6.4 Arc Length of a Curve and Surface Area

Arc Length of a Curve and Surface Area

Arc length and surface area formulas let you measure distances along curves and the areas of surfaces formed by rotating those curves. Both build directly on derivatives and definite integrals, combining them in ways that show up frequently in physics and engineering problems.

Arc Length of $y = f(x)$ Curves

Arc length measures the actual distance traveled along a curve between two points, not just the straight-line distance between them. The idea comes from approximating the curve with many tiny line segments, then taking the limit as those segments get infinitely small.

For a curve defined by $y = f(x)$ on the interval $[a, b]$, the arc length $L$ is:

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

Where does this formula come from? Each tiny segment of the curve has a horizontal piece $dx$ and a vertical piece $dy$. By the Pythagorean theorem, the length of that tiny segment is $\sqrt{(dx)^2 + (dy)^2}$. Factor out $dx$ and you get $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$. Summing all those tiny lengths via integration gives the total arc length.

Steps to calculate arc length:

1. Find $\frac{dy}{dx}$, the derivative of $f(x)$
2. Square it and add 1 inside the square root
3. Set up the definite integral from $a$ to $b$
4. Evaluate the integral (this is often the hardest step)

Example: Find the arc length of $y = \frac{1}{3}x^{3/2}$ from $x = 0$ to $x = 8$.

1. $\frac{dy}{dx} = \frac{1}{3} \cdot \frac{3}{2} x^{1/2} = \frac{1}{2}\sqrt{x}$
2. $\left(\frac{dy}{dx}\right)^2 = \frac{x}{4}$
3. $L = \int_{0}^{8} \sqrt{1 + \frac{x}{4}} \, dx = \int_{0}^{8} \sqrt{\frac{4 + x}{4}} \, dx = \int_{0}^{8} \frac{1}{2}\sqrt{4 + x} \, dx$
4. Using substitution with $u = 4 + x$: $L = \frac{1}{2} \cdot \frac{2}{3}(4+x)^{3/2} \Big|_0^8 = \frac{1}{3}\left(12^{3/2} - 4^{3/2}\right) = \frac{1}{3}\left(24\sqrt{3} - 8\right)$

Arc Length of $x = g(y)$ Curves

Sometimes a curve is easier to express as $x$ in terms of $y$. The formula works the same way, just with the roles of $x$ and $y$ swapped. For $x = g(y)$ on the interval $[c, d]$:

$L = \int_{c}^{d} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$

Steps to calculate:

1. Find $\frac{dx}{dy}$, the derivative of $g(y)$
2. Square it, add 1, and place under the square root
3. Integrate from $c$ to $d$ with respect to $y$

Example: Find the arc length of $x = \frac{1}{4}y^2$ from $y = 0$ to $y = 2$.

1. $\frac{dx}{dy} = \frac{1}{2}y$
2. $\left(\frac{dx}{dy}\right)^2 = \frac{y^2}{4}$
3. $L = \int_{0}^{2} \sqrt{1 + \frac{y^2}{4}} \, dy$

This integral requires a trigonometric substitution (let $y = 2\tan\theta$) or a table/CAS. Many arc length integrals don't simplify neatly, so don't be surprised if the algebra gets heavy.

Surface Area of Rotational Solids

When you rotate a curve around an axis, the resulting surface has a measurable area. Think of it like wrapping a thin ribbon along the curve and spinning it around. Each tiny arc length segment traces out a circle, and the surface area is the sum of all those thin circular bands.

Rotation around the x-axis: For $y = f(x)$ on $[a, b]$:

$S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Here, $f(x)$ is the radius of each circular band (the distance from the curve to the x-axis), and the square root term is the arc length element. Multiplying radius by $2\pi$ gives the circumference of each band.

Rotation around the y-axis: For $y = f(x)$ on $[a, b]$:

$S = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Now $x$ is the radius, since each point on the curve is a distance $x$ from the y-axis.

Steps to calculate surface area:

1. Identify the curve, the interval, and which axis you're rotating around
2. Find $\frac{dy}{dx}$
3. Determine the radius: use $f(x)$ for rotation about the x-axis, or $x$ for rotation about the y-axis
4. Plug into the appropriate formula and evaluate the integral

Example: Find the surface area when $y = \sqrt{x}$ from $x = 0$ to $x = 1$ is rotated around the x-axis.

1. $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$, so $\left(\frac{dy}{dx}\right)^2 = \frac{1}{4x}$
2. $S = 2\pi \int_{0}^{1} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx = 2\pi \int_{0}^{1} \sqrt{x + \frac{1}{4}} \, dx$
3. Substituting $u = x + \frac{1}{4}$: $S = 2\pi \cdot \frac{2}{3}\left(x + \frac{1}{4}\right)^{3/2}\Big|_0^1 = \frac{4\pi}{3}\left(\frac{5}{4}\right)^{3/2} - \frac{4\pi}{3}\left(\frac{1}{4}\right)^{3/2} = \frac{\pi}{6}\left(5\sqrt{5} - 1\right)$