(14 points) Show that the vector field F = x sin 2z i + y sin 2z j + cos 2z k is solenoidal in R3 and find a vector potential H = P (y, z) i + Q(x, z) j satisfying H(1, 1, 0

1002微微微甲甲甲01-05班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準 1. Let a > 0 be a positive constant.

(a) (10 points) Given a function f continuous on [0, a], reduce the iterated integral ˆ a

0

du ˆ u

0

dv ˆ v

0

f (w)dw

to a one-dimensional ordinary Riemann integral ˆ a

0

H(w)dw. Find H in terms of f . (b) (4 points) Evaluate

ˆ a 0

du ˆ u

0

dv ˆ v

0

e^(w−a)3dw.

Solution:

(a)

Answer is ˆ a

0

f (w)dw ˆ a

w

du ˆ u

w

dv = ˆ a

0

1

2(w − a)²f (w)dw

or ˆ a

0

f (w)dw ˆ a

w

dv ˆ a

v

du = ˆ a

0

1

2(w − a)²f (w)dw So

H(w) = 1

2(w − a)²f (w)

(10 points) (If you draw the right graph, you obtain 2 point. And integral range is right only for u or v , you can get 3 points)

(b) ˆ a

0

du ˆ u

0

dv ˆ v

0

f (w)dw = ˆ a

0

1

2(w − a)²e^(w−a)3dw

(1 points) Let t = (w − a)³⇒ dt = 3(w − a)²dw

It implies

1 6

ˆ 0

−a³

e^tdt = 1

6(1 − e^−a3)

(3 points)

2. (14 points) Show that the vector field F = x sin 2z i + y sin 2z j + cos 2z k is solenoidal in R³ and find a vector potential H = P (y, z) i + Q(x, z) j satisfying H(1, 1, 0) = j and H(0, 0, 0) = 0.

Solution:

divF = sin2z + sin2z − 2sin2z = 0 ⇒ solenoidal (2 points)

vector potential means that

curlH = F and curlH = (−Qz(x, z), Pz(y, z), Qx(x, z) − Py(y, z)) So we have that







Qz(x.z) = −xsin2z Pz(y, z) = ysin2z

Qx(x, z) − Py(y, z) = cos2z

(3 points)

By integration we obtain

Q(x, z) = x

2cos2z + A(x) P (y, z) = −y

2 cos2z + B(y)

(2 points) Q_x(x, z) − P_y(y, z) =1

2cos2z + A⁰(x) −−1

2 cos2z + B⁰(y) = cos2z

(2 points)

So we know that

A⁰(x) = B⁰(y) = constant = c ∀x, y so we can assume A(x) = cx + a and B(y) = cy + b

(2 points) use the condition of H to find a,b,c

H(1, 1, 0) = P (1, 0)i + Q(1, 0)j = j ⇒ Q(1, 0) = 1, P (1.0) = 0 H(0, 0, 0) = P (0, 0)i + Q(0, 0)j = 0 ⇒ Q(0, 0) = 0, P (0.0) = 0 So from above arguments, we can get

a = 0, b = 0, c =1 2 It implies

H(x, y, z) = (−ycos2z + y

2 ,xcos2z + x 2 , 0)

(3 points)

3. (15 points) Find the volume of the region lying inside all three of the circular cylinders x²+ y²= a², x²+ z²= a², and y²+ z²= a². [Hint: See the figure for the first octant part of the region, and use symmetry whenever possible.]

Solution:

One eighth of the required volume lies in the first octant. The eighth is divided into two equal parts by the plane x = y. One of these parts lies above the circular sector D in the xy−plane specified in polar coordinate by 0 ≤ r < a, 0 ≤ θ <π

4, and beneath the cylinder z =p

a²− x². Thus, the total volume lying inside all three cylinders is

V = 16

¨

D

pa²− x²dxdy

where 16 : 1% , p

a²− x² : 6% . Then we use the polar coordinate to get

V = 16 ˆ π/4

0

dθ ˆ a

0

pa²− r²cos²θ rdr

where 0 ≤ θ <π

4 : 2% , 0 ≤ r < a : 2% , dxdy = rdrdθ : 2%

Let u = a²− r²cos²θ, du = −2r cos²θdr.

V = 8 ˆ π/4

0

dθ cos²θ

ˆ a² a²sin²θ

u^1/2du = 16a³ 3

ˆ π/4 0

1 − sin³θ cos²θ dθ

= 16a³ 3

ˆ π/4 0

sec²θ −1 − cos²θ cos²θ sin θ

!

dθ =16a³

3 tan θ − 1

cos θ− cos θ

!    

π/4

0

= 16a³

3 1 − 0 −√

2 + 1 − 1

√2 + 1

!

= 16 1 − 1

√2

!

a³ cu. units.

where 1 − 1

√2

! a³: 2%

4. Consider the vector fields

F = (1 + x)e^x+yi + (xe^x+y+ 2y)j − 2zk, G = (1 + x)e^x+yi + (xe^x+y+ 2z)j − 2yk.

(a) (7 points) Show that F is conservative by finding a potential for it.

(b) (8 points) Evaluate ˆ

C

G · dr, where C is given by

r = (1 − t)e^ti + tj + 2tk, (0 ≤ t ≤ 1) by taking advantage of the similarity between F and G.

Solution:

(a) F = (1 + x)e^x+yi + (xe^x+y+ 2y)j − 2zk = ∇(xe^x+y+ y²− z²) (7points).

