9.3 Surface Area of a Solid of Revolution

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In Chapter 6, we used a definite integral to find the volume of a solid of revolution. To obtain the solid of revolution, we revolved a region around an axis. Here, we are interested in finding the surface area of the solid of revolution. To obtain the surface, we revolve a smooth curve about an axis.

Consider a line segment of length $L$ that lies above the $x$-axis. See Figure 19(a). If the region bounded by the line segment and the $x$-axis from $a$ to $b$ is revolved about an axis, the resulting solid of revolution is a frustum. The frustum of a right circular cone has slant height $L$ and base radii $r_{1}$ and $r_{2}.$ See Figure 19(b).

The surface area $S$ of the frustum is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ S=2\pi \left( \dfrac{r_{1}+r_{2}}{2}\right)\, L }}$$

That is, the surface area $S$ equals the product of $2\pi ,$ the average radius of the frustum, and the slant height of the frustum. This equation forms the basis for the formula for the surface area of a solid of revolution.

Suppose $C$ is a smooth curve represented by the parametric equations $$\begin{equation*} x=x (t)\qquad y=y(t)\quad a\leq t\leq b \end{equation*}$$

where $y=y(t)\geq 0$ on the closed interval $[a,b],$ as shown in Figure 20. Revolving $C$ about the $x$-axis generates a solid of revolution with surface area $S$. To find $S,$ we partition the interval $[ a,b]$ into $n$ subintervals: $$\begin{equation*} \lbrack a ,t_{1}], [t_{1},t_{2}], \ldots , [t_{i-1},t_{i}], \ldots , [t_{n-1},b] \end{equation*}$$

each of length $\Delta t=\dfrac{b-a}{n}$. Corresponding to each number $a, t_{1}, t_{2}, \ldots , t_{i-1}, t_{i}, \ldots , t_{n-1}, b$, there is a point $P_{0}, P_{1}, \ldots , P_{i-1}, P_{i}, \ldots , P_{n-1}, P_{n}$ on the curve $C$. We join each point $P_{i-1}$ to the next point $P_{i}$ with a line segment and focus on the line segment joining the points $P_{i-1}$ and $P_{i}$. See Figure 21(a). When this line segment of length $d(P_{i-1},P_{i})$ is revolved about the $x$-axis, it generates a frustum of a right circular cone whose surface area $S_{i}$ is $$\begin{equation*} S_{i}=2\pi \left[ \frac{y(t_{i-1})+y(t_{i})}{2}\right] \, \left[d\,(P_{i-1},P_{i})\right] \tag{1} \end{equation*}$$

See Figure 21(b).

We follow the same reasoning used for finding the arc length of a smooth curve, (Section 9.2). Using the distance formula, the length of the $i$th line segment is $$\begin{equation*} d(P_{i-1},P_{i})= \sqrt{[ x(t_{i}) -x(t_{i-1}) ] ^{2}+[ y(t_{i}) -y(t_{i-1}) ] ^{2}} \end{equation*}$$

Now we apply the Mean Value Theorem to $x(t)$ and $y(t)$. There are numbers $u_{i}$ and $v_{i}$ in each open interval $( t_{i-1},t_{i})$ for which $$\begin{equation*} x(t_{i}) -x ( t_{i-1} ) =\left[ \dfrac{dx}{dt} (u_{i} ) \right] \, \Delta t\qquad\hbox{and}\qquad y(t_{i}) -y ( t_{i-1} ) =\left[ \dfrac{dy}{dt} ( v_{i} ) \right] \, \Delta t \end{equation*}$$

So, $$\begin{eqnarray*} d(P_{i-1},P_{i})&=& \sqrt{\left\{ \left[ \dfrac{dx}{dt}( u_{i}) \right] \, \Delta t\right\} ^{2}+ \left\{ \left[ \dfrac{dy}{dt}(v_{i}) \right] \, \Delta t\right\} ^{2}}\\[5pt] &=& \sqrt{\left[ \frac{dx}{dt}(u_{i})\right] ^{2}+ \left[ \frac{dy}{dt}(v_{i})\right] ^{2}}\,\Delta t \end{eqnarray*}$$

where $u_{i}$ and $v_{i}$ are numbers in the $i$th subinterval.

Now we replace $d(P_{i-1},P_{i})$ in equation (1) with $\sqrt{\left[ \dfrac{dx}{dt}(u_{i})\right] ^{2}+\left[ \dfrac{dy}{dt}(v_{i})\right] ^{2}}\,\Delta t$. Then the surface area generated by the sum of the line segments is $$\begin{equation*} \sum_{i=1}^{n}S_{i}=\sum_{i=1}^{n}2\pi \left[ \frac{y(t_{i-1})+y(t_{i})}{2} \right] \, \sqrt{\left[ \frac{dx}{dt}(u_{i})\right] ^{2}+\left[ \frac{dy}{dt} (v_{i})\right] ^{2}}\,\Delta t \end{equation*}$$

These sums approximate the surface area generated by revolving $C$ about the $x$-axis and lead to the following result.

1 Find the Surface Area of a Solid of Revolution Obtained from Parametric Equations

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To help remember the formula for the surface area of a solid of revolution, think of the integrand as the product of the slant height $\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}}$ and the circumference $2\pi y$ of the circle traced by a point $(x,y)$ on the corresponding sub-arc. Also keep in mind that the limits of integration are parameter values, not $x$ values.

If the curve is revolved about the $y$-axis, we have a similar formula for the surface area of the solid of revolution.

2 Find the Surface Area of a Solid of Revolution Obtained from a Rectangular Equation

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Let $C$ be a smooth curve represented by a rectangular equation $y=f(x),$ $\ a\leq x\leq b$, where $f(x)\geq 0$ on $[a,b]$. A set of parametric equations for this curve is $x(t) =t$ and $y(t) =f(t)$. Then $\dfrac{dx}{dt}=1,$ $dx=dt,$ and $\ \dfrac{dy}{dt}= \dfrac{dy}{dx}\dfrac{dx}{dt}=\dfrac{dy}{dx}\cdot 1=f’ ( x)$. The surface area $S$ of the solid of revolution obtained by revolving $C,$ $a\leq x\leq b$, about the $x$-axis is $S=2\pi \int_{a}^{b}\,y(t) \sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}}\,dt.$ So in terms of $y=f( x) ,$ $a\leq x\leq b,$ we have $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ S=2\pi \int_{a}^{b}\,f( x) \sqrt{1+[ f’ ( x) ] ^{2}}\,dx }} \tag{3} \end{equation*}$$