P (x, y)i + Q(x, y)j and ∂P∂y = ∂Q∂x on an open connected region D, then F is conservative on D

1032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. (5%) Determine the statement is true (◯) or false (⨉).

(a) If f (x, y) is continuous on the rectangle R = {(x, y)∣ a ≤ x ≤ b, c ≤ y ≤ d} except for finitely many points, then f (x, y) is integrable on R and

∬R

f (x, y)dA =∫

d c ∫

b a

f (x, y)dxdy =∫

b a ∫

d c

f (x, y)dydx.

(b) If F(x, y) = P (x, y)i + Q(x, y)j and ^∂P_∂y = ^∂Q_∂x on an open connected region D, then F is conservative on D.

(c) If curl F=curl G on R³, then∫CF ⋅ dr =∫CG ⋅ dr for all closed path C.

(d) If F and G are vector fields and curl F=curl G, div F=div G, then F − G is a constant vector field.

(e) Let B be a rigid body rotating about the z-axis with constant angular speed ω. If v(x, y, z) is the velocity at point (x, y, z) ∈ B, then curl v is parallel to k.

Answer. (每小題各 1 分) (a) (b) (c) (d) (e)

⨉ ⨉ ⨉ ⨉ ◯

2. (10%) Write the integral ∫

1 0 ∫

1

√x∫

1−y 0

f (x, y, z)dzdydx in 5 other orders.

Answer. (每小題錯一格扣 1 分，錯兩格以上全錯)

(a) ∫ 1

0 ∫

y²

0 ∫

1 − y

0 f (x, y, z)dzdxdy

(b) ∫ 1

0 ∫

1 − z

0 ∫

y²

0 f (x, y, z)dxdydz

(c) ∫ 1

0 ∫

1 − y

0 ∫

y²

0 f (x, y, z)dxdzdy

(d) ∫ 1

0 ∫

1 −√ x

0 ∫

1 − z

√x f (x, y, z)dydzdx

3. (15%) Evaluate the integrals.

(a) ∫

1 0 ∫

π 4

tan⁻¹y

cos x ⋅ tan(cos x)dxdy.

(b) ∫

√ 2

1 ∫

√ 2−y² 0

x + y

x²+y²dxdy +∫

1 0 ∫

1 1−y

x + y

x²+y²dxdy +∫

√ 2

1 ∫

√ 2−x² 0

x + y x²+y²dydx.

Solution:

(a)

∫

1 0 ∫

π 4

tan⁻¹y

cos x tan (cos x)dxdy = ∫

π 4

0 ∫

tan x 0

cos x tan (cos x)dydx (3pt)

= ∫

π 4

0

sin x tan (cos x)dx (Let u = cos x, du = − sin xdx)

= − ∫

√1 2

1

tan udu

= ln (cos u)∣

√1 2

1 (2pt)

= ln (cos 1

√2) −ln (cos 1) (1pt)

(b)

∫

√ 2

1 ∫

√

2−y² x + y

x²+y²dxdy +∫

1 0 ∫

1 1−y

x + y

x²+y²dxdy +∫

√ 2

1 ∫

√ 2−x² 0

x + y x²+y²dydx

= ∬

D

x + y

x²+y²dA, where D is bounded by x²+y²=2 and x + y = 1.

By polar coordinate, we have

∬D

x + y

x²+y²dA = ∫

π 2

0 ∫

√ 2

cos θ+sin θ1

r cos θ + r sin θ

r² ⋅rdrdθ (4pt)

= ∫

π 2

0 ∫

√ 2

cos θ+sin θ1

(cos θ + sin θ)drdθ

= ∫

π 2

0

√

2(cos θ + sin θ) − 1dθ

=

√

2(sin θ − cos θ)∣

π 2

0 − π 2 (3pt)

= 2√ 2 −π

2 (2pt)

Page 2 of 9

4. (10%) Let S be the surface x²+y²+z²=a², x ≥ 0, y ≥ 0, z ≥ 0 (a > 0), and let C be the boundary of S. Find the centroid of C.

