Surface Area of Revolution

Formula for Rotation around x-axis

When $y = f(x)$ is rotated around the x-axis from $x = a$ to $x = b$:

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

where $ds = \sqrt{1 + (dy/dx)^2} \, dx$ is the arc length element.

Formula for Rotation around y-axis

When $y = f(x)$ is rotated around the y-axis from $x = a$ to $x = b$:

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

General Pattern

$S = 2\pi \int (\text{radius})(\text{arc length element})$

• Radius is the distance from the curve to the axis of rotation
• Arc length element is $ds = \sqrt{1 + (dy/dx)^2} \, dx$

Parametric Form

For $x = x(t)$, $y = y(t)$ rotated around x-axis:

$S = 2\pi \int_\alpha^\beta y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

As a Function of y

For $x = g(y)$ rotated around x-axis from $y = c$ to $y = d$:

$S = 2\pi \int_c^d y \sqrt{1 + [g'(y)]^2} \, dy$

Examples

Example 1: Cone

Rotate $y = x$ from $x = 0$ to $x = h$ around the x-axis.

$\frac{dy}{dx} = 1$

$ds = \sqrt{1 + 1} \, dx = \sqrt{2} \, dx$

$S = 2\pi \int_0^h x \cdot \sqrt{2} \, dx = 2\pi\sqrt{2} \left[\frac{x^2}{2}\right]_0^h = \pi\sqrt{2} \cdot h^2$

Verify: For a cone with radius $r = h$ and slant height $l = h\sqrt{2}$:

$S = \pi rl = \pi h \cdot h\sqrt{2} = \pi h^2\sqrt{2}$ ✓

Example 2: Sphere

Rotate $y = \sqrt{r^2 - x^2}$ from $x = -r$ to $x = r$ around x-axis.

$\frac{dy}{dx} = \frac{-x}{\sqrt{r^2 - x^2}}$

$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^2}{r^2 - x^2} = \frac{r^2}{r^2 - x^2}$

$ds = \frac{r}{\sqrt{r^2 - x^2}} \, dx$

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

$= 2\pi r [x]_{-r}^r = 2\pi r \cdot 2r = 4\pi r^2$

Example 3: Paraboloid

Rotate $y = x^2$ from $x = 0$ to $x = 1$ around the y-axis.

$\frac{dy}{dx} = 2x$

$ds = \sqrt{1 + 4x^2} \, dx$

$S = 2\pi \int_0^1 x \cdot \sqrt{1 + 4x^2} \, dx$

Let $u = 1 + 4x^2$, $du = 8x \, dx$

$= 2\pi \cdot \frac{1}{8} \int_1^5 \sqrt{u} \, du = \frac{\pi}{4} \cdot \frac{2}{3} [u^{3/2}]_1^5$

$= \frac{\pi}{6} [5^{3/2} - 1] = \frac{\pi}{6} [5\sqrt{5} - 1] \approx 5.33$

Example 4: Using y as parameter

Rotate $x = \sqrt{y}$ from $y = 0$ to $y = 4$ around x-axis.

$\frac{dx}{dy} = \frac{1}{2\sqrt{y}}$

$ds = \sqrt{1 + \frac{1}{4y}} \, dy = \sqrt{\frac{4y + 1}{4y}} \, dy$

$S = 2\pi \int_0^4 y \cdot \sqrt{\frac{4y + 1}{4y}} \, dy = 2\pi \int_0^4 \sqrt{y} \cdot \frac{\sqrt{4y + 1}}{2} \, dy$

$= \pi \int_0^4 \sqrt{4y^2 + y} \, dy$

Let $u = 4y + 1$, then $y = \frac{u-1}{4}$:

$= \frac{\pi}{8} \int_1^{17} \sqrt{\frac{(u-1)u}{4}} \, du = \frac{\pi}{16} \int_1^{17} \sqrt{u^2 - u} \, du$

Using substitution and formula: $\approx 30.85$

Example 5: Gabriel's Horn

Rotate $y = \frac{1}{x}$ from $x = 1$ to $x = \infty$ around x-axis.

$\frac{dy}{dx} = -\frac{1}{x^2}$

$ds = \sqrt{1 + \frac{1}{x^4}} \, dx$

$S = 2\pi \int_1^\infty \frac{1}{x} \sqrt{1 + \frac{1}{x^4}} \, dx$

Since $\sqrt{1 + \frac{1}{x^4}} > 1$:

$S > 2\pi \int_1^\infty \frac{1}{x} \, dx = 2\pi [\ln x]_1^\infty = \infty$

The surface area is infinite!

Note: Interestingly, the volume is finite:

$V = \pi \int_1^\infty \frac{1}{x^2} \, dx = \pi \left[-\frac{1}{x}\right]_1^\infty = \pi$

This is the famous Gabriel's Horn paradox.

Example 6: Zone of sphere

Find surface area of sphere of radius $r$ between $y = a$ and $y = b$ (where $-r \leq a < b \leq r$).

$x = \sqrt{r^2 - y^2}$

$\frac{dx}{dy} = \frac{-y}{\sqrt{r^2 - y^2}}$

$ds = \frac{r}{\sqrt{r^2 - y^2}} \, dy$

$S = 2\pi \int_a^b \sqrt{r^2 - y^2} \cdot \frac{r}{\sqrt{r^2 - y^2}} \, dy = 2\pi r \int_a^b dy = 2\pi r(b - a)$

The surface area depends only on the height of the zone, not its position!