Now that we know what double integrals are,
we can start to compute them. The key idea is to work with
one variable at a time.
In order to integrate over a rectangle [a,b]×[c,d], we
first integrate with respect to one variable (say, y) for each
fixed value of x. That's an ordinary integral, which we
can compute using the fundamental theorem of calculus. We
then integrate the result over the other variable (in this case
x), which we can also compute using the fundamental theorem of
calculus. So a 2-dimensional double
integral becomes two ordinary 1-dimensional integrals,
one inside the other. We call this an iterated integral.
There are two ways to see the relation between double integrals
and iterated integrals. In the bottom-up approach, we
evaluate the sum m∑i=1n∑j=1f(x∗i,yj)∗ΔxΔy=m∑i=1(n∑j=1f(x∗i,y∗j)Δx)Δy,
by first summing over all of the boxes with
a fixed i to get the contribution of a column (as indicated with
the parentheses on the second sum), and then adding up the
columns. (We could do this in the other order, by reversing
the summations.)
f(x∗i,y∗j)ΔyΔx is the contribution of the volume
of a tower over a single rectangle to our
double integral, which approximates the volume under the
surface $f(x,y) and over the rectangle.
(n∑j=1f(x∗i,y∗j)Δy)Δx
is the sum of volumes of towers
over all the rectangles in a single column,
approximating the volume under the surface over the
column. Here, Δx is the width of the
column. As n→∞, the sum over n turns into
an integral, and we get (∫dcf(x,y)dy)Δx.
Adding up the columns then gives
m∑i=1(∫dcf(x∗i,y)dy)Δx. Taking a limit as
m→∞ turns the sum into an iterated integral,
giving us the actual volume
over the large rectangle, under the surface:∫ba(∫dcf(x,y)dy)dx.
This approach is explained in the following video, and an example
is worked out. (Video Fix? However, there
is a small error. At the beginning it says that we're going to
integrate over the rectangle [0,1]×[0,2], but for the
rest of the video the region R is actually the rectangle [0,2]×[0,1].)