Disks and Washers Examples

Let R be the region bounded by y = x and y = x². Find the volume of the solid obtained when R is rotated around the

(a) x-axis

(b) y-axis

The region R looks like

(a) When R is rotated around the x-axis we get the solid

A slice perpendicular to the x-axis is a washer. The radius of the washer is x and the radius of its hole is x².

This means the area of the side of the washer is

π(r_outer)² – π(r_inner)² = π(x)²-π(x²)² = π[x² – x⁴].

The volume of the washer is

π[x² – x⁴] Δ x.

Since x goes from 0 to 1 in this region, the volume of the solid is

(b) Rotating R around the y-axis gives us the solid

A slice perpendicular to the y-axis is a washer with inner radius y and outer radius .

The area of the side of the washer is

and the volume of the washer is

π[y – y²] Δ y.

Since y goes from 0 to 1 in the region R, the volume of the solid is

Let R be the region bounded by y = x and y = x². Find the volume of the solid obtained by rotating R around

(a) the line y = -1

(b) the line x = 4

We're getting pretty familiar with this region by now:

(a) When we rotate R around the line y = -1, we get a solid with a hole in its middle. This means a slice of this solid will look like a washer. The outside of the solid is bounded by the line y = x. However, since the center axis of the solid is not the x-axis, the outer radius is not just x. If we look at the picture we can see that the outer radius is

r_outer = x + 1.

Similarly, the inside of the solid is bounded by the line y = x². Since the center axis of the solid is at y = -1, which is a distance of 1 away from the x-axis, we have to add 1 to get the inner radius

r_inner = x² + 1.

We have r_outer = x + 1 and r_inner = x² + 1, so the volume of the washer is

π[(r_outer)² – (r_inner)²] Δ x = π [(x + 1)² – (x² + 1)² ] Δ x.

The variable x goes from 0 to 1 in the region, so the volume of the solid is

(b) When we rotate R around the line x = 4 we get the solid

Slices perpendicular to the line x = 4 are washers. The outer boundary of the solid is at the line y = x. However, the center line of the solid is at x = 4. By looking at the picture we can see that the outer radius of the slice at height y is

r_outer = 4 – x = 4 – y.

The inner boundary of the solid is the line y = x², or . By looking at the picture, we can see that the inner radius of the slice at height y is .

The volume of the slice is

The variable y goes from 0 to 1 in the original region R, so the volume of the solid is

We're being bombarded with a multitude of pitched frisbee disks because it's important. Even the most experienced volume-making, cake baking, frisbee tossing expert draws pictures to avoid making mistakes.