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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