1. (11%) Evaluate the iterated integral Z 1
0
Z 1
√x
sin (y3) dy dx.
Sol:
Z 1 0
Z 1
√x
sin(y3) dy dx = Z 1
0
Z y2 0
sin(y3) dx dy
= Z 1
0
y2sin y3dy
= −1
3cos y3dy
1 0
= 1 3 −1
3cos 1 2. (11%) Evaluate
I
C
(4y − ecos x) dx + (13x +p
y2+ y + 1) dy, where C is the circle x2+ y2 = 4.
Sol:
I
C
(4y − ecos x) dx + (13x +p
y2+ y + 1) dy = Z
A
(13 − 4) dx dy, where A is the region enclosed by C.
= 9 × 4π = 36π.
3. (11%) Find the average value of the potential function f (x, y, z) = mM G
px2+ y2+ (z − 2)2 on the spherical shell r ≤px2+ y2+ z2 ≤ 1, where 0 < r < 1.
Sol:
faverage = RRR
E
f (x, y, z)dV
V , where
V = Z Z Z
E
1dV
= Z 2π
0
Z π 0
Z 1 r
ρ2sin φdρdφdθ
= 4π
3 (1 − r3)
And the numerator is Z Z Z
E
f (x, y, z)dV = Z Z Z
E
mM G
px2+ y2+ (z − 2)2dV
= Z 2π
0
Z π 0
Z 1 r
mM G
pρ2 − 4ρ cos φ + 4ρ2sin φdρdφdθ
= 2π · Z π
0
Z 1 r
1
pρ2− 4ρ cos φ + 4ρ2sin φdρdφ
= (2πmM G) Z 1
r
Z π 0
1
pρ2− 4ρ cos φ + 4ρ2sin φdφdρ
= (2πmM G) Z 1
r
[ρ 2
pρ2− 4ρ cos φ + 4] |φ=πφ=0 dρ
= (2πmM G) Z 1
r
ρ
2[(ρ + 2) − |ρ − 2|]dρ
= (2πmM G) Z 1
r
ρ
2· 2ρdρ
= (2πmM G) Z 1
r
ρ2dρ
= (2
3πmM G)(1 − r3)
∴ faverage = (23πmM G)(1 − r3)
4π
3 (1 − r3)
= mM G 2 4. (11%) Evaluate the integral
Z Z
R
cos (x − y) dA, where R is the region bounded by |x| + |y| = π 2. Sol:
I ≡ Z Z
R
cos(x − y)dA R : |x| + |y| ≤ π 2 method1:
Let u = x + y, v = x − y, then |J | =
∂(x, y)
∂(u, v)
= 1 2 Therefore,
I =
Z π/2
−π/2
Z π/2
−π/2
cos(v) 1
2 dvdu = π sin(v)
π/2
−π/2
1 2 = π
method2:
R1 =
Z π/2 0
Z π/2−x 0
cos(x − y) dydx = 1
R2 =
Z π/2 0
Z 0
−π/2+x
cos(x − y) dydx = π 2 − 1
R3 = Z 0
−π/2
Z π/2+x 0
cos(x − y) dydx = π 2 − 1 R4 =
Z 0
−π/2
Z 0
−π/2−x
cos(x − y) dydx = 1
I = R1+ R2+ R3+ R4 = π
5. (11%) Evaluate Z Z
S
y dS, where S is the surface z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.
Sol:
Let r = (x, y, x + y2), then rx = (1, 0, 1), ry = (0, 1, 2y) rx× ry = (−1, −2y, 1) and |rx× ry| =p2 + 4y2 Thus, we have
Z Z
S
y dS = Z 1
0
Z 2 0
yp
2 + 4y2dy dx
= Z 1
0
1 8· 2
3(2 + 4y2)32
2 0
dx
= Z 1
0
1 12
h√
183−√ 23i
dx
= 13 3
√ 2
6. (11%) Let T be the surface of the tetrahedron bounded by x = 0, y = 0, z = 0 and x
1 + y 2 + z
3 = 1, given with outward orientation. Let F = y2
z + 2i + (xz + 1)j + ( x2 y + 2)k be a vector field. Evaluate
Z Z
S
curlF · dS, where S is the surface when the bottom surface (z = 0) is removed from T .
