(1%), F has a potential on D and is conservative.

1042微微微甲甲甲01-04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. (10%) Evaluate the double integral∫

2

1 ∫

x

√x

sin y

y dydx +∫

4

2 ∫

2

√x

sin y y dydx

Solution:

= ∫

2

1 ∫

y²

y

sin y

y dxdy (4pts)

= ∫

2

1

y sin ydy −∫

2

1

sin ydy

= −y ⋅ cos y∣²_y=1+ ∫

2

1

cos ydy + cos 2 − cos 1

=sin 2 − sin 1 − cos 2 (6pts)

2. (10%) Find the area of the region in the first quadrant enclosed by the curves xy = a, xy = b, xy^1.4=c and xy^1.4=d where 0 < a < b and 0 < c < d.

Solution:

Let R ∶= {(x, y) ∶ a ≤ xy ≤ b, c ≤ xy^1.4≤d}, then define u = xy and v = xy^1.4⇒x = (u^1.4

v )

5 2

and y = (v u)

5 2

. So we have R^′∶= {(u, v) ∶ a ≤ u ≤ b, c ≤ v ≤ d},

and ∣J ∣ = ∣ RR RR RR RR RR RR RR RR R

7

2u⁵²v⁻⁵² − 5 2u⁷²v⁻⁷²

− 5

2u⁻⁷²v⁵² 5 2u⁻⁵²v³²

RR RR RR RR RR RR RR RR R

∣ = ( 35

4 − 25

4 )v⁻¹= 5 2v Area =∬

R

1dA =∬

R^′

1 × ∣J ∣dA^′= ∫

d

c ∫

b

a

5

2vdudv =5

2(b − a) lnd c.

3. (10%) Find the mass of the solid S bounded by the paraboloid z = x²+2y² and the plane z = 2 + 4y if S has density function ρ(x, y, z) = ∣x∣.

Solution:

There are two ways:

Method 1: The projection of S onto xy-plane is the ellipse

D ∶ {(x, y)∣x²+2(y − 1)²=4} (2%)

Since D is symmetric with respect to yz-plane, the mass of S is twice of the part in the half space x ≥ 0. Hence the mass is

m =∭

S

ρ dV = 2∭

S_x≥0

ρ dV

=2∬

D_x≥0∫

2+4y

x²+2y²

x dzdA (3%)

=2∫

π/2

−π/2∫

1

0

2r cos θ(4 − 4r²)2√ 2 rdrdθ

=32

√ 2∫

π/2

−π/2cos θ∫

1

0 (r²−r⁴)dr =128 15

√

2 (2%)

where we use the change of variables:

{

x = 2r cos θ y = 1 +√

2r sin θ , 0 ≤ r ≤ 1, −π 2≤θ ≤π

2

with Jacobian

∣

2 cos θ −2r sin θ

√

2r sin θ √

2r cos θ ∣ =2√

2 (3%)

Method 2: The mass is given by

m = 2∫

1+√ 2

1−√ 2 ∫

2+4y

2y² ∫

√ z−2y²

0

x dxdzdy (3%)

= ∫

1+√ 2

1−√ 2 ∫

2+4y

2y²

(z − 2y²)dzdy (3%)

= ∫

1+√ 2 1−√

2 2y⁴−8y³+4y²+8y + 2 dy (3%)

= 128

15

√

2 (1%)

4. (10%) Let S be a cone has radius a and height h without base. Evaluate the integral of the distance of the points to its axis over S.

Solution:

Solution 1. Under Cartesian coordinate system Z = a

h

√ x²+y²

∬S

F ⋅ dS =∬

D

F ⋅ ∣ rx×ry∣dA r =< x, y, h

a

√

x²+y²>

rx=<1, 0, h a

x

√ x²+y²

>

ry=<0, 1, h a

y

√

x²+y² >

∣r_x×r_y∣=

√ 1 +h²

a²

∬D

F ⋅ ∣ r_x×r_y∣dA =∬

D

√ x²+y²

√ 1 +h²

a² dA

Change the above equation from Cartesian coordinate to polar coordinate, and 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π .

