POINT 2. EVALUATION OF A DOUBLE INTEGRAL IN CARTESIAN COORDINATES
Def. 3. A domain D is called a domain of the first type if it is bounded (see fig. 6):
a) from the left by a straight line ;
b) from the right by a straight line ;
c) below by a line ;
d) above by a line ,
A double integral over a domain of the first type is calculated by a formula
. ( 5 )
Correspondingly to this formula we integrate at first with
■We’ll prove the formula (5) proceeding from the mechanic sense of a double integral. Let the integrand be the surface density of a plate, defi-ned by the figure (fig. 6). Hence the mass of the plate equals the double integral
Now we’ll find the same mass by the other way and compare the results. The mass of the element of the plate between and (fig. 7) equals . Adding all such the masses from to we find the mass of the hatched strip (fig. 7), namely
Adding finally the masses of all such the strips from to we find the mass of the whole plate that is
Comparing of two results of the mass calculating proves validity of the formula (5).■
Note 1. Integral with respect to x from a to b is called that exterior [external]. The right side of the formula (5) is called the repeated integral.
Def. 4. A domain D is called a domain of the se-cond type if it is bounded (see fig. 8):
a) below by a straight line ; b) above by a a straight line ; c) from the left by a line ;
A double integral over a domain D of the second type is calculated by a formula
. ( 6 )
At first we calculate an inner integral
that is integrate with respect to x from to , and then we integrate the result with respect to y from c to d.
■Prove this formula yourselves.■
Ex. 1. Let a domain of integration is a rectangle
with the sides parallel to Ox-, Oy-axes (fig. 9). The rec-tangle is the domain of the first and second types, there-
. ( 7 )
The formula (6) means that in the case of the rectangle we can integrate in any order. But one of orders of integration can lead to easier calculations than the other one.
Ex. 2. Find the mass of a plate
(fig. 10) if the surface density of the plate equals .
The other order of integration isn’t well (verify!).
Ex. 3. Evaluate by two ways the integral if the domain D is defined by inequalities (fig. 11).
The domain D is that of the first and second types.
The first way. We consider D as a domain of the first type,
a) from the left by the straight line ;
b) from the right by the straight line ;
c) below by the line ;
d) above by the line .
Using the formula (5) we have
In the second way we treat D as a domain of the second type,
that is D is bounded
a) below by the straight line ;
b) above by the straight line ;
c) from the left by the line ;
d) from the right by the line .
Therefore with the help of the formula (6) we write
Fulfill evaluation of the integral yourselves.
Ex. 4. Set the limits of integration in a double integral over a triangular domain D with the vertices (fig. 12)
At first we compile the equations of the straight lines OA and AB.
Fig. 12 .
The first way. The domain D is that of the first type because of it’s bounded from the left by a straight line x = 0, from the right by a straight line x = 5, below by a line , above by a line hence on the base of the formula (5)
The second way. To apply the formula (6) we divide the domain D into two domains of the second type by a straight line y = 4 (fig. 12). If we describe them by two double inequalities, namely
Ex. 5. Evaluate the double integral over the domain
The domain D is that of the first type. It can be divided into two do-mains of the second type by the straight line
The integral in question equals the sum of two integrals. It’s well to calculate
the first one over the domain D and the second one as the sum of integrals over and .