## Section 7.4 Arc Length and Surface Area

- Given a region, what is its area?
- Given a solid, what is its volume?

- Given a curve, what is its length? This is often referred to as arc length.
- Given a solid, what is its
*surface area?*

### Subsection 7.4.1 Arc Length

*not*a Riemann Sum. While we could conclude that taking a limit as the subinterval length goes to zero gives the exact arc length, we would not be able to compute the answer with a definite integral. We need first to do a little algebra.

Now pull the \(\dx_i^2\) term out of the square root:

\begin{align*} \amp = \sum_{i=1}^n\sqrt{1 + \frac{\Delta y_i^2}{\dx_i^2}}\,\dx_i.\\ \end{align*}This is nearly a Riemann Sum. Consider the \(\Delta y_i^2/\dx_i^2\) term. The expression \(\Delta y_i/\dx_i\) measures the “change in \(y\)/change in \(x\text{,}\)” that is, the “rise over run” of \(f\) on the \(i\)th subinterval. The Mean Value Theorem of Differentiation (Theorem 3.2.4) states that there is a \(c_i\) in the \(i\)th subinterval where \(\fp(c_i) = \Delta y_i/\dx_i\text{.}\) Thus we can rewrite our above expression as:

\begin{align*} \amp = \sum_{i=1}^n\sqrt{1+\fp(c_i)^2}\,\dx_i.\\ \end{align*}This *is* a Riemann Sum. As long as \(\fp\) is continuous, we can invoke Theorem 5.3.26 and conclude

#### Example 7.4.5. Finding arc length.

## Solution 1.

## Solution 2. Video solution

#### Example 7.4.7. Finding arc length.

## Solution 1.

## Solution 2. Video solution

#### Example 7.4.9. Approximating arc length numerically.

## Solution.

*cannot*be evaluated in terms of elementary functions so we will approximate it with Simpson’s Method with \(n=4\text{.}\)

\(x\) | \(\sqrt{1+\cos^2(x) }\) |

\(0\) | \(\sqrt{2}\) |

\(\pi/4\) | \(\sqrt{3/2}\) |

\(\pi/2\) | \(1\) |

\(3 \pi/4\) | \(\sqrt{3/2}\) |

\(\pi\) | \(\sqrt{2}\) |

### Subsection 7.4.2 Surface Area of Solids of Revolution

*frustum*of a cone) as shown in Figure 7.4.12.(b). The surface area of a frustum of a cone is

#### Theorem 7.4.13. Surface Area of a Solid of Revolution.

- The surface area of the solid formed by revolving the graph of \(y=f(x)\text{,}\) where \(f(x)\geq0\text{,}\) about the \(x\)-axis is\begin{equation*} \text{Surface Area}\, = 2\pi\int_a^b f(x)\sqrt{1+\fp(x)^2}\, dx\text{.} \end{equation*}
- The surface area of the solid formed by revolving the graph of \(y=f(x)\) about the \(y\)-axis, where \(a,b\geq0\text{,}\) is\begin{equation*} \text{Surface Area}\, = 2\pi\int_a^b x\sqrt{1+\fp(x)^2}\, dx\text{.} \end{equation*}

#### Example 7.4.14. Finding surface area of a solid of revolution.

## Solution 1.

## Solution 2. Video solution

#### Example 7.4.16. Finding surface area of a solid of revolution.

- the \(x\)-axis
- the \(y\)-axis.

## Solution 1.

- The integral is straightforward to setup:\begin{align*} SA \amp = 2\pi\int_0^1 x^2\sqrt{1+(2x)^2}\, dx.\\ \end{align*}
Like the integral in Example 7.4.14, this requires Trigonometric Substitution.

\begin{align*} \amp = \left.\frac{\pi}{32}\left(2(8x^3+x)\sqrt{1+4x^2}-\sinh^{-1}(2x)\right)\right|_0^1\\ \amp =\frac{\pi}{32}\left(18\sqrt{5}-\sinh^{-1}(2) \right)\\ \amp \approx 3.81\,\text{units}^2\text{.} \end{align*} - Since we are revolving around the \(y\)-axis, the “radius” of the solid is not \(f(x)\) but rather \(x\text{.}\) Thus the integral to compute the surface area is:\begin{align*} SA \amp = 2\pi\int_0^1x\sqrt{1+(2x)^2}\, dx.\\ \end{align*}
This integral can be solved using substitution. Set \(u=1+4x^2\text{;}\) the new bounds are \(u=1\) to \(u=5\text{.}\) We then have

\begin{align*} \amp = \frac{\pi}4\int_1^5 \sqrt{u}\, du\\ \amp = \left.\frac{\pi}{4}\frac23 u^{3/2}\right|_1^5\\ \amp = \frac{\pi}6\left(5\sqrt{5}-1\right)\\ \amp \approx 5.33\,\text{units}^2\text{.} \end{align*}

## Solution 2. Video solution

#### Example 7.4.18. The surface area and volume of Gabriel’s Horn.

## Solution 1.

Integrating this expression is not trivial. We can, however, compare it to other improper integrals. Since \(1\lt \sqrt{1+1/x^4}\) on \([1,\infty)\text{,}\) we can state that

\begin{align*} 2\pi\int_1^\infty \frac{1}{x}\, dx \amp \lt 2\pi\int_1^\infty \frac{1}{x}\sqrt{1+1/x^4}\, dx \text{.} \end{align*}