MA4006: Exercise Sheet 4: Solutions
                                                                     H          2        2
                     1.   Verify Green's theorem in the plane for C(xy + y )dx + x dy where C is the closed
                                                                        2
                     curve of the region bounded by y = x and y = x .
                     Solution. Now
                                      I (xy+y2)dx+x2dy =I f ·dr=I f ·dr+I f ·dr,
                                       C                           C           C           C
                                                                                1           2
                     where f = (xy+y2,x2) and C and C are given in Figure 1. We parameterise C as x = t,
                                                    1       2                                             1
                     y = t2, 0 ≤ t ≤ 1. Then
                                         2                  ′                                    3    4  2
                              r(t) = (t,t )      =⇒        r (t) = (1,2t)     &      f(r(t)) = (t +t ,t ).
                     Thus            I            Z                      Z
                                                    1                      1                     19
                                                                ′             3    4     3
                                      C f · dr = 0 f(r(t))·r (t)dt = 0 (t +t +2t )dt = 20.
                                        1
                     We parameterise C as r(t) = r +(r −r )t, 0 ≤ t ≤ 1, where r = (1,1) and r = (0,0).
                                         2            0      1    0                       0                1
                     Then
                                                          ′                                              2        2
                     r(t) = (1 − t,1 − t)      =⇒        r (t) = (−1,−1)       &     f(r(t)) = (2(1 − t) ,(1 − t) ).
                     Thus                         I            Z
                                                                 1
                                                      f · dr =     −3(1−t)2dt=−1,
                                                    C           0
                                                     2
                     and so                            I
                                                          f · dr = 19 − 1 = − 1 .
                                                        C          20          20
                     Green's theorem says
                                             I                   ZZ µ∂f         ∂f ¶
                                                 f dx+f dy =               2 −    1   dxdy,
                                                  1       2              ∂x     ∂y
                                               C                    R
                                        2            2
                     where f = xy +y and f = x . So
                             1                  2
                                                     ∂f               ∂f
                                                        2 = 2x ,         1 = x+2y,
                                                      ∂x               ∂y
                     and so By Green's theorem
                     I        2        2      ZZ                        Z 1Z x                  Z 1   4    3          1
                        (xy+y )dx+x dy =          (2x−x−2y)dxdy =             2(x−2y)dydx =         (x −x )dx = −20.
                      C                          R                       0   x                    0
                                                                    1
                                     y
                                                      y = x
                                                C                              2
                                                  2                      y = x
                                                        R       C1
                                                                                    x
                                                   Figure 1: Question 1.
                  2.   Verify the divergence theorem for the vector eld F = 4xi − 2y2j + z2k taken over the
                  region bounded by the cylinder x2 +y2 = 4, z = 0 and z = 3.
                  Solution. The divergence theorem says RRS F·dS = RRRV ∇·FdV. Now
                                    ∇·F= ∂ (4x)+ ∂ (−2y2)+ ∂ (z2)=4−4y+2z.
                                            ∂x        ∂y          ∂z
                  So                      ZZZ              ZZZ
                                              V ∇·FdV =         V(4−4y+2z)dV.
                  Use cylindrical co-ordinates
                          x=rcosθ, y=rsinθ, z=z,              0 ≤ θ ≤ 2π,   0 ≤ r ≤ 2,  0 ≤ z ≤ 3.
                  So
                                ZZZ ∇·FdV =Z 3Z 2πZ 2(4−4rsinθ+2z)rdrdθdz
                                    V             0   0   0
                                                 Z 3Z 2π£                    ¤
                                               =          2r2 −2r2sinθ+zr2 2dθdz
                                                  0   0                       0
                                                 Z 3Z 2π¡               ¢
                                               = 0 0      8−8sinθ+4z dθdz
                                               =Z 3£8θ+8cosθ+4zθ¤2πdz
                                                  0                    0
                                                 Z 3¡          ¢      £           ¤
                                               =      16π+8πz dz = 16πz+4πz3 3 =84π.
                                                  0                                0
                  We now calculate RR F·dS. See Figure 2.
                                      S
                                                             2
                                                     S             z
                                                      2
                                                                               S3
                                                           2
                                              x                                    y
                                                             S1
                                                      Figure 2: Question 2.
                                    b
                    On S (z =0): n = −k. So
                         1
                                               b            2  2                  2
                                           F·n=(4x,−2y ,z )·(0,0,−1)=−z =0,
                                    RR            RR      b
                    since z = 0. So   S F·dS= S F·ndS=0.
                                       1             1
                                    b
                    On S (z =3): n = k. So
                         2
                                                b            2   2              2
                                             F·n=(4x,−2y ,z )·(0,0,1) = z = 9,
                    since z = 3. So      ZZ             ZZ                ZZ
                                                                 b
                                            S F·dS=        S F·ndS =9       S dS = 36π,
                                             2              2                 2
                    since the area of S is 4π.
                                      2
                                     2    2
                    Finally, on S (x +y = 4): Parameterise cylinder as
                                 3
                                  x=2cosθ, y=2sinθ, z=z,                0 ≤ θ ≤ 2π,    0 ≤ z ≤ 3.
                    So r(θ,z) = (2cosθ,2sinθ,z). Also
                                                ¯                      ¯
                                                ¯                      ¯
                                                ¯     i        j    k ¯
                                                ¯                      ¯
                                                ¯                      ¯
                                      r ×rz =¯ −2sinθ 2cosθ 0 ¯=(2cosθ,2sinθ,0),
                                       θ        ¯                      ¯
                                                ¯                      ¯
                                                ¯    0        0     1 ¯
                    and                         ¯                      ¯
                                                                           2    2
                                                F(r(θ,z)) = (8cosθ,−8sin θ,z ).
                                                                 3
                     So
                                           ZZ F·dS=ZZ F(r(θ,z))·(r ×r )dθdz
                                                                             θ    z
                                              S3             R
                                                          Z 3Z 2π¡        2          3 ¢
                                                       = 0 0        16cos θ−16sin θ dθdz
                                                             Z 2π ¡   2        3 ¢
                                                       =48 0       cos θ −sin θ dθ = 48π.
                     Thus
                              ZZ F·dS=ZZ F·dS+ZZ F·dS+ZZ F·dS=36π+48π=84π.
                                 S             S              S              S
                                                1              2              3
                     3.  Verify Stokes' theorem for the vector eld F = zi − 3xj + 2zk, where S is the surface
                               2    2                                     2    2
                     z = 1−x −y , z ≥ 0, C is the boundary circle x +y = 1. Assume that S is oriented in
                     the positive z-direction.
                     Solution. Stokes' Theorem states H F·dr = RR (∇×F)·dS. See Figure 3.
                                                         C             S
                                                                  z
                                                                 1
                                               x                                         y
                                                                          2   2
                                                                      C: x  + y   = 1
                                                         Figure 3: Question 3.
                     Parameterise C as
                                            x=cosθ, y=sinθ, z=0,                 0 ≤ θ ≤ 2π.
                     Thus r(θ) = (cosθ,sinθ,0) and so
                                         ′
                                        r (θ) = (−sinθ,cosθ,0),        F(r(θ)) = (0,−3cosθ,0).
                                                                    4