∬D

√ x²+y²

√ 1 +h²

a² dA =

√ 1 +h²

a²∫

2π

0 ∫

a

0

r² drdθ = 2 3πa²

√ a²+h². Solution 2. Under cylindrical coordinate system

r =< ρ cos θ, ρ sin θ, h aρ >

rρ=<cos θ, sin θ, h a>

r_φ=< −ρ sin θ, ρ cos θ, 0 >

∣r_ρ×r_φ∣=ρ

√ 1 + h²

a² Therefore,∫

2π

0 ∫

√ a²+h²

0 ρ² dρdφ =2 3πa²

√ a²+h² Solution 3. Under Spherical coordinate system Let tan α =a

h ⇒α = tan⁻¹a h.

r =< ρ cos θ sin α, ρ sin θ sin α, cos α >

∣rρ×rφ∣=ρ sin α

∫

2π

0 ∫

√ a²+h²

0

ρ sin α ρ sin α dρ dθ =∫

2π

0

sin α∫

√ a²+h²

0

ρ²dρ = 2 3πa²

√ a²+h²

Grading policies:

(A) 2 points for the corret formula for parametric surface.

(B) Another 3 points were given when the Jacobian term is correct.

(C) The other 2 points were given when integrate formula and regions are correct.

(D) The last 3 points were given when integration is correct.

5. (18%) Consider the vector field defined by G(x, y) = (3x²+y)i + (2x²y − x)j, (x, y) ∈ R². (a) Is G(x, y) conservative?

(b) Find a function µ(x) with µ(1) = 1 such that µ(x)G(x, y) is conservative.

(c) Set F(x, y) to be the conservative vector field in (b). Find the potential function f (x, y) of F with f (1, 0) = 3.

(d) Let C be the curve with defining equation in polar coordinate given by r = sec θ +

√2

π θ, θ ∈ [0,π 4]. Evaluate the integral∫

C

F ⋅ dr.

Solution:

(a) Let P = 3x²+y, Q = 2x²y − x. Since ∂Q

∂x =4xy − 1, ∂P

∂y =1

∴G(x, y) is not conservative (2pts) (b) ∂

∂x(µ(x) ⋅ Q) = ∂

∂y(µ(x) ⋅ P ) = µ(x) ⋅ 1

⇒µ^′(x)(2x²y − x) + µ(x)(4xy − 1) = µ(x)

⇒µ^′(x)(2x²y − x) = µ(x)(2 − 4xy)

⇒ µ^′(x)

µ(x) =

2(1 − 2xy) x(2xy − 1) =

−2 x (5pts)

⇒ln ∣µ(x)∣ = −2 ln x + C0

⇒ ∣µ(x)∣ = C1x⁻² (2pts)

Since µ(1) = 1, then µ(x) = x⁻²(1pt) (c) µ(x)G(x) = (3 + y

x²)i + (2y −1

x)j = ▽f (x, y) (1pt)

⇒f (x, y) = 3x − y

x+y²+C (2pts)

⇒f (1, 0) = 3 + C = 3 ⇒ C = 0

⇒f (x, y) = 3x − y

x+y² (1pt) (d) r(π

4) = (

√ 2 +

√2

4 ) ⋅ (cosπ 4, sinπ

4) = ( 5 4,5

4)(1pt) r(0) = (1 + 0) ⋅ (cos 0, sin 0) = (1, 0) (1pt)

∫C

F ⋅ dr = f (5 4,5

4) −f (1, 0) = 15

4 −1 + 25

16−3 = 21 16 (2pts)

6. (12%) Let F(x, y) = y³ (x²+y²)²

i − xy² (x²+y²)²

j.

(a) Show that F is conservative on the domain D = R²− {(0, y)∣y ≤ 0}.

(b) Compute∫

C

F ⋅ dr, where C is the part of the polar curve r = 1 + sin θ, 0 ≤ θ ≤ π.

Solution:

(a) For P (x, y) = y³ (x²+y²)²

and Q(x, y) = −xy² (x²+y²)²

, we have

∂P

∂y =

3x²y²−y⁴ (x²+y²)³

=

∂Q

∂x (5%)

Since the domain D is simply connected

(1%), F has a potential on D and is conservative.