Solution:

For a quarter circle of radius a (named C^′) on a plane, its centroid can be found to be at (^2a_π,^2a_π)by either way:

(1) Parametrize the curve:

Parametrize C^′by r(t) = ⟨a cos t, a sin t⟩, t ∈ [0,^π₂].

⇒ ∣r^′(t)∣ = ∣⟨−a sin t, a cos t⟩∣ = a.

Arc length s =¹₄⋅2πa = ¹₂πa.

x ⋅ s =∫

C^′

x ⋅ ds =∫

π 2

0

x(t)∣r^′(t)∣dt =∫

π 2

0

a cos(t)adt = a²

∴x = a²

1 2πa=

2a π By the symmetry of the arc, y = x =^2a_π. (2) Pappus’s Theorem:

Knowing that the surface area of a hemisphere of radius a is 2πa² and the arc length of a quarter circle of radius a is ¹₂πa, if the quarter circle is in the first quadrant and is rotated about the x-axis, Pappus’s Theorem gives

A = 2πy ⋅ s

⇒2πa²=2πy ⋅πa 2

⇒y =2a π By the symmetry of the arc, x = y =^2a_π.

(7 points up to this point.)

The curve C is composed of quarter circles C1, C2, and C3 on the xy-, yz-, and xz-planes, respectively. By the above discussion, their centroids are (^2a_π,^2a_π, 0), (0,^2a_π,^2a_π), and (^2a_π, 0,^2a_π), respectively. Since they have equal masses, the centroid of C is the average of them, namely (^4a_3π,^4a_3π,_3π^4a). (3 points)

(Note: if you misunderstood the problem but correctly calculated the centroid of the surface S to be at (^a₂,^a₂,^a₂), you still get 4 points. But no points will be given if you calculated the centroid of the part of the volume inside the sphere in the first octant.)

5. (10%) Let C be the curve of intersection of x²+y²+z²=4, x²+y²=2x, z ≥ 0, oriented C to be counterclockwise when viewed from above. Evaluate∫

C

y²dx + z²dy + x²dz.

Solution:

ˆ Solution 1: Using line integral to solve this problem directly.

r(θ) =< 1 + cos θ, sin θ, 2 sin^θ₂ >, θ ∈ [0, 2π]. (3 points) The original equation =∫

2π

0 [−sin³θ + sin^{2 θ}₂cos θ + (1 + cos θ)²cos^θ₂]dθ (3 points) By symmetry, the first and the third term will be zero in the end.

Therefore, the above equation will change as follows:

4∫

2π

0 sin^{2 θ}₂ dθ

=4∫

2π 0

1−cos θ

2 cos θ dθ

= −2π (4 points)

ˆ Solution 2: Using Stokes’ Theorem to solve this problem.

F =< y², z², x²>

∇ ×F =< −2z, −2x, −2y > (2 points) r(x, y) =< x, y,

√

4 − x²−y²>

rx=<1, 0,^{√ −x}

4−x²−y² >

r_y=<0, 1,^{√ −y}

4−x²−y² >

r_x×r_y=< ^{√ x}

4−x²−y²,^{√ y}

4−x²−y², 1 > (2 points) By Stokes’ Theorem,

∮cF ⋅ dr =∬S(∇ ×F ) ⋅ dS

= ∬D< −2√

4 − x²−y², −2x, −2y > ⋅ <^{√ x}

4−x²−y²,^{√ y}

4−x²−y², 1 > dA

= ∬D(−2x −^{√ 2xy}

4−x²−y² −2y) dA (3 points)

By symmetry, the second and the third term will be zero in the end.

