When a curve \(y=f(x)\) is revolved about the \(x\)-axis, it generates a surface of revolution (Figure 1). We can use the arc length formula to calculate the area of such a surface.
Figure 1
The surface area of the “collar” generated by rotating an element of the curve \(ds\) about the \(x\)-axis is \(2\pi yds\), where \(y\) is the distance of the midpoint of \(ds\) from the \(x\)-axis (Figure 2): \[\bbox[#F2F2F2,5px,border:2px solid black]{dS=\underbrace{2\pi\overbrace{y}^{\text{radius}}}_{\text{circumference}}ds.\tag{i}}\] Therefore, the total surface area between the planes \(x=a\) and \(x=b\) is \[S=\int_{a}^{b}dS=\int_{a}^{b}2\pi yds.\] Because \(ds=\sqrt{(dx)^{2}+(dy)^{2}}=\sqrt{1+(dy/dx)^{2}}dx\), the total surface is \[\bbox[#F2F2F2,5px,border:2px solid black]{S=\int_{a}^{b}2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx.}\] Since \(y=f(x)\) and \(dy/dx=f'(x)\), the above formula can also be written as \[S=\int_{a}^{b}2\pi f(x)\sqrt{1+\left[f'(x)\right]^{2}}dx.\]
Figure 2
Analogously, we can say that if the surface of revolution is generated by revolving a curve about the y-axis, then \[\bbox[#F2F2F2,5px,border:2px solid black]{dS=2\pi x\ ds.\tag{ii}}\] If the curve \(x=g(y)\), \(c\leq y\leq d\), is revolved about the \(y\)-axis, then since \(ds=\sqrt{(dx)^{2}+(dy)^{2}}=\sqrt{(dx/dy)^{2}+1}\,dy\) the surface area is \[S=\int_{c}^{d}2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}\,dy=\int_{c}^{d}2\pi g(y)\sqrt{1+\left[g'(y)\right]^{2}}\,dy.\] We may combine (i) and (ii) and say \[\bbox[#F2F2F2,5px,border:2px solid black]{dS=2\pi\rho\,ds,}\] where \(\rho\) is the distance of \(ds\) from the axis of rotation.
Figure 3
Example 1
Determine the area of the surface formed by rotating the curve \(y=\sqrt{x}\), \(0\leq x\leq2\), about the \(x\)-axis.
Figure 4
Solution
To find the surface area, we evaluate the formula
\[S=\int_{a}^{b}2\pi y\sqrt{1+\left(y’\right)^{2}}\:dx\] with \[a=0,\quad b=2,\quad y=\sqrt{x},\quad\text{and}\quad y’=\frac{1}{2\sqrt{x}}.\] So \[\begin{aligned}
S&=\int_{0}^{2}2\pi\sqrt{x}\sqrt{1+\frac{1}{4x}}\:dx\\
&=2\pi\int_{0}^{2}\sqrt{x}\sqrt{\frac{4x+1}{4x}}\:dx\\
&=\frac{\cancel{2}\pi}{\cancel{\sqrt{4}}}\int_{0}^{2}\sqrt{4x+1}\,dx.
\end{aligned}\]
Figure 5
To find the above integral, let \(u=4x+1\). Then \(du=4dx\).
The lower limit of integration: When \(x=0\), \(u=1\).
The upper limit of integration: When \(x=4\), \(u=9\).
Therefore \[\begin{aligned} S & =\pi\int_{1}^{9}\sqrt{u}\overbrace{\frac{1}{4}du}^{dx}\\ & =\frac{\pi}{4}\cdot\left.\frac{2}{3}u^{3/2}\right|_{1}^{9}\\ & =\frac{\pi}{6}\left(9^{3/2}-1^{3/2}\right)=\frac{\pi}{6}\left(27-1\right)\\ & =\frac{13\pi}{3}.\end{aligned}\]
Example 2
Find the surface area of a sphere of radius \(R\).
Solution
The surface of this sphere can be generated by revolving the semicircle \(y=\sqrt{R^{2}-x^{2}}\) around the \(x\)-axis. Since \[\frac{dy}{dx}=\frac{-2x}{2\sqrt{R^{2}-x^{2}}}=-\frac{x}{\sqrt{R^{2}-x^{2}}},\] and the distance of \(ds\) from the axis of rotation (the \(x\)-axis) is \(y\) and, we have \[\begin{aligned} dS & =2\pi y\sqrt{(dx)^{2}+(dy)^{2}}\\ & =2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\\ & =2\pi\underbrace{\sqrt{R^{2}-x^{2}}}_{y}\sqrt{1+\left(-\frac{x}{\sqrt{R^{2}-x^{2}}}\right)^{2}}dx\\ & =2\pi\sqrt{R^{2}-x^{2}}\sqrt{\frac{R^{2}}{R^{2}-x^{2}}}dx=2\pi R\,dx.\end{aligned}\] Because \(x\) varies between \(-R\) and \(R\), the surface area is \[S=\int_{-R}^{R}dS=\int_{-R}^{R}2\pi R\,dx=2\pi Rx\bigg|_{-R}^{R}=4\pi R^{2}.\] The same result will be obtained, if we revolve the semicircle \(x=\sqrt{R^{2}-y^{2}}\) about the \(y\)-axis. In this case \[ds=\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}dy,\] the distance of \(ds\) from the \(y\)-axis is \(x\), and \(y\) varies between \(-R\) and \(R\). That is \[S=\int_{-R}^{R}2\pi\underbrace{\sqrt{R^{2}-y^{2}}}_{x}\sqrt{1+\frac{y^{2}}{R^{2}-y^{2}}}\ dy=\int_{-R}^{R}2\pi Rdy=4\pi R^{2}.\]
Example 3
Find the lateral surface area of a cone with base radius \(r\) and height is \(h\). (The lateral surface excludes the base area.)
