9.3 Surface Area of a Solid of Revolution

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

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

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 \[ \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ S=2\pi \int_{a}^{b}\,f( x) \sqrt{1+[ f’ ( x) ] ^{2}}\,dx }} \tag{3} \end{equation*} \]