Therefore, the above equation will change as follows:

−2∫

π

−π2 2 ∫

2 cos θ

0 r cos θrdrdθ

−2∫

π

−π2 2

cos θ (¹₃r³) ∣^{2 cos θ}₀ rdrdθ

= ⁻¹⁶₃ ∫

π

−π2 2

cos⁴θ dθ

= ⁻³²₃ ∫

π 2

0 cos⁴θ dθ

= ⁻³²₃ ∫

π 2

0 (^{cos 2θ−1}₂ )² dθ

= ⁻⁸₃ ∫

π 2

0 (cos²2θ − 2 cos 2θ + 1) dθ

By symmetry, the second term will be zero in the end.

Therefore, the above equation will change as follows:

= ⁻⁸₃ ⋅³₂⋅^π₂

= −2π. (3 points)

Page 4 of 9

6. (20%) Let F = _(x^(x−y)2+y²²)^y2i +^−(x−y)_(x2+y²²)^x2j.

(a) Verify that F is conservative on the right half plane x > 0. Find a potential function of F on the right half plane.

(b) Evaluate∮

C1

F ⋅ dr where C1 is the ellipse ^x₄² + (y − 2)²=1.

(c) Evaluate∫

C₂

F ⋅ dr where C₂ is the curve with polar equation r = e^∣θ∣, −^9π₄ ≤θ ≤ ^9π₄.

x y

Solution:

(a) (6%) ∂_yf =^−(x−y)_(x2+y2^2x

)² Ô⇒ f =_(x2^−x_+y²₂

)² −tan⁻¹(^y_x) +g(x)

Ô⇒ ∂xf =_(x^−2xy2+y²²)²+_(x2+y^y2)² +g^′(x) = _(x^(x−y)2+y²²)^y2+g^′(x) =_(x^(x−y)2+y²²)^y2

Ô⇒ g^′(x) = 0 Ô⇒ g is constant Ô⇒ f =_(x2^−x_+y²₂

)² −tan⁻¹(_x^y)or 1 +_(x2^−x_+y²₂

)² −tan⁻¹(^y_x) = ^y

2

(x²+y²)²−tan⁻¹(^y_x) (6%) Other point: Py=Qx=^x

4−4x³y+4xy³−y⁴

(x²+y²)³ (2%); ”Py=Qx” implies f is conservative (1%) (b) (4%) Since {y > 0} is simple connected, F is conservative on {y > 0}.

On the other hand, C1 is closed curve on y > 0 (1%); therefore,∮C₁F ⋅ dr = 0 (3%) (c) (10%) ”method 1”

the integral on C₂is equal to the integral on unit circle times two and integral on the tail.

∫DF ⋅ dr =∫

2π

0 −(cos θ − sin θ)²dθ = −2π (4%), where D is unit circle.

The integral on tail is independent of path, which equals to

f (cos^9π₄ e^9π4 , sin^9π₄ e^9π4) −f (cos^−9π₄ e^−9π4 , sin^−9π₄ e^−9π4 ) = −^π₂(2%), where f is potential function of F Therefore,∫C2F ⋅ dr = −4π −^π₂ = −^9π₂(2%)

”method2”

∫C2F ⋅ dr =∫γ1F ⋅ dr +∫γ2F ⋅ dr, where γ1 is the ”θ ≥ 0” part of C2, γ2is ”θ < 0” part of C2, in which x(θ) and y(θ) is differentiable.

∫C₂F ⋅ dr =∫

9π 4

0 −(cos θ − sin θ)²dθ + ∫

0

−^9π₄ −(cos θ − sin θ)²dθ (4%)

= ∫

9π 4

−^9π₄ −(cos θ − sin θ)²dθ = −^9π₂ (6%) (the answer worth 2 point)

7. (10%) Evaluate∫

C

(y + sin³x)dx + (z²+cos⁴y)dy + (x³+tan⁵z)dz where C is the curve r(t) = sin t i + cos t j + sin 2t k, 0 ≤ t ≤ 2π. [Hint : C lies on the surface z = 2xy.]

