TBIL-Cal Surface Areas of Revolution (AI4)

Section 6.4 Surface Areas of Revolution (AI4)

Learning Outcomes

Compute surface areas of surfaces of revolution.

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.

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.

described in detail following the image — Figure 136. Plot of bounded region rotated about \(x\)-axis.

(a)

Suppose we wanted to estimate the surface area with two frustums with \(\Delta x=2\text{.}\)

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?

\(\displaystyle 2\pi \frac{0+\sqrt{2}}{2}\cdot2\)
\(\displaystyle 2\pi \frac{0+\sqrt{2}}{2}\cdot\sqrt{2^2+\sqrt{2}^2}\)
\(\displaystyle \pi \sqrt{2}^2\cdot2\)
\(\displaystyle \pi \sqrt{2}^2\cdot\sqrt{2^2+\sqrt{2}^2}\)

(b)

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?

\(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{2}\)
\(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6}\)
\(\displaystyle 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6-2\sqrt{2}}\)

(c)

Suppose we wanted to estimate the surface area with four frustums with \(\Delta x=1\text{.}\)

\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.

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?

\(\displaystyle \pi \left(\frac{f(x_0)+f(x_0+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\text{.}\)
\(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\text{.}\)
\(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)

(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.

Subsection 6.4.2 Videos

Figure 141. Video: Compute surface areas of surfaces of revolution

Calculus for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version