Learning Outcomes
- Compute surface areas of surfaces of revolution.
Section 6.4 Surface Areas of Revolution (AI4)
Subsection 6.4.1 Activities
Fact 6.4.1.
A frustum is the portion of a cone that lies between one or two parallel planes. 
Figure 129. Plot of a frustum.
The surface area of the “side” of the frustum is:
\begin{equation*}
2\pi \frac{r+R}{2}\cdot l
\end{equation*}
where \(r\) and \(R\) are the radii of the bases, and \(l\) is the length of the side.
Note that if \(r=R\text{,}\) this reduces to the surface area of a “side” of a cylinder.
Activity 6.4.2.
Suppose we wanted to find the surface area of the the solid of revolution generated by rotating
\begin{equation*}
y=\sqrt{x}, 0\leq x\leq 4
\end{equation*}
about the \(y\)-axis.
Figure 130. Plot of bounded region rotated about \(x\)-axis.
Bounded region rotated about \(x\)-axis.
(a)
Suppose we wanted to estimate the surface area with two frustums with \(\Delta x=2\text{.}\)
Figure 131. Plot of bounded region rotated about \(x\)-axis. What is the surface area of the frustum formed by rotating the line segment from \((0,0)\) to \((2, \sqrt{2})\) about the \(x\)-axis?
Bounded region rotated about \(x\)-axis.
- \(\displaystyle 2\pi \frac{0+\sqrt{2}}{2}\cdot2\)
- \(\displaystyle 2\pi \frac{0+\sqrt{2}}{2}\cdot\sqrt{2^2+\sqrt{2}^2}\)
- \(\displaystyle \pi \sqrt{2}^2\cdot2\)
- \(\displaystyle \pi \sqrt{2}^2\cdot\sqrt{2^2+\sqrt{2}^2}\)
(b)
Bounded region rotated about \(x\)-axis.
- \(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{2}\)
- \(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6}\)
- \(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6-2\sqrt{2}}\)
(c)
Suppose we wanted to estimate the surface area with four frustums with \(\Delta x=1\text{.}\)
Figure 133. Plot of bounded region rotated about \(x\)-axis.
Bounded region rotated about \(x\)-axis.
\begin{equation*}
\begin{array}{|c|c|c|c|c|c|}
\hline
x_i & \Delta x & r_i & R_i & l & \text{Estimated Surface Area}\\
\hline
x_1=0 & 1 & 0 & 1 & \sqrt{1^2+1^2} &\\
\hline
x_2=1 & 1& 1 & \sqrt{2} & \sqrt{1^2+(\sqrt{2}-1)^2} & \\
\hline
x_3=2 & 1& \sqrt{2} & \sqrt{3} & \\
\hline
x_4=3 & 1 & 3 & 2 & \\
\hline
\end{array}
\end{equation*}
(d)
Suppose we wanted to estimate the surface area with \(n\) frustums.
Figure 134. Plot of bounded region rotated about \(x\)-axis. Let \(f(x)=\sqrt{x}\text{.}\) Which of the following expressions represents the surface area generated bo rotating the line segment from \((x_0, f(x_0))\) to \((\Delta x, f(x_0+\Delta x))\) about the \(x\)-axis?
Bounded region rotated about \(x\)-axis.
- \(\displaystyle \pi \left(\frac{f(x_0)+f(x_0+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\text{.}\)
- \(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\text{.}\)
- \(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)
(e)
Which of the following Riemann sums best estimates the surface area of the solid generated by rotating \(y=\sqrt{x}\) over \([0,4]\) about the \(x\)-axis ? Let \(f(x)=\sqrt{x}\text{.}\)
- \(\displaystyle \sum \pi \left(\frac{f(x_i)+f(x_i+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}\text{.}\)
- \(\displaystyle \sum 2\pi\frac{f(x_i)+f(x_i+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}\text{.}\)
- \(\displaystyle \sum 2\pi\frac{f(x_i)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)
Fact 6.4.3.
Recall from Fact 6.2.2 that
\begin{align*}
\lim_{\Delta x\to 0}\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2} & = \lim_{\Delta x\to 0} \sqrt{(\Delta x)^2\left(1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2\right)}\\
&= \lim_{\Delta x\to 0} \sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x\\
&=\sqrt{1+(f'(x))^2}dx,
\end{align*}
and that
\begin{equation*}
\lim_{\Delta x\to 0} \frac{f(x_i)+f(x_i+\Delta x)}{2}=f(x_i).
\end{equation*}
Thus given a function \(f(x)\geq 0\) over \([a,b]\text{,}\) the surface area of the solid generated by rotating this function about the \(x\)-axis is
\begin{equation*}
SA=\int_a^b 2\pi f(x)\sqrt{1+(f'(x))^2}dx.
\end{equation*}
Activity 6.4.4.
Consider again the solid generated by rotating \(y=\sqrt{x}\) over \([0,4]\) about the \(x\)-axis.
(a)
Find an integral which computes the surface area of this solid.
(b)
If we instead rotate \(y=\sqrt{x}\) over \([0,4]\) about the \(y\)-axis, what is an integral which computes the surface area for this solid?
Activity 6.4.5.
Consider again the function \(f(x)=\ln(x)+1\) over \([1,5]\text{.}\)
(a)
Find an integral which computes the surface area of the solid generated by rotating the above curve about the \(x\)-axis.
(b)
Find an integral which computes the surface area of the solid generated by rotating the above curve about the \(y\)-axis.
