# High School Calculus/Area of a Surface of Revolution

### Area of a Surface of Revolution

This section starts by construction surfaces. A curve $y = f(x)$ is revolved around an axis. That produces a "surface of revolution," which is symmeric around the axis we get a cylinder (a pipe). By revolving a curve we might get a lamp or a lamp shade (or even the light bulb).

The key idea is to resolve short straight line segments. Their slope is $\frac {\operatorname {d}y}{\operatorname {d}x}$. They can be the same pieces of lenth $\operatorname {d}s$ or $\Delta s$ that were used to find length. Now we compute area. When revolved, a straight piece produces a thing band. The curved surface, from revolving $y = f(x)$, is close to the bands. The first step is to compute the surface area of a band.

A small comment: Curved surfaces can also be cut into tiny patches. Each patch is nearly flat, like a little square. The sum of those patches leads to a double integral (with $\operatorname {d}x \operatorname {d}y$). Here the integral stays one-dimentional ($\operatorname {d}x$ or $\operatorname {d}y$ or $\operatorname {d}t$). Surface of revolution are special - we approximate them by bands that go all the way around. A band is just a belt with aa slope, and its slope has an effect on its area.

Revolve a small straight piece (length $\Delta s$ not $\Delta x$). The center of the piece goes around a circle of radius r. The band is a slice of cone. When we flatten it out we discover its area. The area is the side length $\Delta s$ times the middle circumference $2 \pi r$:

The surface area of a band is $2 \pi r \Delta s = 2 \pi r \sqrt{1 + \left (\frac {\Delta y}{\Delta x}\right)^2} \Delta x$

For revolution around the y axis, the radius is r = x. For revolution around the x axis, the radius is the height: $r = y = f(x)$. The sum of band areas $2 \pi r \Delta s$ is close to the area S of the curved surface. In the limit we integrate $2 \pi r \operatorname {d}s$:

The surface area generated by revolving the curve $y = f(x)$ between $x = a$ and $x = b$ is

$S = \int_{a}^{b} 2 \pi y \sqrt{1 + \left (\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2} \operatorname {d}x$ around the x axis (r = y) (1)

$S = \int_{a}^{b} 2 \pi x \sqrt{1 + \left (\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2} \operatorname {d}x$ around the y axis (r = x). (2)

Example 1

Revolve a complete semicircle $y = \sqrt{R^2 - x^2}$ around the x-axis.

The surface of revolution is a sphere. Its area is $4 \pi R^2$. The limits on x are -R and R. The slope of $y = \sqrt{R^2 - x^2}$ is $\frac {\operatorname {d}y}{\operatorname {d}x} = \frac {-x}{\sqrt{R^2 - x^2}}$.
area $S = \int_{-R}^{R} 2 \pi \sqrt{R^2 - x^2} \sqrt{1 + \frac {x^2}{R^2 - x^2}} \operatorname {d}x = \int_{-R}^{R} 2 \pi R \operatorname {d}x = 4 \pi R^2$.

Example 2

Revolve a piece of the straight line $y = 2x$ around the x axis.

The surface is a cone with $\left (\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2 = 4$. The band from x = 0 to x = 1 has area $2 \pi \sqrt{5}$:

$S = \int 2 \pi y \operatorname {d}s = \int_{0}^{1} 2 \pi (2x) \sqrt{1 + 4} \operatorname {d}x = 2 \pi \sqrt{5}$.

The answer must agree with the formula $2 \pi r \Delta s$ (which it came from). The line from (0,0) to (1, 2) has length $\Delta s = \sqrt{5}$. Its midpoint is $\left (\frac {1}{2}, 1 \right)$. Around the x axis, the middle radius is r = 1 and the area is $2 \pi \sqrt{5}$.

Example 3

Revolve the same straight line segment around the y axis. Now the radius is x instead of y = 2x. The area in Example 2 is cut in half:

$S = \int 2 \pi x \operatorname {d}s = \int_{0}^{1} 2 \pi x \sqrt{1 + 4} \operatorname {d}x = \sqrt {5}$.

For surfaces with arc length, only a few examples have convenient answers. Watermelons, basketballs and light bulbs are in the exercises. Rather than stretching this section out, the final area formula will be shown.

The formula applies when there is a parameter t. Instead of $(x, f(x))$ the points on the curve are $\left (x(t), y(t) \right)$. As t varies, we move along the curve. The length formula $(\operatorname {d}s)^2 = (\operatorname {d}x)^2 + (\operatorname {d}y)^2$ is expressed in terms of t.

For the surface of revolution around the x axis, the area becomes a t-integral.

The surface area is $\int 2 \pi y \operatorname {d}s = \int 2 \pi y(t) \sqrt{\left (\frac {\operatorname {d}x}{\operatorname {d}t}\right)^2 + \left (\frac {\operatorname {d}y}{\operatorname {d}t}\right)^2}\operatorname {d}t$. (3)

Example 4

The point $x = \cos t, y = 5 + \sin t$ travels on a circle with center at (0, 5). Revolving that circle around the x axis produces a doughnut. Find tis surface area.
SOLUTION

$\left (\frac {\operatorname {d}x}{\operatorname {d}t}\right)^2 + \left(\frac {\operatorname {d}y}{\operatorname {d}t}\right)^2 = \sin^2 t + \cos^2 t = 1$. The circle complete at $t = 2 \pi$:

$\int 2 \pi y \operatorname {d}s = \int_{0}^{2 \pi} 2 \pi(5+\sin t) \operatorname {d}t= \left [2 \pi (5t - \cos t)\right]_{0}^{2 \pi} = 20 \pi^2$

### Exercises

A surface of revolution comes from revolving a a around b. This section computes the c. When the curve is a short straigh piece (length $\Delta s$), the surface is a d. Its area is $\Delta S = \bullet$. In that formula r is the radius of t. The line from (0, 0) to (1, 1) has length g and revolving it produces area h.

When the curve $y = f(x)$ revolves around the a axis, the surface area is the integral i. For $y = x^2$ the integral to compute j. When $y = x^2$ is revolved around the y axis the area is S = k. For the curve given by $x = 2t$, $y = t^2$, change $\operatorname {d}s$ is l $\operatorname {d}t$

Find the surface area when curves 1 - 6 revolve around the x axis.

1) $y = sqrt{x}[itex], [itex]2 \le x \le 6$

2) $y = x^3$, $0 \le x \le 1$

3) $y = 7x$, $-1 \le x \le 1$

4) $y = \sqrt{4 - x^2}$, $0 \le x \le 2$

5) $y = \sqrt{4 - x^2}$, $-1 \le x \le 1$

6) $y = \cosh x$, $0 \le x \le 1$

In 7-10 find the area of the surface of revolution around the y axis.

7) $y = x^2$, $0 \le x \le 2$

8) $y = \frac {1}{2} x^2 + \frac {1}{2}$, $0 \le x \le 1$

9) $y = x + 1$, $0 \le x \le 3[itex]

10) [itex]y = x^{\frac {1}{3}}$

, $\le x \le 1$

11) Slices of a basketball all have the same area of cover, if they have the same thickness.
(a) Rotate $y = \sqrt {1 - x^2}$ around the x axis. Show that $\operatorname {d}S = 2 \pi \operatorname {d}x$.

(b) The area between x = a and x = a + h is __________.