2.4 Arc Length of a Curve and Surface Area

Arc Length and Surface Area

Arc length and surface area formulas let you measure the actual length along a curve and the area of a shape formed by rotating a curve around an axis. Both rely on setting up definite integrals that account for how a curve bends and stretches, building directly on your derivative and integration skills from earlier units.

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

The idea behind arc length is that you're adding up infinitely many tiny straight-line segments along the curve. Each tiny segment has a horizontal piece ($dx$) and a vertical piece ($dy$), and by the Pythagorean theorem, the length of that tiny piece is $\sqrt{(dx)^2 + (dy)^2}$. Factor out $dx$ and you get the formula:

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

Steps to calculate arc length:

1. Find $\frac{dy}{dx}$, the derivative of $f(x)$.
2. Square the derivative and add 1 inside the square root.
3. Set up and evaluate the definite integral from $x = a$ to $x = b$.

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

1. Derivative: $\frac{dy}{dx} = \frac{1}{2}x^{1/2}$
2. Square it: $\left(\frac{dy}{dx}\right)^2 = \frac{x}{4}$
3. Set up the integral:

$L = \int_{0}^{8} \sqrt{1 + \frac{x}{4}} \, dx = \int_{0}^{8} \sqrt{\frac{4 + x}{4}} \, dx = \int_{0}^{8} \frac{1}{2}\sqrt{4 + x} \, dx$

1. Use substitution $u = 4 + x$, $du = dx$. When $x = 0$, $u = 4$; when $x = 8$, $u = 12$.

$L = \frac{1}{2} \int_{4}^{12} u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \Big|_{4}^{12} = \frac{1}{3}\left(12^{3/2} - 4^{3/2}\right) = \frac{1}{3}\left(24\sqrt{3} - 8\right)$

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

Sometimes a curve is easier to express as $x$ in terms of $y$. The logic is identical, but you integrate with respect to $y$:

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

Steps to calculate arc length:

1. Find $\frac{dx}{dy}$, the derivative of $g(y)$.
2. Square the derivative and add 1 inside the square root.
3. Set up and evaluate the definite integral from $y = c$ to $y = d$.

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

1. Derivative: $\frac{dx}{dy} = \frac{1}{2}y$
2. Square it: $\left(\frac{dx}{dy}\right)^2 = \frac{y^2}{4}$
3. Set up the integral:

$L = \int_{0}^{2} \sqrt{1 + \frac{y^2}{4}} \, dy$

This integral requires a trigonometric substitution (let $y = 2\tan\theta$) or a hyperbolic substitution to evaluate. The point is that even "simple" curves can produce integrals that demand the techniques you learned in Unit 1.

Surface Area of Rotated Solids

When you rotate a curve around an axis, the surface it traces out has a measurable area. Think of each tiny arc length piece $ds$ sweeping out a thin band (like a ring). The circumference of that ring is $2\pi r$, where $r$ is the distance from the piece to the axis of rotation. So the surface area is the integral of $2\pi r \, ds$.

Rotation around the x-axis ($r = f(x)$, the y-value):

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

Rotation around the y-axis ($r = x$, the x-value):

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

Steps to calculate surface area:

1. Identify the curve and the axis of rotation.
2. Determine the radius of rotation: use $f(x)$ for rotation about the x-axis, or $x$ for rotation about the y-axis.
3. Find $\frac{dy}{dx}$ and build the integrand.
4. Evaluate the definite integral from $a$ to $b$.

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

1. Derivative: $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$
2. Square it: $\left(\frac{dy}{dx}\right)^2 = \frac{1}{4x}$
3. Set up the integral:

$A = 2\pi \int_{0}^{1} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx = 2\pi \int_{0}^{1} \sqrt{x + \frac{1}{4}} \, dx$

1. Substitute $u = x + \frac{1}{4}$, $du = dx$. When $x = 0$, $u = \frac{1}{4}$; when $x = 1$, $u = \frac{5}{4}$.

$A = 2\pi \cdot \frac{2}{3} u^{3/2} \Big|_{1/4}^{5/4} = \frac{4\pi}{3}\left[\left(\frac{5}{4}\right)^{3/2} - \left(\frac{1}{4}\right)^{3/2}\right] = \frac{4\pi}{3}\left[\frac{5\sqrt{5}}{8} - \frac{1}{8}\right] = \frac{\pi(5\sqrt{5} - 1)}{6}$