(b) There are three ways:

Method 1: Since F is conservative on D, for any curve C^′in D with the same starting point and end point of C, we have

∫C

F ⋅ dr =∫

C^′

F ⋅ dr (3%)

Take C^′ to be the upper half unit circle (cos t, sin t), 0 ≤ t ≤ π, then

∫C

F ⋅ dr =∫

C^′

F ⋅ dr

= ∫

π

0

(sin t)³(−sin t) − (cos t sin²t)(cos t) dt (2%)

= ∫

π

0 −sin²tdt = −π

2 (1%)

Method 2: We find the potential function f with ∇f = F .

∫ y³ (x²+y²)²

dx =∫

−xy² (x²+y²)

dy

=y³∫

y sec θ

(y sec θ)²dθ (x = y tan θ)

= ∫ 1

sec²θdθ =1

2θ + sin 2θ 4

= 1 2tan⁻¹(

x y) +

1 2⋅

xy

x²+y² (5%)

It is easy to see that the function

f (x, y) = 1 2tan⁻¹(

x y) +

1 2⋅

xy x²+y² satisfies fy(x, y) = Q(x, y) and hence is a potential function. Hence

∫CF ⋅ dr = f (−1, 0) − f (1, 0) = 1 2(−

π 2−

π 2) = −

π

2 (1%)

Remark 0.1. A potential function of F on D is given by

f (x, y) =˜

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎩

f (x, y) = f+(x, y), (x, y) ∈ I f (x, y) = f−(x, y), (x, y) ∈ II f−(x, y), (x, y) ∈ III

f+(x, y), (x, y) ∈ IV where

⎧⎪

⎪⎪

⎪

⎨

⎪⎪

⎪⎪

⎩

f+(x, y) = 1 2(

π

2 −tan⁻¹( y x)) +

1 2 ⋅

xy

x²+y², x > 0 f−(x, y) = 1

2(−

π

2 −tan⁻¹( y x)) +

1 2 ⋅

xy

x²+y², x < 0

Method 3: A parametrization of the curve C is

γ(θ) = ((1 + sin θ) cos θ, (1 + sin θ) sin θ), 0 ≤ θ ≤ π, and vector field F,

F(θ) =⎛

⎝

((1 + sin θ) sin θ)³

(1 + sin θ)⁴ , − (1 + sin θ)³sin²θ cos θ (1 + sin θ)⁴

⎞

⎠

(2%)

Since

γ^′(θ) = (cos²θ − (1 + sin θ) sin θ, cos θ sin θ + (1 + sin θ) cos θ) (2%) the line integral is

∫CF ⋅ dr =∫

π

0−sin⁴θ − sin²θ cos²θ dθ (1%)

= ∫

π

0

−sin²θ dθ = −π

2 (1%)

7. (10%) Let C be the curve formed by the intersection of the plane z = x and the surface z = x²+y². C is oriented counterclockwise when viewed from above. Evaluate∮

C

(xyz + tan⁻¹x)dx + (x²+sinh y)dy + (xz + ln z)dz.

Solution:

By Stoke’s Theorem,∮

C

F ⋅ dr =∬

S

curlF ⋅ dS =∬

S

∇ ×F ⋅ dS.

(Method I) Let S1∶= {(x, y, z) ∶ z ≥ x²+y², z = x}.

∇ ×F = RR RR RR RR RR RR RR RR RR

i j k

∂

∂x

∂

∂y

∂

∂z xyz + tan⁻¹x x²+sinh y xz + ln z

RR RR RR RR RR RR RR RR RR

= (0 − 0) i + (xy − z) j + (2x − xz) k

∬S1

∇ ×F ⋅ dS =∬

S₁^′

(0, xy − z, 2x − xz) ⋅ (−1, 0, 1)dA =∬

S^′₁

2x − x²dA, where S₁^′= {(x, y) ∶ x²−x + y²≤0}.

I - 1 : Then by polar coordinate x²−x + y²=0 ⇒ r = cos θ

⇒S₁^′ is the region enclosed by the curve r = cos θ.

