6.1 DOUBLE INTEGRALS
An alternative way of evaluating the integral, however, is first to sum up the
contributions from the elemental rectangles arranged intoverticalstrips and then
to combine these vertical strips to cover the regionR.Wethenwrite
I=
∫x=b
x=a
{∫y=y 2 (x)
y=y 1 (x)
f(x, y)dy
}
dx, (6.4)
wherey=y 1 (x)andy=y 2 (x) are the equations of the curvesSTUandSVU
respectively. In going to (6.4) from (6.3), we have essentially interchanged the
order of integration.
In the discussion above we assumed that the curveCwas such that any line
parallel to either thex-ory-axis intersectedCat most twice. In general, provided
f(x, y) is continuous everywhere inRand the boundary curveChas this simple
shape, the same result is obtained irrespective of the order of integration. In cases
where the regionRhas a more complicated shape, it can usually be subdivided
into smaller simpler regionsR 1 ,R 2 etc. that satisfy this criterion. The double
integral overRis then merely the sum of the double integrals over the subregions.
Evaluate the double integral
I=
∫∫
R
x^2 ydxdy,
whereRis the triangular area bounded by the linesx=0,y=0andx+y=1. Reverse
the order of integration and demonstrate that the same result is obtained.
The area of integration is shown in figure 6.2. Suppose we choose to carry out the
integration with respect toyfirst. Withxfixed, the range ofyis 0 to 1−x.Wecan
therefore write
I=
∫x=1
x=0
{∫y=1−x
y=0
x^2 ydy
}
dx
=
∫x=1
x=0
[
x^2 y^2
2
]y=1−x
y=0
dx=
∫ 1
0
x^2 (1−x)^2
2
dx=
1
60
.
Alternatively, we may choose to perform the integration with respect toxfirst. Withy
fixed, the range ofxis 0 to 1−y, so we have
I=
∫y=1
y=0
{∫x=1−y
x=0
x^2 ydx
}
dy
=
∫y=1
y=0
[
x^3 y
3
]x=1−y
x=0
dx=
∫ 1
0
(1−y)^3 y
3
dy=
1
60
.
As expected, we obtain the same result irrespective of the order of integration.
We may avoid the use of braces in expressions such as (6.3) and (6.4) by writing
(6.4), for example, as
I=
∫b
a
dx
∫y 2 (x)
y 1 (x)
dy f(x, y),
where it is understood that each integral symbol acts on everything to its right,