# Section7.2Volume by Cross-Sectional Area; Disk and Washer Methods¶ permalink

The volume of a general right cylinder, as shown in Figure 7.2.1, is

Area of the base × height.

We can use this fact as the building block in finding volumes of a variety of shapes.

Given an arbitrary solid, we can approximate its volume by cutting it into $n$ thin slices. When the slices are thin, each slice can be approximated well by a general right cylinder. Thus the volume of each slice is approximately its cross-sectional area × thickness. (These slices are the differential elements.)

By orienting a solid along the $x$-axis, we can let $A(x_i)$ represent the cross-sectional area of the $i^\text{ th }$ slice, and let $\dx_i$ represent the thickness of this slice (the thickness is a small change in $x$). The total volume of the solid is approximately: \begin{align*} \text{ Volume } \amp \approx \sum_{i=1}^n \Big[\text{ Area } \ \times\ \text{ thickness } \Big]\\ \amp = \sum_{i=1}^n A(x_i)\dx_i. \end{align*}

Recognize that this is a Riemann Sum. By taking a limit (as the thickness of the slices goes to 0) we can find the volume exactly.

##### Example7.2.3Finding the volume of a solid

Find the volume of a pyramid with a square base of side length 10 in and a height of 5 in.

Solution

An important special case of Theorem 7.2.2 is when the solid is a solid of revolution, that is, when the solid is formed by rotating a shape around an axis.

Start with a function $y=f(x)$ from $x=a$ to $x=b\text{.}$ Revolving this curve about a horizontal axis creates a three-dimensional solid whose cross sections are disks (thin circles). Let $R(x)$ represent the radius of the cross-sectional disk at $x\text{;}$ the area of this disk is $\pi R(x)^2\text{.}$ Applying Theorem 7.2.2 gives the Disk Method.

##### Key Idea7.2.6The Disk Method

Let a solid be formed by revolving the curve $y=f(x)$ from $x=a$ to $x=b$ around a horizontal axis, and let $R(x)$ be the radius of the cross-sectional disk at $x\text{.}$ The volume of the solid is \begin{equation*} V = \pi \int_a^b R(x)^2\ dx. \end{equation*}

##### Example7.2.7Finding volume using the Disk Method

Find the volume of the solid formed by revolving the curve $y=1/x\text{,}$ from $x=1$ to $x=2\text{,}$ around the $x$-axis.

Solution

While Key Idea 7.2.6 is given in terms of functions of $x\text{,}$ the principle involved can be applied to functions of $y$ when the axis of rotation is vertical, not horizontal. We demonstrate this in the next example.

##### Example7.2.11Finding volume using the Disk Method

Find the volume of the solid formed by revolving the curve $y=1/x\text{,}$ from $x=1$ to $x=2\text{,}$ about the $y$-axis.

Solution

We can also compute the volume of solids of revolution that have a hole in the center. The general principle is simple: compute the volume of the solid irrespective of the hole, then subtract the volume of the hole. If the outside radius of the solid is $R(x)$ and the inside radius (defining the hole) is $r(x)\text{,}$ then the volume is \begin{equation*} V = \pi\int_a^b R(x)^2 \ dx - \pi\int_a^b r(x)^2\ dx = \pi\int_a^b \left(R(x)^2-r(x)^2\right)\ dx. \end{equation*}

One can generate a solid of revolution with a hole in the middle by revolving a region about an axis. Consider Figure 7.2.15(a), where a region is sketched along with a dashed, horizontal axis of rotation. By rotating the region about the axis, a solid is formed as sketched in Figure 7.2.15(b). The outside of the solid has radius $R(x)\text{,}$ whereas the inside has radius $r(x)\text{.}$ Each cross section of this solid will be a washer (a disk with a hole in the center) as sketched in Figure 7.2.16(c). This leads us to the Washer Method.

##### Key Idea7.2.17The Washer Method

Let a region bounded by $y=f(x)\text{,}$ $y=g(x)\text{,}$ $x=a$ and $x=b$ be rotated about a horizontal axis that does not intersect the region, forming a solid. Each cross section at $x$ will be a washer with outside radius $R(x)$ and inside radius $r(x)\text{.}$ The volume of the solid is \begin{equation*} V = \pi\int_a^b \Big(R(x)^2-r(x)^2\Big)\ dx. \end{equation*}

Even though we introduced it first, the Disk Method is just a special case of the Washer Method with an inside radius of $r(x)=0\text{.}$

##### Example7.2.18Finding volume with the Washer Method

Find the volume of the solid formed by rotating the region bounded by $y=x^2-2x+2$ and $y=2x-1$ about the $x$-axis.

Solution

When rotating about a vertical axis, the outside and inside radius functions must be functions of $y\text{.}$

##### Example7.2.23Finding volume with the Washer Method

Find the volume of the solid formed by rotating the triangular region with vertices at $(1,1)\text{,}$ $(2,1)$ and $(2,3)$ about the $y$-axis.

Solution

This section introduced a new application of the definite integral. Our default view of the definite integral is that it gives “the area under the curve.” However, we can establish definite integrals that represent other quantities; in this section, we computed volume.

The ultimate goal of this section is not to compute volumes of solids. That can be useful, but what is more useful is the understanding of this basic principle of integral calculus, outlined in Key Idea 7.0.1: to find the exact value of some quantity,

• we start with an approximation (in this section, slice the solid and approximate the volume of each slice),

• then make the approximation better by refining our original approximation (i.e., use more slices),

• then use limits to establish a definite integral which gives the exact value.

We practice this principle in the next section where we find volumes by slicing solids in a different way.

# Subsection7.2.1Exercises

Terms and Concepts

In the following exercises, a region of the Cartesian plane is shaded. Use the Disk/Washer Method to find the volume of the solid of revolution formed by revolving the region about the $x$-axis.