⇒ ∫

π

0 ∫

cos θ

0

(2r cos θ − r²cos²θ) rdrdθ =∫

π

0

2

3r³cos θ −1

4r⁴cos²θ∣

cos θ

0

dθ

= ∫

π

0

2

3cos⁴θ −1

4cos⁶θdθ =∫

π

0

2 3(

1 + cos 2θ

2 )

2

− 1 4(

1 + cos 2θ

2 )

3

dθ

= ∫

π

0

1

6(1 + 2 cos 2θ + cos²2θ) − 1

32(1 + 3 cos 2θ + 3 cos²2θ + cos³2θ) dθ

= ( 1 6 −

1

32)π + (1 6−

3 32) ∫

π

0

cos²2θdθ =13 96π +25

96∫

π

0

1 + cos 4θ

2 dθ

= 13 96π + 7

96⋅ 1 2π = 33

192π =11 64π

I - 2 : Let

⎧⎪

⎪

⎨

⎪⎪

⎩

x = r cos θ +1 2 y = r sin θ ⇒ ∬

S^′₁2x − x²dA =∫

2π

0 ∫

1 2

0 (2 (r cos θ +1

2) − (r cos θ +1 2)

2

)rdrdθ

= ∫

2π

0 ∫

1 2

0

2r cos θ + 1 − (r²cos²θ + r cos θ +1

4)rdrdθ =∫

2π

0 ∫

1 2

0

3

4r − r³cos²θdrdθ

=2π ⋅3 8 ⋅ (

1 2)

2

− 1 4(

1 2)

4

∫

2π

0

1 + cos 2θ

2 dθ =6π 32 −

π 64 =

11 64π (Method II) Let S2∶= {(x, y, z) ∶ z = x²+y², z ≥ x},

⇒r(u, v) = (u, v, u²+v²) ⇒r_u×r_v= (1, 0, 2u) × (0, 1, 2v) = (−2u, −2v, 1)

∬S₂

∇ ×F ⋅ dS =∬

S₂^′

(0, uv − (u²+v²), 2u − u (u²+v²)) ⋅ (−2u, −2v, 1)dA

= ∬S^′₂−2uv²+2v (u²+v²) +2u − u (u²+v²)dA =∬

S₂^′2v³−3uv²+2u²v + 2u − u³dA

= ∬S^′₂

−3uv²+2u − u³,

where S₂^′= {(u, v) ∶ u²−u + v²≤0}.

II - 1 : Then by polar coordinate u²−u + v²=0 ⇒ r = cos θ

⇒S₂^′ is the region enclosed by the curve r = cos θ.

⇒ ∫

π

0 ∫

cos θ

0

(−3r³cos θ sin²θ + 2r cos θ − r³cos³θ) rdrdθ

= ∫

π

0

− 3

5r⁵cos θ sin²θ + 2

3r³cos θ −1

5r⁵cos³θ∣

cos θ

0

dθ

= ∫

π

0

− 3

5cos⁶θ sin²θ + 2

3cos⁴θ −1

5cos⁸θdθ

= ∫

π

0

− 3 5(

1 + cos 2θ

2 )

3

+ 2 3(

1 + cos 2θ

2 )

2

+ 2 5(

1 + cos 2θ

2 )

4

dθ

= ∫

π 0

1

6(1 + 2 cos 2θ + cos²2θ) − 3

40(1 + 3 cos 2θ + 3 cos²2θ + cos³2θ) +

1

40(1 + 4 cos 2θ + 6 cos²2θ + 4 cos³2θ + cos⁴2θ) dθ

= ( 1 6 −

3 40+

1

40)π + (1 6−

9 40+

6 40) ∫

π

0

cos²2θdθ + 1 40∫

π

0

cos⁴2θdθ

= 7

60π + 11 120∫

π

0

1 + cos 4θ 2 dθ + 1

40∫

π

0

(

1 + cos 4θ

2 )

2

dθ

= 7

60π + 11 120⋅

1 2π + 1

160∫

π

0

1 + 2 cos 4θ + cos²4θdθ

= 39 240+

1 160π + 1

160∫

π

0

1 + cos 8θ

2 dθ = 27 160π + 1

160⋅ 1 2π = 55

320π =11 64π II - 2 : Let

⎧⎪

⎪

⎨

⎪⎪

⎩

u = r cos θ +1 2 v = r sin θ ⇒ ∬

S^′₂

2v³−3uv²+2u²v + 2u − u³dA

= ∫

2π

0 ∫

1 2

0

(−3 (r cos θ +1

2)r²sin²θ + 2 (r cos θ +1

2) − (r cos θ +1 2)

3

)rdrdθ

= ∫

2π

0 ∫

1 2

0

−3r⁴cos θ sin²θ +7 8r −3

2r³drdθ

= ∫

2π

0

7 16⋅

1 4−

3 8⋅

1

16dθ =11 64π

8. (10%) Let S be the surface x²+y²+z²=1, x, y, z ≥ 0, an eighth of a sphere, and F = x²i + y²j + z²k. Find the outward flux of F across S.

