1092 Calculus4 01-05 Final Exam Solution

1092 Calculus4 01-05 Final Exam Solution

June 19, 2021 1. (10 pts) Let D be any square in R²not containing the origin with side length a > 0. Let C be the positively oriented

boundary of D. Evaluate

∫C

⎛

⎝

2x²+y²

√

x²+y²+y⎞

⎠ dx +⎛

⎝ xy

√

x²+y²−x⎞

⎠ dy

Solution:

By Green’s Theorem

∫C

⎛

⎝

2x²+y²

√

x²+y² +y⎞

⎠ dx +⎛

⎝ xy

√

x²+y²−x⎞

⎠ dy =∬

D+

∂

∂x

⎛

⎝ xy

√

x²+y² −x⎞

⎠

−

∂

∂y

⎛

⎝

2x²+y²

√

x²+y²+y⎞

⎠ dA 至此正確 6分。沒有寫 ”Green’s Theorem”，扣 1分。

公式寫成 −∂x^∂ ( ) +_∂y^∂ ( )，扣 2分。寫成 +_∂x^∂ ( ) +_∂y^∂ ( ) 或 −_∂x^∂ ( ) −_∂y^∂ ( )，整題只給 1分，下面不必看。

= ∬D

∂

∂x

⎛

⎝ xy

√ x²+y²

⎞

⎠

−

∂

∂y

⎛

⎝

2x²+y²

√ x²+y²

⎞

⎠

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

=0

計算過程不必看，得到 0，得 2分

+

∂

∂x(−x) − ∂

∂yy dA

= ∬D

−2 dA 正負號上方已考慮了。算出 2，得 1分

= −2Area(D) = −2a² 此處的 2, a² 為獨立給分，不管另一塊偏導數計算是否算出 0。

寫出 a²，得 1分

使用 1個_::::特定的正方形，用 Green’s Theorem 或 line integral 算, 得到正確答案, 6分. 算錯, 0分.

2. (14 pts) Evaluate∬

S

z dS, where S is the part of the paraboloid z = x²+y² that lies under the plane z = 4.

Solution:

Method 1. Notice that S is the graph of the function f (x, y) = x²+y² above the disc D of radius 2 centered at origin. (2 points)

In this case, we have

dS =

√

1 + f_x²+f_y²dA =

√

1 + (2x)²+ (2y)²dA. (4 points) Therefore,

∬S

z dS =∬

D

(x²+y²)

√

1 + (2x)²+ (2y)²dA. (∗)

Using polar coordinates (x, y) = r(cos θ, sin θ), we have dA = rdrdθ, (3 points) (∗) =2π∫

2 0

r²

√

1 + 4r²⋅r dr. (1 points)

Here,

∫

2 0

r³

√

1 + 4r²dr = r²

12(1 + 4r²)^3/2∣

2 0

− 1 6∫

2 0

r(1 + 4r²)^3/2dr

=17^3/2 3 − 1

120(1 + 4r²)^5/2∣

2 0

= 17^3/2

3 − 17^5/2

120 + 1

120. (4 points; allowing small mistake) All together, one obtains∬Sz dS = 2π[¹⁷₃^3/2 −¹⁷

5/2

120 +₁₂₀¹ ].

Method 2. One uses the parametrization of S:

r(ρ, θ) = ⟨x = ρ cos θ, y = ρ sin θ, z = x²+y²=ρ²⟩, 0 ≤ ρ ≤ 2, 0 ≤ θ ≤ 2π. (2 points)

In this case,

rρ×rθ= ⟨−2ρ²cos θ, −2ρ²sin θ, ρ⟩,

(3 points; it is OK if there is sign error) dA = ∥r_ρ×r_θ∥dρdθ = ρ

√

1 + 4ρ²dρdθ. (1 points) Therefore,

∬Sz dS =∫

2π

0 ∫

2 0 ρ²⋅ρ

√

1 + 4ρ²dρdθ

=2π∫

2 0

ρ³

√

1 + 4ρ²dρ. (4 points)

Here,

∫

2 0

ρ³

√

1 + 4ρ²dr =ρ²

12(1 + 4ρ²)^3/2∣

2 0

− 1 6∫

2 0

ρ(1 + 4ρ²)^3/2dρ

= 17^3/2

3 − 1

120(1 + 4ρ²)^5/2∣

2 0

= 17^3/2

3 − 17^5/2

120 + 1

120. (4 points; allowing small mistake) All together, one obtains∬Sz dS = 2π[¹⁷₃^3/2 −¹⁷

5/2

120 +₁₂₀¹ ].

