In this section, we will learn two methods for computing the volumes of solids of revolution. A solid of revolution is a solid figure obtained by revolving a region in the plane around some straight line, called the axis of revolution.

• Some examples of solids of revolution in engineering and manufacturing are pistons, axles, and funnels. Solids of revolution are also widely used in architecture (Figure 1).

Disk Method

Consider the region bounded by the curve $$y=f(x)$$, the $$x$$-axis, and the vertical lines $$x=a$$ and $$x=b$$ (Figure 2(a)). If this region is revolved around the $$x$$-axis, it generates a three-dimensional body called a solid of revolution (Figure 2(b)).

 (a) (b)

Figure 2

Figure 3(a) shows the region itself together with a typical infinitesimally thin rectangle of height $$f(x)$$ and width $$dx$$. When this rectangle is revolved, we obtain a circular disk in the form of a thin right circular cylinder with base radius $$f(x)$$ and thickness is $$dx$$ (Figure 3(b)). The volume of this disk, which is our element of volume $$dV$$, is then $dV=\pi[f(x)]^{2}dx.$ [Recall that the volume of a right circular cylinder of radius $$R$$ and height (or thickness) $$h$$ is area of the base $$\times$$ thickness = $$\pi R^{2}h$$.]

 (a) (b)

Figure 3

The solid is composed of infinitely many infinitesimally thin disks like this. So, its total volume is obtained by adding up all the elements of volume as the typical disk sweeps through the solid from left to right. That is, $V=\int_{a}^{b}dV=\int_{a}^{b}\pi[f(x)]^{2}dx.$ This method of finding volume is called the disk method.

Volume of a Solid of Revolution: Disk Method:

If the region bounded by the curve $$y=f(x)$$, the $$x$$-axis, $$x=a$$, and $$x=b$$ is revolved about the $$x$$-axis, the volume of the solid generated this way is $V=\pi\int_{a}^{b}[f(x)]^{2}dx.$

If the region bounded by the curve $$x=h(y)$$, the $$y$$-axis, and the horizontal lines $$y=c$$ and $$y=d$$ ($$c<d$$) is revolved about the $$y$$-axis, we can compute the volume of the resulting solid in a similar way (Figure 4).

Figure 4: The element of volume is $$dV=\pi[h(y)]^{2}dy$$ and the total volume is $$V=\int_{a}^{b}dV=\pi\int_{c}^{d}[h(y)]^{2}dy=\pi\int_{c}^{d}x^{2}dy$$.

Example 1
The region under $$y=2x$$ on $$[0,1]$$ is revolved about the $$x$$-axis (Figure 5). Find the volume of the resulting solid.

Solution 1
The element of volume is \begin{aligned} dV & =\pi[f(x)]^{2}dx\\ & =\pi2x^{2}dx=4\pi x^{2}dx\end{aligned} So the volume of the solid is $V=\int dV=\int_{0}^{1}4\pi x^{2}dx=\left.4\pi\frac{x^{3}}{3}\right|_{0}^{1}=\frac{4\pi}{3}$ This is the result that we expected, because the solid is a cone whose radius is $$r=2$$ and whose height is $$h=1$$. So $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(2)^{2}(1)=\frac{4}{3}\pi.$

Figure 6

Example 2
The region bounded by $$y=2x$$, $$y=2$$, and $$x=0$$ is rotated about the $$y$$-axis (Figure 7). Compute the volume of the resulting solid.

Solution 2
The element of volume is $dV=\pi x^{2}dy$ To integrate, we have to express $$x$$ in term of $$y$$. So $dV=\pi\left(\frac{y}{2}\right)^{2}dy,$ and the total volume of a circular cone whose radius is $$r=1$$ and whose height is $$h=2$$ is $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(1)^{2}(2)=\frac{2}{3}\pi.$

Figure 8

Example 3
Compute the volume of the solid obtained by revolving the region under the curve $$y=x^{3}$$ on $$[1,2]$$ about $$x$$-axis (Figure 9).

Solution 3
The element of volume is $dV=\pi y^{2}dx=\pi[f(x)]^{2}dx=\pi(x^{3})^{2}dx=\pi x^{6}dx$ and the total volume is \begin{aligned} V= & \int dV=\int_{1}^{2}\pi x^{6}dx=\frac{\pi}{7}x^{7}\big|_{1}^{2}\\ = & \frac{\pi}{7}\left(2^{7}-1\right)\approx56.9975.\end{aligned}

Figure 10

Example 4
Compute the volume of the solid obtained by revolving the region bounded by $$y=x^{3}$$, $$y=8$$ and $$x=0$$ about the $$y$$-axis (Figure 11).

