{2y}}\,dy$

This form may be easier than the $dx$ version.

Piecewise curves: Split into smooth segments and add lengths.

Curves traversed multiple times: A parametrization might trace a curve more than once. Check before integrating.

For example, $x = \cos(2t)$, $y = \sin(2t)$ from $0$ to $2\pi$ traces the unit circle twice: total $= 4\pi$, but the arc length of the circle itself is $2\pi$.

Practice Problems

Concept Checks

Computation

Summary

• Choose $ds$ based on how the curve is given (Cartesian, parametric, polar)
• Perfect-square problems are designed for hand computation
• Multiple traversals multiply the arc length
• On the AP exam: show setup first, then evaluate

Next: Part 6 — Problem-Solving Workshop.

Arc Length & Surface Area — Workshop

Part 6 of 7 — Problem-Solving Workshop

Mixed problems covering all forms of arc length and surface area.

Workshop Overview

Problem 1 — Cartesian

Find the arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$ from $x = 1$ to $x = 3$.

$y' = x^2 - \frac{1}{4x^2}$

$(y')^2 = x^4 - \frac{1}{2} + \frac{1}{16x^4}$

$1 + (y')^2 = x^4 + \frac{1}{2} + \frac{1}{16x^4} = \left(x^2 + \frac{1}{4x^2}\right)^2$

$L = \int_1^3 \left(x^2 + \frac{1}{4x^2}\right)dx = \left[\frac{x^3}{3} - \frac{1}{4x}\right]_1^3 = \left(9 - \frac{1}{12}\right) - \left(\frac{1}{3} - \frac{1}{4}\right) = \frac{108 - 1}{12} - \frac{1}{12} = \frac{106}{12} = \frac{53}{6}$

Workshop Questions

Form Selection Practice

Workshop Computation

Workshop Summary

• Recognize perfect-square arc length problems: $y = ax^n + bx^{-m}$
• Choose polar form for polar curves, parametric for parametric curves
• Surface area: always $2\pi\int(\text{radius})\,ds$

Next: Part 7 — Comprehensive Review.

Arc Length & Surface Area — Comprehensive Review

Part 7 of 7 — Full Topic Review

Master Formula Sheet

Key Fact: All formulas derive from $ds = \sqrt{dx^2 + dy^2}$ and $S = 2\pi\int(\text{radius})\,ds$.

Exam Checklist

✅ Arc Length:

• Identify the curve form (Cartesian/parametric/polar)
• Compute $ds$ correctly
• Check for perfect squares
• If no closed form: calculator + show integral setup

✅ Surface Area:

• Identify axis of revolution
• Determine the radius (distance to axis)
• Set up $S = 2\pi\int(\text{radius})\,ds$
• Verify with known shapes when possible

✅ Common Errors to Avoid:

• Forgetting the $2\pi$ in surface area
• Using $y$ when revolving about $y$-axis (should be $x$)
• Not taking absolute value when curve dips below axis
• Confusing arc length with displacement

Review Questions

Final Concept Checks

Final Computation

Topic Complete!

You've mastered arc length and surface area:

• Arc length in Cartesian, parametric, and polar forms
• Surface area of revolution about both axes
• Perfect-square trick for hand computation
• AP exam strategies and partial credit optimization

$\boxed{L = \int ds \qquad S = 2\pi\int(\text{radius})\,ds}$

Up next: Infinite Sequences — the foundation of series and convergence.