6. Note for (a) N = y 2 + x, M = −x2 . For (b), the
vector field is conservative.
Final Review Answer
1. M = 3x2 y − 2y 3 + 5, N = x3 − 6xy 2 + 2y
(a)
(b)
Z
2xydx + x2 dy =
F · dr = f (r(2π)) − f (r(0)) = 0
F · dr = f (r(π)) − f (r(0))
which is = f (−1, −1) − f (5, −1)
Alternatively you could choose a straight
line path from (5, −1) to (−1, −1) and
evaluate the line integral of F over it.
−3(x − 2) − 2(y − 3) + (z − 6) = 0
(b) ru = h4 cos v, 4 sin v, 1i and rv =
h−4u sin v, 4u cos v, 0i, so N = ru × rv =
h−4u cos v, −4u sin v, 16ui. Hence
2. The pot. fn. is f (x, y, z) = 12 (x2 + y 2 ) − z 3 .
ZZ
SA =
Using the FTOLI,
||N|| dA
S
F · dr = f (−1, 0, π) − f (1, 0, 0) = −π 3
Z 2π Z 3 q
=
√ 1 2 3
= 2π 4 17 u 2
0
√
√
= 9(4π) 17 = 36π 17
3.
Z 4D
E
4t − t2 , −t · h1, 4 − 2ti dt = 21
1
C
4. The pot. fn. is f (x, y) = x2 y + K.
(c) ru = k and rv = −3 sin vi−3 cos vj, so N =
ru × rv = h−3 cos v, −3 sin v, 0i. Hence
Using the FTOLI,
Z
ZZ
SA =
F · dr = f (8, 5) − f (0, 0) = 320
ZZ
R
=
0
4x3 dydx
=
(b) Green’s thm. for area, (NOT Fig. 2)
=
Z
C1
=
3
2
+
3
=
2
F · dr
ZC
F · dr +
C2
Z π/2
Z
(1 + r2 )r dr dθ =
0
3π
2
(b) Using the Divergence Theorem
Z
ZZ
F · dr
C3
cos4 t sin2 t + cos2 t sin4 t dt
=
F · NdS
S
ZZZ
div(F) dV
Q
0
Z 1
1 + x2 + y 2 dA
R
2π Z 1
0
Z
F · dr +
=
F · NdS
ZZS
0
area(R) =
9π
2
8. (a) N = h2x, 2y, 1i, So
4x3 dA
4
Z 1Z √
1−x4
3 du dv =
0
0
2y 3 dx + (x4 + 6y 2 x) dy
C
ZZ
=
Z
=
5. (a) Using Green’s thm.,(see Fig. 2)
||N|| dA
S
π/2Z 3
C
Z
16(17u2 ) du dv
0
0
C
F · dr =
0 dA = 0.
R
7. (a) ru = h1, 0, vi and rv = h0, 1, ui, so N =
ru × rv = h−v, −u, 1i. At the point
(2, 3, 6), N = h−3, −2, 1i.
Equation of tangent line is
C1
Z
ZZ
C
(c) If you know pot. fn., use FTOLI and
Z
1 dA
R
= area(R) = 81 π
C1
Z
ZZ
(y 2 + x)dy − x2 dx =
C
(a) Nx = 3x2 − 6y 2 = My hence cons.
(b) By FTOLI
Z
Z
0 dt +
0
Z π/2
Z 1
0 dt
2 + 3 + 4 dV
2
4
= 9 · vol(Q) = 9 π33
3
Q
0
2
ZZZ
=
cos t sin t dt
0
1
(c) Using the Divergence Theorem
ZZ
ZZZ
F·NdS =
S
div(F) dV =
ZZZ
Q
0 dV
Q
(d) Using Gauss Theorem
ZZ
F · NdS
S
ZZZ
=
Z
=
div(F) dV
Q
4Z 5 Z 4−x
0 + 2y + 0 dzdydx = 200
0
0
0
9. Use Stokes theorem to evaluate
Z
F · dr
C
(a) We take as S the plane whoseZ boundary
F · dr =
is C. Using Stokes’ Theorem
C
ZZ
curl(F) · NdS. Use (i + j + k) × 2j
for normal to the plane and find eqn of
plane to be z = x. So that N = h−1, 0, 1i.
The region, D, to integrate over is the triangle in the xy-plane whose vertices are
(0, 0), (0, 2) and
Z (1, 1). Since curl(F) =
S
F · dr
h−3, −1, −2i,
C
=
=
ZZ
curl(F) · NdS
ZZS
h−3, −1, −2i · h−1, 0, 1i dA
D
1
= (2)(1) = 1
2
(b) curl(F) = x2 i − 2xyj + y 2 k If we use
g(x, y) = 6 − x, then N = (i + k)and the
region, D, to integrate over is inside the
circle in the xy-plane of radius 2.
= area(D)
Z
=
=
=
F · dr
C
ZZ
curl(F) · NdS
ZZS D
E
x2 , −2xy, y 2 · h1, 0, 1i dA
ZZD
x2 + y 2 dA
D
=
Z 2πZ 2
0
= 2π
r3 dr dθ
0
r4
4
!2
=
8π
0
2