Hello! We’ve discussed using integrals to find areas of 2-dimensional shapes and now we’re ready to discuss how to use integrals to find volumes of 3-dimensional shapes.

Lesson 23: Determining Volumes by Slicing

#### Learning goals:

- Determine the volume of a solid by integrating the area of a cross-section
- Determine the volume of a solid of revolution using the disk/washer method.

**Topic**:

- Volume 2, Section 2.2 Determining Volumes by Slicing (link to textbook section)

**WeBWorK**:

- Applications – Volumes by Slicing

**Motivating question**

How can we use integration to find volumes of irregular 3-dimensional shapes?

#### Warmup exercise 1

Evaluate the integral:

\[\int_0^5 \pi x^2 dx\]

#### Show answer 1

\[\int_0^5 \pi x^2 dx = \left. \frac{ \pi x^3}{3} \right|_0^5 = \frac{\pi \cdot 5^3}{3} – \frac{\pi \cdot 0^3}{3} = \frac{125 \pi}{3} \]

## Main idea

Recall from Lesson 21 (link here) that the main idea of Riemann sums is that you can approximate the area of a region by chopping it into bits and then adding up the approximate areas of each bit. The more bits you have, the smaller they are, and the better the approximation. Integration is just the limit of this: you are “adding up” the areas of infinitely many bits, where each bit is infinitely small.

This idea works not just for areas but for other calculations too. In this lesson and in Lesson 24 (link here), we’ll apply this idea to volume calculations. In Lesson 25 (link here), we’ll apply this idea to length (the version of length that we want is called arc length) and surface area calculations. We’ll chop our thing up into small bits, do our calculation on the small bit, add it up over all the small bits, and take a limit. This means that in addition to area, the calculations for volume, arc length, and surface area each involve an integral.

## Volume of 3D shapes by areas of cross-sections

There are different ways to chop up a three-dimensional solid. In today’s lesson, we chop the solid into thin slabs, which are stacked next to each other. The thickness of the slabs is denoted $\Delta x$ and the area of the end of the slab at position $x$ is denoted $A(x)$. This means that the volume of the slab at position $x$ is then $A(x) \Delta x$. When the number of slabs grows, their thickness shrinks; in the limit, we replace the volume of the slab at position $x$ by $A(x)dx$ and the sum with the integral.

Video 1 below explains how areas of cross-sections of a three-dimensional shape can be integrated in this way to find the volume of a shape; it then shows one basic example and one more interesting example.

The next few videos show more examples of using integration to determine volumes of solids using the area $A(x)$ of a generic cross-section at position $x$.

## Special cases: 3D solids of revolution

There is a special class of 3-dimensional objects whose volumes are easy to calculate using these integration techniques: so-called “solids of revolution.” For these shapes, we start with a 2-dimensional region in the plane, usually bounded by the graphs of some functions. Then we rotate this region around some line to obtain a 3-dimensional shape. For us, this line—*the axis of rotation*—will always be horizontal or vertical.

We’ll see these types of solids again in Lesson 24 (link here). The difference between this lesson and Lesson 24 is the shape of the cross-section whose area we are integrating. In this lesson, the cross-section will be a flat *disk* or *washer* (which is a disk with a smaller disk removed from the middle, also called an annulus). In Lesson 24, this “cross-section” will be an infinitely thin cylindrical shell.

The disk method is actually a special case of the the washer; the only difference between the two methods is whether the region being rotated is bounded by the axis of rotation. That’s actually the easiest case, so we start with that one.

## The disk method

We’ll start with rotation around the $x$-axis; this method generalizes to rotation about other axes that bound the region being rotated. Assume that a function $f(x)$ is continuous and nonnegative on the interval $[a,b]$. We will rotate the region bounded by the graph $y=f(x)$, $x=a$, $x=b$, and the $x$-axis around the $x$-axis.

Let $x$ be any value in $[a,b]$. Consider the vertical segment joining the point $(x, 0)$ to the point $(x, f(x))$. When we rotate this vertical segment around the $x$-axis, it sweeps out a flat disk whose radius is $f(x)$, which is a cross-section of the 3-dimensional shape. We have one such disk for every $x$ value between $a$ and $b$ and we denote its area by $A(x)$. To determine the volume, we’ll “add up” the areas of these disks by integrating $A(x)$ from $x=a$ to $x=b$.

**Disk method **

Consider the solid obtained by rotating the region bounded by $y=f(x)$, $x=a$, $x=b$, and the $x$-axis around the $x$-axis. The volume of this solid is given by

\[V = \int_a^b \pi (f(x))^2 dx\]

*Video 1 *explains this formula and shows a simple example rotating around the $x$-axis. It also shows how to apply this formula when the axis of rotation is the $y$-axis.

*Videos 2 *and *3* show more examples applying the disk method where the axis of rotation is the $x$-axis.

*Video 4* shows how to apply the disk method when the axis of rotation is a horizontal line other than the $x$-axis.

## The washer method

As mentioned above, the disk method is just a special case of the washer method. In the case that the axis of rotation is not a boundary of the region being rotated, then that segment that sweeps out a disk actually sweeps out a washer: a disk with a concentric disk removed. This means that there is an inner radius and an outer radius.

We’ll assume first that there are two nonnegative functions $f(x)$ and $g(x)$ whose graphs on $[a, b]$ are rotated about the $x$-axis. Again, the method generalizes to other axes of rotation.

**Washer method **

Assume that $f(x) \leq g(x)$ on $[a,b]$. Consider the solid obtained by rotating the region bounded by $y=f(x)$, $y=g(x)$, $x=a$, and $x=b$ around the $x$-axis. The volume of this solid is given by

\[V = \int_a^b \pi \left((g(x))^2- (f(x))^2 \right) dx\]

*Video 5* explains the method with pictures and shows a simple example where the region is rotated around the $x$-axis.

*Video 6* shows how the method can be applied when the axis of rotation is the $y$-axis.

*Video 7* explains how to determine whether to integrate with respect to $x$ or to $y$ when using the washer method and it shows two examples where the axis of rotation is neither the $x$-axis or the $y$-axis.

Drawing these representative thin rectangles (or segments) is a good tip when figuring out whether to use the disk/washer method, as in this lesson, or the method of cylindrical shells, as in Lesson 24 (link here). Notice that when the representative rectangle is perpendicular to the axis of rotation, the disk/washer method is the appropriate one.

### Additional Video Resources

#### Exit ticket

Find the volume of the solid obtained by rotating the region bounded by the graphs of $y = x^2 + 2$, $y = -x^2 + 10$, $x \geq 0$ about the $y$-axis.