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

Answer

Here's the region *R*:

(a) When *R* is rotated around the *y*-axis it gives us the solid

Since *R* lies along the *y*-axis for the whole height of *R*, slices perpendicular to the *y*-axis are disks.

The disk at height *y* has radius *x* = *y*^{1/3} and volume

π(*y*^{1/3})^{2} Δ *y* = π *y*^{2/3} Δ *y*.

Since *y* goes from 0 to 8 in the region *R*, the volume of the entire solid is

(b) When *R* is rotated around the line *y* = 8 we get a solid volume

*R* lies along the line *y* = 8 for the entire *x*-interval [0,2], so slices perpendicular to *y* = 8 are disks. The radius of the disk at position *x* is

8 – *y* = 8 – *x*^{3}.

The volume of this disk is

π(8 – *x*^{3})^{2} Δ *x*.

Since *x* goes from 0 to 2 in the region *R*, the volume of the solid is

When the region doesn't lie along the axis of rotation, a slice perpendicular to the axis of rotation usually looks like a washer, which is just a disk with a hole in the middle. Lucky for us, a frisbee with a hole in it doesn't fly nearly as far. Our faces should be safe from lofted washers.

The volume of a washer is the area of a side multiplied by the thickness. A washer has an outer radius and an inner radius.

To find the area of the side we find the area using the outer radius and subtract out the area of the hole.s

Since the area using the outer radius is

π (*r*_{outer})^{2}

and the area of the hole is

π (*r*_{inner})^{2},

we get a simple equation for the area of the side of the washer:

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

Then we multiply by whatever thickness is appropriate (Δ *x* or Δ *y*) to find the volume of the washer.