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Section 6.4 Surface Areas of Revolution (AI4)

Subsection 6.4.1 Activities

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.
Bounded region rotated about \(x\)-axis.
Figure 130. Plot of bounded region rotated about \(x\)-axis.
(a)
Suppose we wanted to estimate the surface area with two frustums with \(\Delta x=2\text{.}\)
Bounded region rotated about \(x\)-axis.
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?
  1. \(\displaystyle 2\pi \frac{0+\sqrt{2}}{2}\cdot2\)
  2. \(\displaystyle 2\pi \frac{0+\sqrt{2}}{2}\cdot\sqrt{2^2+\sqrt{2}^2}\)
  3. \(\displaystyle \pi \sqrt{2}^2\cdot2\)
  4. \(\displaystyle \pi \sqrt{2}^2\cdot\sqrt{2^2+\sqrt{2}^2}\)
(b)
Bounded region rotated about \(x\)-axis.
Figure 132. Plot of bounded region rotated about the \(x\)-axis.
What is the surface area of the frustum formed by rotating the line segment from \((2,\sqrt{2})\) to \((4, 2)\) about the \(x\)-axis?
  1. \(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{2}\)
  2. \(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6}\)
  3. \(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6-2\sqrt{2}}\)
(c)
Suppose we wanted to estimate the surface area with four frustums with \(\Delta x=1\text{.}\)
Bounded region rotated about \(x\)-axis.
Figure 133. Plot of 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.
Bounded region rotated about \(x\)-axis.
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?
  1. \(\displaystyle \pi \left(\frac{f(x_0)+f(x_0+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\text{.}\)
  2. \(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\text{.}\)
  3. \(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)
(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 \(f(x)=\sqrt{x}\text{.}\)
  1. \(\displaystyle \sum \pi \left(\frac{f(x_i)+f(x_i+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}\text{.}\)
  2. \(\displaystyle \sum 2\pi\frac{f(x_i)+f(x_i+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}\text{.}\)
  3. \(\displaystyle \sum 2\pi\frac{f(x_i)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)

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.

Subsection 6.4.2 Videos

Figure 135. Video: Compute surface areas of surfaces of revolution