TBIL-Cal Surface Areas of Revolution (AI4)

Section 6.4 Surface Areas of Revolution (AI4)

Learning Outcomes

Compute surface areas of surfaces of revolution.
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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.

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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.

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Note that if \(r=R\text{,}\) this reduces to the surface area of a “side” of a cylinder.

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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.

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(a)

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

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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?

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

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

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\(\displaystyle \pi \sqrt{2}^2\cdot2\)

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

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(b)

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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?

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

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

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

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(c)

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

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\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*}

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(d)

Suppose we wanted to estimate the surface area with \(n\) frustums.

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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?

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

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

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\(\displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)

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(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 \(f(x)=\sqrt{x}\text{.}\)

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

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\(\displaystyle \sum 2\pi\frac{f(x_i)+f(x_i+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}\text{.}\)

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\(\displaystyle \sum 2\pi\frac{f(x_i)+f(x_0+\Delta x)}{2}\Delta x\text{.}\)

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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*}

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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*}

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