Solution:

First observe that r(t) = sin t i + cos t j + sin 2t k is negative oriented.

Thus by Stoke’s theorem:

∫C

(y + sin³x) dx + (z²+cos⁴y) dy + (x³+tan⁵z) dz = −∬

S

∇F ⋅ dS (2%)

where S is the surface z = 2xy bounded by D = {x²+y²≤1}

∇F = RR RR RR RR RR RR RR

i j k

∂

∂x

∂

∂y

∂

∂z

y + sin³x z²+cos⁴y x³+tan⁵z RR RR RR RR RR RR RR

= −2z i − 3x² j − k (2%)

n = (−⃗ ∂z

∂x, −∂z

∂y, 1)/∥(−∂z

∂x, −∂z

∂y, 1)∥ = (−2y, −2x, 1)/∥(−2y, −2x, 1)∥ (2%)

∬S

∇F ⋅ dS =∬

D

(−2z, −3x², −1) ⋅ (−2y, −2x, 1) dA

= ∬D

4yz + 6x³−1 dA

= ∬D

8x²y + 6x³−1 dA

= ∫

2π

0 ∫

1 0

(8r³cos θ sin²θ + 6r³cos³θ − 1)r dr dθ

⋮

= −π (4%)

∫C

(y + sin³x) dx + (z²+cos⁴y) dy + (x³+tan⁵z) dz = π

Page 6 of 9

8. (10%) Evaluate∬

S

xdS where S is the part of the cone z =√

2(x²+y²)that lies below the plane z = 1 + x.

Solution:

Step1.

Find the projection onto the xy-plane of the curve of intersection of the cone z =

√

2(x²+y²) and the plane z = 1 + x.

{ z =

√

2(x²+y²) z = 1 + x

⇒ 2(x²+y²) = (x + 1)²

⇒ (

x − 1

√2 )

2

+y²=1 (1pt)

Step2.

If we regard x and y as parameters, then we can write the parametric equations of S as x = x y = y z =

√

2(x²+y²) (1pt) where

1 −

√

2(1 − y²) ≤x ≤ 1 +

√

2(1 − y²) , −1 ≤ y ≤ 1 (1pt) and the vector equation is

r(x, y) = xi + yj +

√

2(x²+y²)k Step3.

Find ∣rx×ry∣.

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

rx=1i + 0j +

√

√ 2x x²+y²k ry=0i + 1j +

√

√ 2y x²+y²k

⇒ r_x×r_y=

−

√2x

√ x²+y²

i + −

√2y

√ x²+y²

j + 1k (2pts)

⇒ ∣r_x×r_y∣ =

√

3 (1pt) Step4.

Evaluate xdS.

9. (10%) Let S be the surface of the solid bounded by x²+y²+z²=1 and z ≥ ¹₂. Find the total flux of F(x, y, z) = x²i + y²j + z²k across S.

Solution:

(Method I)

Let V = {(x, y, z) ∈ R³∣x²+y²+z²≤1, z ≥¹₂}, then by Divergence Theorem, Flux of F =∬

S

F ⋅ dS =∭

V

divF dS =∭

V

2(x + y + z) dS (3%)

From the symmetry of V , we have∭

V

x dS =∭

V

y dS = 0.

Therefore,∭

V

divF dS =∭

V

2z dS = 2∫

2π 0 ∫

π 3

0 ∫

1

1

2sec φρ cos φρ²sin φdρdφdθ

∫

2π

0 ∫

π 3

0 ∫

1

1 2sec φ

ρ cos φρ²sin φdρdφdθ = 1 4∫

2π

0 ∫

π 3

0

ρ⁴cos φ sin φ∣

1

1 2sec φ

dφdθ

= 2π

4 ∫

π 3

0

cos φ sin φ − 1

16tan φ sec²φ dφ = 2π

4 ( 1

2sin²φ − 1

32tan²φ)

