Consider a two dimensional material. Of course there is no such thing but a flat plate might be modeled as
one. The density ρ is a function of position and is defined as follows. Consider a small chunk of area dA
located at the point whose Cartesian coordinates are

(x,y)

. Then the mass of this small chunk of material
is given by ρ

(x,y)

dA. Thus if the material occupies a region in two dimensional space U, the total mass of
this material would be

∫
ρdA
U

In other words you integrate the density to get the mass. Now by letting ρ depend on position, you can
include the case where the material is not homogeneous. Here is an example.

Example 14.1.6Let ρ

(x,y)

denote the density of the plane region determined by the curves

13

x + y = 2,x = 3y^{2}, and x = 9y. Find the total mass if ρ

(x,y)

= y.

You need to first draw a picture of the region R. A rough sketch follows.

PICT

This region is in two pieces, one having the graph of x = 9y on the bottom and the graph of x = 3y^{2} on
the top and another piece having the graph of x = 9y on the bottom and the graph of

13

x + y = 2 on the
top. Therefore, in setting up the integrals, with the integral with respect to x on the outside, the double
integral equals the following sum of iterated integrals.

You notice it is not necessary to have a perfect picture, just one which is good enough to figure out what
the limits should be. The dividing line between the two cases is x = 3 and this was shown in the picture.
Now it is only a matter of evaluating the iterated integrals which in this case is routine and gives
1.