Sol:
Method 1:
curlF = (−x2(y + 2)−2− x, −y2(z + 2)−2− 2x(y + z)−1, z − 2y(z + 2)−1) Z Z
S
curlF · dS = Z Z
D
curlF(x, y, 0) · (0, 0, 1)dA = Z Z
D
−ydA
= Z 1
0
Z 2−2x 0
−ydydx = Z 1
0
−1
2(2 − 2x)2dx = −2 3 Method 2:
By stoke’s theorem, Z Z
S
curlF · dS = I
C
F · dr
C = C1 ∪ C2∪ C3 = {(t, 0, 0) ∪ (1 − t, 2t, 0) ∪ (0, 2 − 2t, 0)}, t ∈ [0, 1]
Z
C1
F · dr = Z 1
0
0dt = 0 Z
C2
F · dr = Z 1
0
−2t2+ 2dt = 4 3 Z
C3
F · dr = Z 1
0
−2dt = −2
⇒ Z Z
S
curlF · dS = 0 +4
3 − 2 = −2 3
7. (12%) (a) Let F1(x) = x
|x|3. Calculate Z Z
S1
F1· dS, where S1 is the sphere centered at (0, 0, 0) with radius a. (You must show the details.)
(b) Let F2(x) = x
|x|3 − x − v
|x − v|3 where v = h2, 0, 0i. Find Z Z
S2
F2· dS, where S2 is the sphere
centered at (0, 0, 0) with radius 1 2. (c) Find
Z Z
S3
F2· dS, where S3 is the sphere centered at (0, 0, 0) with radius 3.
Sol:
(a) n = x
|x| for any x ∈ S1 Z Z
S1
F1· dS = Z Z
S1
x
|x|3 · x
|x|dS = Z Z
S1
1
|x|2dS = 1
a2 · Area(S1) = 4π
(b) Let F3(x) = F1(x) − F3(x) = x − v
|x − v|3 divF1 = divh x
(x2+ y2+ z2)32
, y
(x2+ y2+ z2)32
, z
(x2+ y2+ z2)32 i
= 3 (x2+ y2+ z2)32 − x · 32(x2+ y2+ z2)12 · 2x (x2+ y2+ z2)3
+−y · 32(x2+ y2+ z2)12 · 2y − z ·32(x2+ y2+ z2)12 · 2z (x2+ y2+ z2)3
= 3 (x2+ y2+ z2) − 3x2− 3y2− 3z2 (x2+ y2+ z2)52
= 0 on R3\ {(0, 0, 0)}.
Similiarly, divF3 = 0 on R3\ {(2, 0, 0)}.
Therefore, we get Z Z
S2
F2· dS = Z Z
S2
F1· dS − Z Z Z
E
divF3dV = Z Z
S2
F1· dS = 4π by divergence theorem where E is the ball bounded by S.
(c) Let S1 be the sphere centered at (0,0,0) with radius 12. Let S2 be the sphere centered at (2,0,0) with radius 12. Then by divergence theorem,
Z Z
S3
F2· dS = Z Z
S1∪S2
F2· dS.
Z Z
S1
F2· dS = 4π as computed in case (b).
Similiarly, Z Z
S2
F2 · dS = −4π.
Therefore, Z Z
S3
F2· dS = 4π − 4π = 0.
8. (11%) Find the general solution of the differential equation y00− 2y0 + y = xex. Sol:
yhomo = c1ex+ c2xex where c1, c2 are constants.
sub y = Ax4ex
we can obtain : A = 1 6
hence , the general solution is: y = c1ex+ c2xex+ 1 6x4ex
9. (11%) Find all power series solutions to the (special case of) Bessel equation x2y00+xy0+x2y = 0.
Can y(0) and y0(0) be given arbitrarily?
Sol:
Let y = P∞
n=0cnxn, then y0 =P∞
n=1ncnxn−1, y00 =P∞
n=2n(n − 1)cnxn−2. x2y00+ xy0+ x2y =
∞
X
n=2
n(n − 1)cnxn+
∞
X
n=1
ncnxn+
∞
X
n=0
cnxn+2
=
∞
X
n=2
n(n − 1)cnxn+
∞
X
n=1
ncnxn+
∞
X
n=2
cn−2xn
=c1x +
∞
X
n=2
(n(n − 1)cn+ ncn+ cn−2)xn= 0.
By comparing the coefficients, c1 = 0, n2cn+ cn−2 = 0, ∀n ≥ 2, i.e., cn= −cn−2n2 , ∀n ≥ 2.
Thus we have for all k ∈ N,
c2k = (−(2k)1 2) · (−(2k−2)1 2) · · · (−212)c0 = (−1)k22kc(k!)0 2
c2k+1= (−(2k+1)1 2) · (−(2k−1)1 2) · · · (−312)c1 = 0
.
Hence y =P∞
n=0cnxn=P∞ k=0
(−1)kx2k 4k(k!)2 .
y(0) = c0, can be given arbitrarily, y0(0) = c1 = 0 is fixed, can’t be given arbitrarily.