Solution:

Method 1:

Calculate flux directly.

∬ F ⋅ n ds =∬ (x

2, y², z²) ⋅ (x, y, z) ds =∬ x³+y³+z³ds = 3∬ z³ds By symmetry.

∬ z³ds =∫

π 2

0 ∫

π 2

0

cos³φ sin φ dφdθ (5pts)

= π 2(−

cos⁴φ 4 ∣

π 2

0) = π 8 (5pts)

Thus∬ F ⋅ n ds = 3∬ z³ds =3π 8 Method 2:

Use Divergence Theorem.

∬S

F ⋅ n ds +∬

S₁∪S₂∪S₃

F ⋅ n ds =∭ divF dE (2pts) where S1 is {(x, y, z) ∈ R³∣x = 0, y ≥ 0, z ≥ 0, y²+z²≤1}

S2and S3have the same shape as S1but in y = 0 and z = 0 . First, take a look at∬

S1

F ⋅ n ds, F = (0, y², z²)while the normal vector is (−1, 0, 0) We have∬

S₁

F ⋅ n ds = 0.

For similar reason∬

S2

F ⋅ n ds =∬

S3

F ⋅ n ds = 0 (3pts) Now we have∬

S

F ⋅ n ds =∭ divF dE =∭ 2x + 2y + 2z dE = 6∭ z dE by symmetry.

∭ z dE =∫

π 2

0 ∫

π 2

0 ∫

1

0

ρ cos φ ⋅ ρ²sin φ dρdφdθ (2pts)

= 1 4⋅

π 2 ⋅

sin²φ 2 ∣

π 2

φ=0= π 16 (3pts) Thus we have∬

S

F ⋅ n ds = 6∭ z dE = 3π 8

9. (10%) Solve the initial value problem

⎧⎪

⎪

⎨

⎪⎪

⎩

y^′′+y = xe^x+sec x, −π

2 <x < π 2 y(0) = 1, y^′(0) = −3.

Solution:

The auxiliary equation is r²+1 = 0, whose roots are ±i, so the solution of y^′′+y = 0 is y_c(x) = C₁cos x + C₂sin x (2pts)

For a particular solution of y^′′+y = xe^x, we try y_p1(x) = (Ax + B)e^x. Then y_p^′′

1+y_p1= (2Ax + (2A + 2B))e^x, and then we have

⎧⎪

⎪⎪

⎪

⎨

⎪⎪

⎪⎪

⎩ A = 1

2 B = −1

2 Thus a particular solution is

yp₁(x) =1

2(x − 1)e^x (3pts)

For a particular solution of y^′′+y = sec x, we use variation of parameters to seek a solution of the form yp₂(x) = u1(x) cos x + u2(x) sin x.

If we set u^′₁cos x + u^′₂sin x = 0, then u^′₁( d

dxcos x) + u^′₂( d

dxsin x) = sec x.

⇒ { u^′₁= −tan x u^′₂=1

⇒ { u1=ln(cos x) u2=x Then we obtain

yp₂(x) = cos x ln(cos x) + x sin x (3pts)

Therefore, the general solution of y^′′+y = xe^x+sec x is y(x) = yc(x) + yp₁(x) + yp₂(x).

⇒y(x) = C1cos x + C2sin x +1

2(x − 1)e^x+cos x ln(cos x) + x sin x

To satisfy the initial conditions we require that

⎧⎪

⎪

⎨

⎪⎪

⎩

y(0) = C1− 1 2 =1 y^′(0) = C₂= −3

⇒

⎧⎪

⎪

⎨

⎪⎪

⎩ C₁=

3 2 C2= −3 Thus the required solution of the initial-value problem is

y(x) =3

2cos x − 3 sin x +1

2(x − 1)e^x+cos x ln(cos x) + x sin x (2pts)