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

Answer

The region *R* looks like

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

Slicing perpendicular to the *x*-axis gives us disks of radius *y* = ln *x*.

The volume of the disk at position *x* is

π (ln *x*)^{2} Δ *x*.

Since *x* goes from 1 to 2 in the region *R*, the volume of the entire solid is

(b) Rotating *R* around the line *x* = 2 produces the solid

Slices perpendicular to the line *x* = 2 are horizontal disks.

Looking at the disk at height *y*, we see that *e*^{y} + *r* must equal 2, so the radius of the disk is

*r* = 2* – e*^{y}.

The volume of the disk is

π(2 – *e*^{y})^{2} Δ *y*.

The variable *y* ranges from 0 to ln 2 in the region *R*, so the volume of the entire solid is