# Area, Volume, and Arc Length

### Topics

## Introduction to Area, Volume, And Arc Length - At A Glance:

If you've ever been confused whether it's spelled disk or disc, you are not alone. "Disk" is from the Greek *diskos* and "disc" is from the Latin *discus*, so go with whichever language you like better. The Greek or Latin to English lesson being over, let's toss around a few disks and see what we volumes we get.

We are going to begin with a solid of revolution, and we can slice it perpendicular to the axis of rotation. We get slices that look like disks or washers. If there's no hole in the middle of the slice, we call the slice a **disk** because it resembles a frisbee disk.

If there's a hole in the middle of the slice, we call the slice a **washer** because it looks like a washer used with a nut and bolt.

Here's a video that shows a solid of revolution with a washer-shaped slice taken out.

We make a disk by beginning with a region that touches the axis of rotation along the whole length of the solid. There won't be any holes in the middle of the solid, so all the slices will look like frisbees. Frisbee coming our way. Heads up!

To find the volume of the solid we first, we find the volume of a single disk. Then we figure out the appropriate limits of integration and write down an integral.

Because a disk is shaped like a circular, we can find the volume of the by finding the area of its circular side and multiplying by its thickness. It's thin disk, so it shouldn't hurt too much if we miss the disk and it beans us in the face.

### Sample Problem

Let *R* be the region bounded by the curve *y* = *x*^{2}, the *x*-axis, and the line *x* = 4. Write an integral expression for the volume of the solid obtained by rotating *R* around

(a) the *x*-axis

(b) the line *x* = 4

Answer.

The region *R* looks like

(a) When we rotate *R* around the *x*-axis we get a solid that looks like

*R* lies along the *x*-axis for the entire length of *R*. If we take slices perpendicular to the axis of rotation, each slice is a disk. The radius of the disk at position *x* is

*y* = *x*^{2}.

This means the volume of that disk is

π(*x*^{2})^{2} Δ *x* = π *x*^{4} Δ *x*.

The variable *x* goes from 0 to 4 in the region *R*, which means we have disks from *x* = 0 to *x* = 4. We integrate from 0 to 4 to find the volume of the solid:

(b) Rotating *R* around the line *x* = 4 produces this solid:

Slices perpendicular to the line *x* = 4 are horizontal. Since *R* lies along the line *x* = 4, each slice will be a disk. By looking at the slice at height *y*, we see from the line *x* = 0 to the edge of the slice is a distance of . From the edge of the slice to the center of the slice is a distance of *r*. Since the distance from *x* = 0 to the center of the slice is 4, we get

so the radius of the slice at height *y* is

The volume of the slice is

The variable *y* ranges from 0 to 16 in the region *R*, so we have disks from *y* = 0 to *y* = 16. This means the volume of the entire solid is

**Be Careful: **square the radii separately and then subtract. You want

π[(*r*_{outer})^{2} – (*r*_{inner})^{2}],

*not*

π(*r*_{outer }– *r*_{inner})^{2}.

To figure out what the outer and inner radii are, we can draw pictures if it helps. We don't need any complicated formulas.

#### Example 1

Let (a) (b) |

#### Example 2

Let (a) the line (b) the line |

#### Exercise 1

Let *R* be the region bounded by *y* = ln *x* and the *x*-axis on the interval [1,2]. Find an integral expression for the volume of the solid obtained by rotating *R* around

(a) the *x*-axis

(b) the line *x* = 2

#### Exercise 2

Let *R* be the region bounded by *y* = *x*^{3}, the *y*-axis, and the line *y* = 8. Find an integral expression for the volume of the solid obtained by rotating *R* around

(a) the *y*-axis

(b) the line *y* = 8

#### Exercise 3

Let *R* be the region bounded by the graphs *y* = 1 – *x*, *y* = *x*^{2} + 1, and the line *x* = 1.

Find an integral expression for the volume of the solid obtained by rotating *R* around the line

(a) *y* = 0

(b) *y* = 2

(c) * y* = -2

(d) *y* = 5

#### Exercise 4

Let *R* be the portion of the unit circle that falls in the first quadrant. Find an integral expression for the volume of the solid obtained by rotating *R* around the line

(a) *y* = 0

(b) *y* = 2

(c) *x* = 1

(d) *x* = -1