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
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
\begin{align*} \amp = \int_a^b\sqrt{1+\fp(x)^2}\, dx\text{.} \end{align*}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.
\(x\) | \(\sqrt{1+\cos^2(x) }\) |
\(0\) | \(\sqrt{2}\) |
\(\pi/4\) | \(\sqrt{3/2}\) |
\(\pi/2\) | \(1\) |
\(3 \pi/4\) | \(\sqrt{3/2}\) |
\(\pi\) | \(\sqrt{2}\) |
Subsection 7.4.2 Surface Area of Solids of Revolution
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*}