MATH2020A Advanced Calculus II, 2013-14

Assignment 8 Suggested Solution

P.1142, Ex. 7 Solution:

Let S be the surface bounded by C.

The upward unit normal of S is

n =(−1, 0, 1)√

2and ∇× F =

∣∣∣∣∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

2z x 3y

∣∣∣∣∣∣∣∣∣∣∣= (3, 2, 1).

By Stoke’s theorem, we have∮C

F ·Tds =∫∫

S

(∇× F) ·ndS

=

∫∫{x2+y2≤4}

(3, 2, 1) · (−1, 0, 1)√2

(√2)dxdy

= −2∫∫{x2+y2≤4}

dxdy = −8π.

P.1142, Ex. 10 Solution:

Let S be the surface bounded by C.

The upward unit normal of S is

n =(0,−1, 1)√

2and ∇× F =

∣∣∣∣∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

y2 z2 x2

∣∣∣∣∣∣∣∣∣∣∣= (−2z,−2x,−2y).

By Stoke’s theorem, we have∮C

F ·Tds =∫∫

S

(∇× F) ·ndS

=

∫∫{x2+(y−1)2≤1}

(−2y,−2x,−2y) · (0,−1, 1)√2

(√2)dxdy

= 2

∫∫{x2+(y−1)2≤1}

(x− y) dxdy

= 2

∫ 1

0

∫ 2π

0

(r cos θ − (r sin θ + 1)) rdθdr

= −4π∫ 1

0

rdr = −2π.

1

P.1142, Ex. 12 Solution:

F = (3y3 − 10xz2, 9xy2,−10x2z).

∇× F =

∣∣∣∣∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

3y3 − 10xz2 9xy2 −10x2z

∣∣∣∣∣∣∣∣∣∣∣= (0,−(−20xz − (−20xz)), 9y2 − 9y2) = (0, 0, 0).

Hence, F is irrotational. Let (u, v, w) ∈ R3 and C is the line segment from (0, 0, 0) to

(u, v, w). That is, r(t) = t(u, v, w) 0 ≤ t ≤ 1. Then a potential function φ of F is given

by

φ(u, v, w) =

∫C

F ·Tds

=

∫ 1

0

(3v3t3 − 10uw2t3, 9uv2t3,−10u2wt3) · (u, v, w)dt

=

∫ 1

0

(12uv3t3 − 20u2w2t3

)dt = 3uv3 − 5u2w2.

2