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 arc length element is denoted by $$ds$$ with lower case $$s$$ and the surface area element is often denoted by $$dS$$ with capital $$S$$.

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}