π 3

0

= 9π 64

⇒ ∬

S

F ⋅ dS =∭

V

2z dS = 2 ⋅^9π₆₄ = ₃₂⁹π (7 %) (Method II)

Let S1= {(x, y, z) ∈ R³∣x²+y²+z²=1, z ≥¹₂}and S2= {(x, y, z) ∈ R³∣x²+y²≤³₄, z = ¹₂} Flux of F = ∬

S

F ⋅ dS =∬

S₁

(x², y², z²) ⋅dS +∬

S₂

(x², y², z²) ⋅dS

= ∬

S1

(x², y², z²) ⋅ (x, y, z) dS +∬

S2

(x², y², z²) ⋅ (0, 0, −1) dS.

= ∬

S1

x³+y³+z³ dS −∬

S2

z² dS =∬

S1

x³+y³+z³ dS −1

4Area(S2). (3%) From the symmetry of S1, we have∬

S₁

x³dS =∬

S₁

y³dS = 0.

Thus, we only need to calculate∬

S₁

z³ dS, then by Spherical coordinate

∬

S₁

z³ dS =∫

2π

0 ∫

π 3

0

cos φ³sin φ dφdθ = 2π ⋅ (−1 4 cos⁴φ)

π 3

0

=

−π 2 [(

1 2)

4

−1] = 15 32π.

⇒ ∬

S

F ⋅ dS =∬

S₁

z³ dS −¹₄Area(S2) = ^15π₃₂ −¹₄⋅ (

√ 3 2 )

2

π = ₃₂⁹π (7 %)

Page 8 of 9

10. (10%) Solve the differential equation

y^′′+y = x²e^x+tan x, x ∈ (−π 2,π

2).

Solution:

Complementary equation: y^′′+y = 0.

Auxiliary equation: r²+1 = 0 ⇒ r = ±i ⇒ y_c=c₁sin x + c₂cos x. (2 points) For the particular solution:

(1) For y^′′+y = x²e^xit’s a better idea to use the method of undetermined coefficients:

Let yp₁= (Ax²+Bx + C)e^x,

y^′_p1= [Ax²+ (2A + B)x + (B + C)]e^x, y^′′_p1= [Ax²+ (4A + B)x + (2A + 2B + C)]e^x.

y^′′_p1+yp1= [2Ax²+ (4A + 2B)x + (2A + 2B + 2C)]e^x≡x²e^x

⇒A =¹₂, B = −1, C = ¹₂

∴y_p1= (¹₂x²−x + ¹₂)e^x.

(2 points for the formulation, 2 points for solving the coefficients.) (2) For y^′′+y = tan x we use the method of variation of parameters:

Let yp₂=u1sin x + u2cos x, y_p^′₂ = (u^′₁sin x + u^′₂cos x) + u1cos x − u2sin x.

Setting u^′₁sin x + u^′₂cos x = 0 (equation 1), we have y^′′_p2=u^′₁cos x − u^′₂sin x − u1sin x − u2cos x.

⇒y_p^′′₂+yp₂=u^′₁cos x − u^′₂sin x ≡ tan x (equation 2).

Solving the system of equations 1 and 2, we have

{u^′₁sin x + u^′₂cos x = 0 u^′₁cos x − u^′₂sin x = tan x

⇒ {

u^′₁(x) = sin x,

u^′₂(x) = −tan x sin x = −^sin

2x cos x = ^cos

2x−1

cos x =cos x − sec x

⇒ {u1(x) = −cos x,

u2(x) = sin x − ln ∣ sec x + tan x∣ =sin x − ln(sec x + tan x) for x ∈ (−^π₂, −^π₂)

⇒yp₂(x) = u1sin x + u2cos x = −(cos x) ln(sec x + tan x) (2 points for the system of equations, 2 points for solving and integrating them.) Combining the above results, we have the general solution

y(x) = c1sin x + c2cos x + (1

2x²−x +1

2)e^x− (cos x) ln(sec x + tan x)