Solution 7
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MATH2020A Advanced Calculus II, 2013-14 Assignment 7 Suggested Solution P.1114, Ex. 24 Solution: If F =(M,N ). By Green’s theorem, Φ= I C F · nds = Z R (M x + N y ) dA where R is the region bounded by C. Hence, for this question, Φ= ZZ {x 2 +y 2 ≤4,x≥0} ( (3y 2 + 4) + (3x 2 - 4) ) dA =3 ZZ {x 2 +y 2 ≤4,x≥0} ( x 2 + y 2 ) dA =3 Z π 0 Z 2 0 r 3 drdθ = 12π. P.1125, Ex. 15 Solution: F = (0, 2y, 3z), n = (-3, 0, 1) √ 10 . ZZ S F · ndS = ZZ {x 2 +y 2 ≤4} (0, 2y, 3(3x + 2)) · (-3, 0, 1) √ 10 ( √ 10)dA =3 Z 2π 0 Z 2 0 ( 3r 2 cos θ +2r ) drdθ = 24π. P.1125, Ex. 21 Solution: F =(x, -y, 0), n = (3, 4, 1) √ 26 on the plane 3x +4y + z = 12. It is easy to check that on coordinate planes, F · n = 0. Hence, the outward flux is given Φ by Φ= ZZ S F · ndS = ZZ {3x+4y≤12,x≥0,y≥0} (x, -y, 0) · (3, 4, 1) √ 26 ( √ 26)dA = Z 4 0 Z (12-3x)/4 0 (3x - 4y)dydx = Z 4 0 3x(12 - 3x) 4 - (12 - 3x) 2 8 dx =0. 1
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Transcript of Solution 7
MATH2020A Advanced Calculus II, 2013-14
Assignment 7 Suggested Solution
P.1114, Ex. 24 Solution:
If F = (M,N). By Green’s theorem,
Φ =
∮C
F · nds =
∫R
(Mx +Ny) dA
where R is the region bounded by C.
Hence, for this question,
Φ =
∫∫{x2+y2≤4,x≥0}
((3y2 + 4) + (3x2 − 4)
)dA
= 3
∫∫{x2+y2≤4,x≥0}
(x2 + y2
)dA
= 3
∫ π
0
∫ 2
0
r3drdθ = 12π.
P.1125, Ex. 15 Solution:
F = (0, 2y, 3z), n =(−3, 0, 1)√
10.
∫∫S
F · ndS =
∫∫{x2+y2≤4}
(0, 2y, 3(3x+ 2)) · (−3, 0, 1)√10
(√
10)dA
= 3
∫ 2π
0
∫ 2
0
(3r2 cos θ + 2r
)drdθ = 24π.
P.1125, Ex. 21 Solution:
F = (x,−y, 0), n =(3, 4, 1)√
26on the plane 3x + 4y + z = 12. It is easy to check that on
coordinate planes, F · n = 0. Hence, the outward flux is given Φ by
Φ =
∫∫S
F · ndS
=
∫∫{3x+4y≤12,x≥0,y≥0}
(x,−y, 0) · (3, 4, 1)√26
(√
26)dA
=
∫ 4
0
∫ (12−3x)/4
0
(3x− 4y)dydx
=
∫ 4
0
(3x(12− 3x)
4− (12− 3x)2
8
)dx = 0.
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