# Volumes

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

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