Solution 4
The element of volume is \begin{aligned} dV= & \pi x^{2}dy=\pi\left(y^{1/3}\right)^{2}dy\\ = & \pi y^{2/3}dy\end{aligned} The total volume is \begin{aligned} V & =\int dV=\int_{1}^{8}\pi y^{2/3}dy=\pi\frac{3}{5}\left.y^{5/3}\right|_{1}^{8}\\ & =\frac{3\pi}{5}(2^{5}-1)\approx58.4336.\end{aligned}

Washer Method

The washer method is a generalization of the disk method. Actually, a washer is a disk with a hole. Suppose $$R$$ is a region enclosed by the curves $$y=f(x)$$ and $$y=g(x)$$ (with $$f(x)\geq g(x)$$) on $$[a,b]$$ and $$R$$ is revolved about the $$x$$-axis (Figure 12(a)). To compute the volume of this solid, consider an infinitely thin vertical rectangle. When this rectangle is revolved about the $$x$$-axis, it generates an infinitely thin circular hollow cylinder whose outer radius is $$f(x)$$ and whose inner radius is $$g(x)$$ (Figure 12(b)). The volume of this hollow cylinder, which is our element of volume, is $dV=\pi\left([f(x)]^{2}-[g(x)]^{2}\right)dV$ [Recall that the volume of a right circular hollow cylinder of outer radius $$R$$, inner radius $$r$$, and height (or thickness) $$h$$ is area of the base $$\times$$ thickness = $$\pi(R^{2}-r^{2})h$$.] and the total volume is $V=\int_{a}^{b}dV=\pi\int_{a}^{b}\left([f(x)]^{2}-[g(x)]^{2}\right)dx.$

 (a) (b)

Figure 12

In fact, the volume of this solid is the volume of the solid generated by revolving the region bounded by the curve $$y=f(x)$$ and the $$x$$-axis minus the volume of the solid generated by revolving the region bounded by the curve $$y=g(x)$$ and the $$x$$-axis (Figure 13). That is,

$V=\pi\int_a^b[f(x)]^2\,dx-\pi\int_a^b[g(x)]^2\,dx.$

Figure 13

Example 5
Let $$R$$ be the region in the first quadrant enclosed by the curve $$y=4-x^{2}$$ and $$y=3x$$ (Figure 14). If $$R$$ is revolved about the $$x$$-axis, compute the volume of the resulting solid.

Solution 5

If a vertical rectangle is revolved about the $$x$$-axis, it will generate a thin hollow disk with outer radius $$4-x^{2}$$ and inner radius $$3x$$. The volume of the thin hollow disk (= washer) is $dV=\pi\left[(4-x^{2})^{2}-(3x)^{2}\right]dx.$ To find the limit of the integral, we have to find where the curves $$y=4-x^{2}$$ and $$y=3x$$ intersect. $4-x^{2}=3x\Rightarrow x^{2}+3x-4=0$ $\Rightarrow(x+4)(x-1)=0$ $\Rightarrow x=1,\,x=-4.$ Because the point $$(-4,-12)$$ does not lie in the first quarter, the limits of integration are $$x=0$$ and $$x=1$$ (Figure 15) and \begin{aligned} \text{total volume} & =\int_{0}^{1}dV\\ & =\pi\int_{0}^{1}\left[(4-x^{2})^{2}-(3x)^{2}\right]dx\\ & =\pi\int_{0}^{1}(16-8x^{2}+x^{4}-9x^{2})dx\\ & =\pi\int_{0}^{1}(x^{4}-17x^{2}+16)dx\\ & =\pi\left[\frac{x^{5}}{5}-\frac{17}{3}x^{3}+16x\right]_{x=0}^{x=1}\\ & =\pi\left[\frac{1}{5}-\frac{17}{3}+16\right]\end{aligned}

Example 6
Find the volume of the solid generated by revolving the region in the first quadrant bound between the circle $$x^{2}+y^{2}=9$$ and the ellipse $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ as shown in Figure 16.

Solution 6
If we revolve an infinitesimally thin vertical rectangle as shown in Figure 17, it will generate a washer with the outer radius MP and the inner radius NP.

To find the length of MP, we solve $$x^{2}+y^{2}=9$$ for $$y(y>0)$$ $x^{2}+y^{2}=9\Rightarrow y=\sqrt{9-x^{2}}$ $\therefore MP=\sqrt{9-x^{2}}.$ For the length of NP, we solve $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ for $$y\ (y>0)$$: $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\Rightarrow y^{2}=4\left(1-\frac{x^{2}}{9}\right)$ $\Rightarrow y=2\sqrt{1-\frac{x^{2}}{9}}$ $\therefore NP=2\sqrt{1-\frac{x^{2}}{9}}.$ The volume of the infinitesimally thin washer is \begin{aligned} dV & =\pi(MP^{2}-NP^{2})dx\\ & =\pi\left(9-x^{2}-4(1-\frac{x^{2}}{9})\right)dx\\ & =\pi\left[\frac{-5}{9}x^{2}+5\right].\end{aligned} And the total volume is: $V=\int_{0}^{3}dV=\pi\int_{0}^{3}(5-\frac{5}{9}x^{2})dx=\pi\left[5x-\frac{5}{27}x^{3}\right]_{x=0}^{x=3}$ $=\pi\left[15-\frac{5}{27}27\right]=10\pi.$