14.2 Double Integrals and Volume
Riemann Sum of a Function of Two Variables
If we divide a plane region R in the xy plane into small rectangular pieces with lines that are parallel to the coordinate axes, we are making a partition P of the region as the figure shows:
Now if we imagine that there is a function f(x,y) >= 0 of two variables defined on the region, we can approximate the volume of the solid formed by the region and the surface above it. We use f(xi,yj) as the height of a rectangular solid above the small rectangle xi<=x<=xi+1, yj<=y<=yj+1. Its volume is DxiDyjf(xi*,yj*) where (xi*,yj*) is some point in the rectangle. For simplicity's sake, just using (xi,yj), the sum of all of those rectangular solids is called a Riemann Sum and is an approximation to the volume we are seeking.
Ex 1 Sketch the region above the xy plane and under the
paraboloid 
.
Hopefully, you can see the solid region under the graph of f and above the xy plane. We can find its volume or a portion of its volume above a region in the xy plane in a rather simple way using a double integral. We obtain the integral by allowing the largest of the Dxi and Dyj to approach zero. Hence the area of the rectangle DA approaches zero.
Let's try to approximate the volume under this graph but above the square -1<= x <=1, -1 <= y <= 1 using the Riemann sum with Dx and Dy fixed at length 1. Can you see that there are 4 squares used here? The actual volume is somewhere between the Upper Sum (illustrated here) and the Lower Sum (illustrated here). The upper sum uses the largest possible value of f for the height of the rectangular solid representing the region. You can figure what the lower sum is.
Obviously with a finer mesh we could get better approximations. Even using an interior point for evaluating the function would improve our approximation. This example was merely instructive. The exact answer may be obtained from a double integral.
Def Double Integral
If f is defined on a closed bounded region R in the xy plane then the double integral of f over R is given by:
provided the limit exists. If the limit exists, then f is integrable over R.
The statement DA approached zero means the size of the small rectangles is approaching zero, or Dxi and Dyj are both getting infinitely small.
The Volume of a Solid Region
If f is integrable over R and f(x,y) >=0 for all (x,y) in R then the volume of the solid that lies above R and below the graph of f is defined as
Ex 2 Calculate with a double integral the exact
volume under the graph of 
and
above the square 
. Everyone
should work this out!
We got an answer between the upper and lower sums given above, as expected. Here's a sketch of the solid whose volume we found:
Ex 3 Sketch the region R and evaluate the double integral of f over R.
Check with the drawing above to see if the answer is reasonable.
Just as in single variable calculus, there are properties of integration that make the computations simpler. Similar rules were shown in Calc I.
Properties of Double Integrals
Theorem 14.1 Properties of Double Integrals
Let f and g be continuous over a closed, bounded plane region R and let c be any constant.
and finally if the region R is made up of two non-overlapping sub-regions R1 and R2, then we can find the double integral over the sub-regions and add them up to get the integral over R:
Evaluation of Double Integrals
The first step in evaluating a double integral is to rewrite it as an iterated
integral. The hardest part of this procedure is often finding the limits of
integration that describe the region R.
Ex 4 Evaluate the double integral over the region R:
Theorem 14.2 Fubini's Theorem
Let f be continuous over a plane region R.
1. If R is defined by 
and 
where 
and 
are continuous on [a,b], then
2. If R is defined
by 
and
where
and
are
continuous on [c,d], then
Ex 5 Set up (then evaluate) the integral that will give the volume of the solid bounded by x2 + y2 + z2 = r2.
R can be defined by 
,
(which defines
a circle of radius r and 
is the function we integrate. The 2 will double the volume of the hemisphere
above the xy plane, hence find the volume of the entire sphere.
Using further symmetry we can simplify this to
The program I'm using to create this file will do this integration symbolically. The expected result follows:
The Average Value of a Function
The average value of the function f over the region R can be calculated
by dividing the double integral of the function over the region by the area
of the region.
Ex 6 Find the average value of the function over
the region R: 
R: the rectangle with vertices (0,0), (4,0), (4,2), (0,2).
To get the average value, divide the answer above
by the area (area = 8) , so
the average value of xy on the rectangle is 2.
Sometimes it becomes necessary to switch the order of integration in order to produce a result from a double integral. This example shows how to alter the limits of integration based on such a switch.
Assignment Pg 997 #1, 7,13, 17,21,25,33,39,49, 55,57,61,73,75