January 2014
Intermediate to advanced
144 pages
4h 2m
English
Content preview from Mathematical Formulas for Industrial and Mechanical Engineering
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Using the fact that the radius r ranges from r 5 0tor 5 2 and that θ ranges from θ 5 0
to θ 5 2π and also that in polar coordinates, the conversion equation is r
2
5 x
2
1 y
2
, the
iterated integral becomes
ð
2
22
ð
ffiffiffiffiffiffiffiffi
42x
2
p
2
ffiffiffiffiffiffiffiffi
42x
2
p
½4 2 ðx
2
1 y
2
Þdy dx 5
ð
2π
0
ð
2
0
ð4 2 r
2
Þr dr dθ
Evaluating this integral in polar coordinates, we obtain
ð
2π
0
ð
2
0
ð4 2 r
2
Þr dr dθ 5
ð
2π
0
ð
2
0
ð4r 2 r
3
Þdr dθ ðdistribute rÞ
5
ð
θ52π
θ50
2r
2
2
1
4
r
4
r52
r50
dθ
0
@
1
A
ðintegrateÞ
5
ð
θ52π
θ50
2ð2Þ
2
2
1
4
ð2Þ
4
0
@
1
A
2 0
2
4
3
5
dθ ðsub in limits of integrationÞ
5
ð
θ52π
θ50
4dθ ðsimplifyÞ
5 4θ
θ52π
θ50
ðintegrateÞ
5 4ð2πÞ2 4ð0Þðsub in limits of integrationÞ
5 8π
Thus, the volume of E is 8π.
5.36 First-Order Differential ...
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