(b) ˆ

c

G · dr = ˆ

c

F · dr + ˆ

c

2(z − y)(j + k) · dr (2points)

= (xe^x+y+ y²− z²)   

(0,1,2) (1,0,0)+ 2

ˆ 1 0

(2t − 1)(1 + 2)dt (4 points)

= −3 − e + 3 = −e (2points)

5. (14 points) Let F =



px²+ y²− x 1 + y²



i + e^x+ tan⁻¹y
j and C be the positively oriented cardioid r = 1+cos θ.

Find

˛

C

F · n ds.

Solution:

Let

P =p

x²+ y²− x

1 + y², Q = e^x+ tan⁻¹y

By Green’s Theorem,

˛

C

F · n ds =

¨

D

(Px+ Qy) dA (4%)

=

¨

D

[( x

px²+ y²− 1

1 + y²) + ( 1

1 + y²)] dA

=

¨

D

x

px²+ y² dA (2%)

= ˆ 2π

0

ˆ 1+cosθ 0

rcosθ

r rdrdθ (4%)

= ˆ 2π

0

cosθ [r²

2 ]^1+cosθ₀ dθ

=1 2

ˆ 2π 0

(cosθ + 2cos²θ + cos³θ)dθ

= ˆ 2π

0

(1 + cos2θ

2 )dθ

= π (4%)

ps. 1 2

ˆ 2π 0

(cosθ + cos³θ)dθ = 0

6. Let F(x, y, z) = yi + zj + xk.

(a) (4 points) Find curl F (b) (10 points) Evaluate

˛

C

ydx + zdy + xdz, where C is the intersection curve of the surface x²+ y²+ z²= a² and x + y + z = 0 oriented counterclockwise when viewed from positive z-axis.

Solution:

(a.) (4 points)

curl( ~F ) =        

~i ~j ~k

∂

∂x

∂

∂y

∂

∂z

y z x

= −~i − ~j − ~k

(b.) Method 1:

C is the boundary of surfaces: x²+ y²+ z²= a² and S : x + y + z = 0 the normal vector of S is ~n =~i + ~j + ~k

√3 (3 points), by stokes theorem ,

˛

c

ydx + zdy + xdz = ˆ

s

curl( ~F ) · ~nds = −√ 3

ˆ

s

ds (4 points,one equal sign is 2 points )

S is a circular disk whose radius is a. Area of S=πa².

−√ 3

ˆ

s

ds=−√

3(area of S)=−√ 3πa² Thus,

˛

c

ydx + zdy + xdz = −√

3πa² (3 points)

Method 2:

C is projected on x-y plane to get the new cuve ˜C C: x˜ ²+ y²+ (−x − y)²= a² , also, x²+ y²+ xy = 1

2a²

C is oriented clockwisely in x-y plane.˜ ( the equation and orientation of ˜C, 3points)

˛

c

ydx + zdy + xdz =

˛

C˜

[ydx − (x + y)dy + x(−dx − dy)] =

˛

C˜

[ydx − 2xdy] + [−xdx − ydy] (1 point)

However, xdx + ydy = 0 has potential 1

2(x²+ y²) ,thus,

˛

C˜

[−xdx − ydy] = 0 (1 point)

By Green theorem, ˛

c

ydx + zdy + xdz =

˛

C˜

[ydx − 2xdy]

by Green theorem, ˛

C˜

[ydx − 2xdy] = ˆ

x²+y²+xy≤¹₂a²

[−2 − 1]dxdy (2 points)

=−3(area of x²+ y²+ xy ≤ 1

2a²)=−3 1

√3πa²=−√

3πa²(3points)

7. (14 points) Evaluate the flux of F(x, y, z) = x²+ sin(y³+ 2z²)
i + (e^x2+ y²)j + (3 + x)k upward across the surface S defined by x²+ y²+ z²= 2az + 3a², z ≥ 0, where a > 0 is a constant.

Solution:

We call the area inside this boundary S.

S = {x²+ y²+ (z − a)²≤ 4a², z ≥ 0} (1pt) Then the boundary of S is

∂S = S₁+ S₂,

where S1is the range we want to calculate on, and S2= {x²+ y²= 3a², z = 0}. Note that the normal direction of S2is downward. Then by Divergence theorem,

˚

S

∇ · F dV = The flux over ∂S (2pts)

=

‹

S1

F · ndS +

‹

S2

F · ndS (3pts) ,

∇ · F = 2x + 2y.

Thus ˚

S

∇ · F dV = 2(¯x + ¯y)

˚

S

dV = 0 (3pts)

(Note that if you really calculate the integration, you must use the correct range of S (2pts), sphere coordinate may cause some problems about the range of φ and will easily omit the region {x²+ y²≤ 3

4(z − a)², 0 ≤ z ≤ a}.)

Hence ‹

S₁

F · ndS = −

‹

S₂

F · ndS.

Note that the normal vector of S2is downward, that is, n = (0, 0, −1). Thus

−

‹

S2

F · ndS = −

‹

S2

F · (0, 0, −1)dS

= −

‹

S₂

−(3 + x)dS

= (3 + ¯x)

‹

S₂

dS

= (3 + 0) ∗ Area of S₂

= 3 ∗ 3a²π

= 9a²π (3pts) Thus the answer we want to calculate is 9a²π.