If R is a region in the x-y plane and is a function, the * volume* of the solid lying above R and below the
graph of the function is given by

The picture gives a heuristic justification for this:

The region R is partitioned into boxes, each by . Above a box we construct a rectangular parallelepiped (i.e. a "tall box") up to the surface. The height of the box is , the height of the surface. The volume of a "tall box" is . The double integral "adds up" the volumes of the "tall boxes" over R to get the total volume.

A careful justification would use Riemann sums for the double integral.

This is really a * signed volume*: If the
function is zero or negative on R, the integral may not represent the
physical volume.

* Example.* Find the volume of the solid lying
below the graph of and above the following region in
the x-y-plane:

The region is

The volume is

* Example.* Find the volume of the solid lying
below the graph of and above the following region in
the x-y-plane:

First, I'll find the equation of the line, which has x-intercept 3 and y-intercept 2. Suppose the line is . The x-intercept is , so plugging this in, I get

The y-intercept is , so plugging this in I get

Thus,

The triangular region is

The volume is

* Example.* Find the volume of the solid lying
below the graph of and above the following region in
the x-y-plane:

If for in a region R, the volume bounded above by the graph of f and bounded below by the graph of g, and lying inside the cylinder determined by R, is given by

* Example.* Find the volume of the solid bounded
above by and bounded below by , and lying inside the triangular cylinder

The volume is

Copyright 2018 by Bruce Ikenaga