Solution
Consider \(y=\frac{r}{h}x,\) \(0\leq x\leq h\). If this line segment is revolved about the \(x\)-axis, it generates a cone whose base radius is \(r\) and whose height is \(h\). By the surface area formula, we have \[\begin{aligned} S & =\int_{0}^{h}2\pi\left(\overbrace{\frac{r}{h}x}^{y}\right)\sqrt{1+\left(\underbrace{\frac{r}{h}}_{y’}\right)^{2}}dx\\ & =\frac{2\pi r}{h}\int_{0}^{h}\sqrt{\frac{r^{2}+h^{2}}{h^{2}}}xdx\\ & =\frac{2\pi r}{h^{2}}\sqrt{r^{2}+h^{2}}\left[\frac{x^{2}}{2}\right]_{0}^{h}\\ & =2\pi r\sqrt{r^{2}+h^{2}}.\end{aligned}\] Notice that \(\sqrt{r^{2}+h^{2}}\) is the slant height of the cone. This means that z
\[\text{lateral surface area of a cone}=\text{circumference of the base}\times\text{slant height}\]
Example 4
Find the area of the surface generated by revolving the curve \(y=x^{2}\) (\(1\leq x\leq2\)) about the \(y\)-axis.
Solution
The distance from
\(ds\) to the
\(y\)-axis is
\(x\). So
\(dS=2\pi x\ ds\).
Method 1: We can use \(x\) as the variable of integration. Then \[\begin{aligned} dS & =2\pi x\,ds=2\pi x\sqrt{1+(y’)^{2}}dx\\ & =2\pi x\sqrt{1+(2x)^{2}}dx\end{aligned}\] and \[S=\int_{1}^{2}dS=\int_{1}^{2}2\pi x\sqrt{1+4x^{2}}\ dx\] Now let \(u=1+4x^{2}\). Then \(du=8xdx\). The upper limit of integration: When \(x=1\), \(u=5\).
The lower limit of integration: When \(x=2\), \(u=17\).
Therefore, \[S=\int_{5}^{17}2\pi\sqrt{u}\overbrace{\frac{1}{8}du}^{xdx}=\frac{\pi}{4}\cdot\frac{2}{3}u^{3/2}\bigg|_{5}^{17}=\frac{\pi}{6}\left(17^{3/2}-5^{3/2}\right)\approx30.8465.\]
Method 2: We can use \(y\) as the variable of integration. Then \(x=\sqrt{y}\) and since \(dx/dy=1/(2\sqrt{y})\) we have \[\begin{aligned} dS & =2\pi x\sqrt{(dx)^{2}+(dy)^{2}}\\ & =2\pi\overbrace{\sqrt{y}}^{x}\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}dy\\ & =2\pi\sqrt{y}\sqrt{1+\frac{1}{4y}}dy=2\pi\sqrt{y}\frac{\sqrt{1+4y}}{2\sqrt{y}}dy\\ & =\pi\sqrt{1+4y}\,dy.\end{aligned}\] Because \(1\leq x\leq2\), \(y\) varies between \(1\) and \(4\) and thus \[S=\int_{1}^{4}\pi\sqrt{1+4y}\,dy.\] Let \(u=1+4y\). Then \(du=4dy\).
The upper limit of integration: When \(y=1\), \(u=5\).
The lower limit of integration: When \(y=1\), \(u=17\).
Therefore, \[S=\int_{5}^{17}\pi\sqrt{u}\overbrace{\frac{1}{4}du}^{dy}=\frac{\pi}{4}\cdot\frac{2}{3}u^{3/2}\bigg|_{5}^{17}=\frac{\pi}{6}\left(17^{3/2}-5^{3/2}\right)\approx30.8465,\] as before.
Example 5
Find the area of the surface formed by revolving the loop \(9x^{2}=y(3-y)^{2}\) about the \(y\)-axis (the following figure).
Solution
For the right part of the curve, when \(0\leq y\leq3\), we have \[x=\frac{1}{3}(3-y)\sqrt{y}.\] Since \[\frac{dx}{dy}=-\frac{1}{3}\sqrt{y}+\frac{3-y}{6\sqrt{y}}=\frac{3(1-y)}{6\sqrt{y}}=\frac{1-y}{2\sqrt{y}}\] the differential of the arc is
\[\begin{aligned} ds & =\sqrt{(dx)^{2}+(dy)^{2}}\\
&=\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\
& =\sqrt{\frac{(1-y)^{2}}{4y}+1}\ dy\\
&=\sqrt{\frac{(1-2y+y^{2})+4y}{4y}}dy\\
& =\sqrt{\frac{1+2y+y^{2}}{4y}}dy\\
&=\sqrt{\frac{(1+y)^{2}}{4y}}dy\\
& =\frac{1+y}{2\sqrt{y}}dy
\end{aligned}\] and
\[\begin{aligned} dS & =2\pi\overbrace{x}^{\text{radius}}\ ds\\
&=2\cdot\overbrace{\frac{1}{3}(3-y)\sqrt{y}}^{x}\cdot\overbrace{\frac{1+y}{2\sqrt{y}}dy}^{ds}\\
& =\frac{\pi}{3}(3-y)(1+y)dy\\
&=\frac{\pi}{3}(3+2y-y^{2})dy.\end{aligned}\] Because \(y\) varies between \(0\) and \(3\), the surface area is \[\begin{aligned} S & =\int_{0}^{3}\frac{\pi}{3}(3+2y-y^{2})dy\\ & =\frac{\pi}{3}\left[3y+y^{2}-\frac{1}{3}y^{3}\right]_{0}^{3}\\ & =3\pi.\end{aligned}\]
