082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
882 Chapter 27 Line integrals and multiple integralsThe result is then integrated fromx = 1 tox = 4 effectively adding up contributionsfrom each vertical strip:∫ x=4x=12x+4dx=[x 2 +4x] 4 1 =32−5=27as expected. We note that∫ y=2 ∫ x=4y=0x=1x+2ydxdy=∫ x=4 ∫ y=2x=1y=0x +2ydydx (27.1)In particular, note the order of dx and dy in Equation (27.1). The simple interchange oflimits isonly possible because inthis example the regionRisarectangle.WhenevaluatingadoubleintegraloverarectangularregionR,theintegrationmaybecarried outinany order, thatis∫ ∫∫ ∫f(x,y)dxdy= f(x,y)dydxR RThe double integral in Example 27.10 could have the following interpretation. Considerthe surfacez = x +2y, for 0 y 2, 1 x 4, as shown in Figure 27.11. Inthis example the surface is a plane but the theory applies to more general surfaces. Thevariable 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 withdimensionsdxanddyandareadxdy.Hencezdxdyrepresentsthevolumeofarectangularcolumn of base dxdy. As we integrate with respect toxfor 1 x4and integratewith respect toyfor 0 y2weareeffectively summing all such volumes. Thus∫ y=2 ∫ x=4y=0x=1x+2ydxdyrepresents the volume bounded by the surfacez = x + 2y and the regionRin thex--yplane.Example27.11 Sketch the regionRover which wewould evaluate the integral∫ y=1 ∫ x=2−2yy=0x=0f(x,y)dxdySolution Firstconsidertheouterintegral.Therestrictiononymeansthatinterestcanbeconfinedto the horizontal strip 0 y 1. Then examine the inner integral. The lower limit onx means that we need only consider values ofxgreater than or equal to 0. The upperxlimit depends upon the value ofy. Ify = 0 this upper limit isx = 2 −2y = 2. Ify = 1theupperlimitisx = 2 −2y = 0.Atanyotherintermediatevalueofywecancalculatethecorrespondingupperxlimit.Thisupperlimitwilllieonthestraightlinex = 2−2y.With this information the region of integration can be sketched. The region is shown inFigure 27.12 and isseen tobe triangular.
27.6 Double and triple integrals 883z876543210 01234xdx dyFigure27.11Thesurfacez=x+2y,0y2,1x4;thequantityzdxdyis the volume ofthe rectangular columnwith base area dxdy.zyy1dy2–xx = 2 – 2y, y = ––— 20 2Figure27.12Theregion over which the integration in Example 27.12isperformed.xExample27.12 Evaluate∫ y=1 ∫ x=2−2yy=0 x=04x+5dxdyover the region described inExample 27.11.Solution We first perform the inner integral∫ x=2−2yx=04x+5dxintegrating with respect tox. This gives[2x 2 +5x] x=2−2yx=0=2(2−2y) 2 +5(2−2y)−0=8y 2 −26y+18Next weperform the outer integral∫ y=1y=0[ 8y8y 2 3−26y+18dy=3 − 26y22= 8 3 − 26 2 +18= 7.667] 1+18y0The region of integration is shown in Figure 27.12. The first integral, with respect tox,corresponds to integrating along the horizontal strip fromx = 0 tox = 2 − 2y. Thenasyvaries from 0 to 1 in the second integral, the horizontal strips will cover the entireregion.
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27.6 Double and triple integrals 883
z
8
7
6
5
4
3
2
1
0 0
1
2
3
4
x
dx dy
Figure27.11
Thesurfacez=x+2y,0y2,1x4;the
quantityzdxdyis the volume ofthe rectangular column
with base area dxdy.
z
y
y
1
dy
2–x
x = 2 – 2y, y = ––— 2
0 2
Figure27.12
Theregion over which the integration in Example 27.12
isperformed.
x
Example27.12 Evaluate
∫ y=1 ∫ x=2−2y
y=0 x=0
4x+5dxdy
over the region described inExample 27.11.
Solution We first perform the inner integral
∫ x=2−2y
x=0
4x+5dx
integrating with respect tox. This gives
[2x 2 +5x] x=2−2y
x=0
=2(2−2y) 2 +5(2−2y)−0
=8y 2 −26y+18
Next weperform the outer integral
∫ y=1
y=0
[ 8y
8y 2 3
−26y+18dy=
3 − 26y2
2
= 8 3 − 26 2 +18
= 7.667
] 1
+18y
0
The region of integration is shown in Figure 27.12. The first integral, with respect tox,
corresponds to integrating along the horizontal strip fromx = 0 tox = 2 − 2y. Then
asyvaries from 0 to 1 in the second integral, the horizontal strips will cover the entire
region.
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