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

27.6 Double and triple integrals 881
y
4
y
4
3
R
3
R
2
2
1
dy
1
dx
0
1 2
3 4 x
Figure27.9
Therectangular region over which the
integration in Example 27.10 is
performed.
0
1 2
3 4 x
Figure27.10
Theintegral could have been foundbyfirst
integrating over a vertical stripsuchasthat
shown.
Example27.10 Evaluate
∫ y=2 ∫ x=4
y=0 x=1
x+2ydxdy
over the region,R, given inExample 27.9 and defined by 1 x4 and 0 y2.
Solution The inner integral is found first by integrating with respect tox, keepingyfixed, that is
constant.
∫ x=4
[ ] x
2 x=4
x+2ydx=
2 +2xy =8+8y− 1 2 −2y=15 2 +6y
x=1
x=1
Then the outer integral isfound by integrating the resultwith respecttoy:
∫ y=2
y=0
[ ]
15 15 2
2 +6ydy= 2 y+3y2 =15+12=27
0
Now consideragain Example 27.10. Byperformingthe innerintegral first,
∫ x=4
x=1
x+2ydx
weareeffectivelysinglingoutahorizontalstrip,suchasthatshowninFigure27.9,and
integrating along it fromx = 1 tox = 4. When we then integrate the result fromy = 0
toy = 2 weare effectively adding upcontributions fromall suchhorizontalstrips.
Theintegraloverthesameregioncouldalsohavebeenperformedbyintegratingfirst
over vertical stripsfor0 y2(Figure 27.10)yielding
∫ y=2
y=0
x+2ydy =[xy+y 2 ] y=2
y=0 =2x+4