082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

882 Chapter 27 Line integrals and multiple integrals
The result is then integrated fromx = 1 tox = 4 effectively adding up contributions
from each vertical strip:
∫ x=4
x=1
2x+4dx=[x 2 +4x] 4 1 =32−5=27
as expected. We note that
∫ y=2 ∫ x=4
y=0
x=1
x+2ydxdy=
∫ x=4 ∫ y=2
x=1
y=0
x +2ydydx (27.1)
In particular, note the order of dx and dy in Equation (27.1). The simple interchange of
limits isonly possible because inthis example the regionRisarectangle.
WhenevaluatingadoubleintegraloverarectangularregionR,theintegrationmay
becarried outinany order, thatis
∫ ∫
∫ ∫
f(x,y)dxdy= f(x,y)dydx
R R
The double integral in Example 27.10 could have the following interpretation. Consider
the surfacez = x +2y, for 0 y 2, 1 x 4, as shown in Figure 27.11. In
this example the surface is a plane but the theory applies to more general surfaces. The
variable z represents the height of the surface above the point (x,y) in the x--y plane.
For example, consider a point in thex--y plane, say (2, 3), that isx = 2,y = 3. Then at
(2,3),z =x +2y = 2 +2(3) = 8. Thus the point on the surface directly above (2,3)
is8unitsfrom (2,3).
Let us now consider dxdy. We call dxdy an element of area. It is a rectangle with
dimensionsdxanddyandareadxdy.Hencezdxdyrepresentsthevolumeofarectangular
column of base dxdy. As we integrate with respect toxfor 1 x4and integrate
with respect toyfor 0 y2weareeffectively summing all such volumes. Thus
∫ y=2 ∫ x=4
y=0
x=1
x+2ydxdy
represents the volume bounded by the surfacez = x + 2y and the regionRin thex--y
plane.
Example27.11 Sketch the regionRover which wewould evaluate the integral
∫ y=1 ∫ x=2−2y
y=0
x=0
f(x,y)dxdy
Solution Firstconsidertheouterintegral.Therestrictiononymeansthatinterestcanbeconfined
to the horizontal strip 0 y 1. Then examine the inner integral. The lower limit on
x means that we need only consider values ofxgreater than or equal to 0. The upperx
limit depends upon the value ofy. Ify = 0 this upper limit isx = 2 −2y = 2. Ify = 1
theupperlimitisx = 2 −2y = 0.Atanyotherintermediatevalueofywecancalculate
thecorrespondingupperxlimit.Thisupperlimitwilllieonthestraightlinex = 2−2y.
With this information the region of integration can be sketched. The region is shown in
Figure 27.12 and isseen tobe triangular.