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

Answer

The region *R* looks like this:

The variables *x* and *y* both go from 0 to 1 in this region, so the limits of integration will be 0 and 1 regardless of whether we're integrating with respect to *x* or *y*.

(a) Rotating *R* around the line *y* = 0 gets us this solid:

Slices perpendicular to the axis of rotation are disks (no hole) with thickness *dx* and radius

The volume of a disk is

and the volume of the solid is

(b) Rotating *R* around the line *y* = 2 produces this solid:

Slices perpendicular to the axis of rotation are washers with thickness *dx*. The outer radius is constant: *r*_{outer} = 2.

The inner radius is the distance from the axis of rotation *y* = 2 to the line .

We get

The volume of a disk is

and the volume of the solid is

(c) Rotating *R* around the line *x* = 1 gets us this solid:

The outer radius is constant: *r*_{outer} = 1.

The inner radius is the distance from the axis of rotation to the graph .

So

The volume of a disk is

The variable *y* goes from 0 to 1, so the volume of the solid is

(d) Rotating *R* around the line *x* = -1 produces this solid:

We can see that the inner radius is constant, and the outer radius is the distance from the axis of rotation to the curve .

So

*r*_{inner} = 1.

The volume of a disk is

The volume of the solid is