3. Let ̃C be the circle which is the intersection of x² +y²+z² = a², a > 0, and the plane 2x + 2y + z = 0. Let C be the upper part of ̃C that is above the xy-plane from (−^√a

2,^√a

2, 0) to (^√a

2, −^√a

2, 0). Consider F(x, y, z) =

⟨x³+y + 1, sin(y³) +z, x + e^z3⟩.

(a) (5 pts) Find curl F (b) (15 pts) Evaluate∫

CF ⋅ dr

Solution:

(a) curl F = RR RR RR RR RR RR RR

i j k

∂x ∂y ∂z

x³+y + 1 sin(y³) +z x + e^z3 RR RR RR RR RR RR RR

= ⟨−1, −1, −1⟩

定義正確 2分, 再看每個分量, 各 1分。若沒寫定義, 直接寫答案, 每個分量 2分, 全對 5分。

(b) ∫CF ⋅ dr, where C ∶ x²+y²+z²=a², 2x + 2y + z = 0 and z ≥ 0.

方法 1. 硬算 line integral.

1 C 的參數式7分(檢查是否 5x²+5y²+8xy = a², z = −2x − 2y) 2 ∫CF ⋅ dr = −⁵₆πa²+

√2a,8分。不必檢查計算過程, 直接看答案。正負號錯扣 1分, 其他錯誤, 就是 0分。

方法 2. 利用 Stokes’ Theorem

1 將 C 再加上 1條 curve L (不必是下方建議解法中的直線段) 構成一個 surface D (不一定是下方的半圓盤) 的 closed boundary, 正確寫下 Stokes’ Thm 的公式, 不管接下來計算正確與否, 給7分。 小錯, 如 curve L 的方向, orientation等, 扣 1分。

2 D上的 surface integral, 先看 dS = ndS, n 正確給3分, 方向錯只給 2分。計算過程不看, 答案對不管正負號給 1分。此部分共4分。

3 L上的 line integral. 只檢查最後答案, 正確給4分。視個別情形小錯扣 1分。

1 2 3 獨立計分。即使 1 中的 L 方向錯了, 3 中只檢查在這個 L 上的 line integral 的數值。

利用 Stokes; Theorem 的參考解法 1 Let L be the line segment from (^√a

2, −^√a

2, 0) to (−^√a

2,^√a

2, 0), and let D be the semi-disc oriented such that ∂D = C ∪ L

Stoke’s Theorem implies

∫C∪L

F ⋅ dr =∬

D

curl F ⋅ dS (正確給 7分) 2 D is oriented with n = ⟨²₃,²₃,¹₃⟩ 正確給3分

∬D

curl F ⋅ dS =∬

D

⟨1, −1, −1⟩ ⋅ ⟨2 3,2

3,1

3⟩dS = −5

3Area(D) = −5 3(

π 2a²)

= − 5

6πa² 正確給 1分, 2 部份共 4分

3 Let r(t) = ⟨−^√a

2t,^√a

2t, 0⟩, t ∈ [−1, 1] be a parametrization of L. Then

∫L

F ⋅ dr =∫

1

−1

⟨(−

a

√2t)

3

+ ( a

√2t) + 1, sin⎛

⎝ (

a

√2t)

3

⎞

⎠ , − a

√2t + 1⟩ ⋅ ⟨− a

√2, a

√2, 0⟩ dt

= ∫

1

−1

− a

√

2dt + 0 Since the integral of an odd function on [−1, 1] is 0

= −

√

2a. 正確給 4分

∴ ∫CF ⋅ dr = −⁵₆πa²+

√ 2a

4. Consider the vector field

F (x, y, z) = ( x

x²+y², y

x²+y², 1) . Surface S1= {(x, y, z) ∶ x²

9 + y²

4 ≤1, x²+y²≥1, z = 1} is a plane region with n = (0, 0, 1) . Surface S2= {(x, y, z) ∶ x²

9 + y²

4 ≤1, x²+y²≥1, z = 0} is a plane region with n = (0, 0, −1) . Surface S₃= {(x, y, z) ∶ x²+y²=1, 0 ≤ z ≤ 1} is a cylinder with n pointing away from the z-axis.

Surface S4= {(x, y, z) ∶ x² 9 +

y²

4 =1, 0 ≤ z ≤ 1} has n pointing away from the z-axis.

(a) (4 pts) Compute the flux of F across S1 and S2, respectively.

(b) (6 pts) Compute the flux of F across S₃. (c) (6 pts) Compute the flux of F across S4.

x

y

z

1 S1

S2

S3

S4

Solution:

(a) The flux of F across S1 is

∬S1

F ⋅ n dS =∬

S1

( x

x²+y², y

x²+y², 1) ⋅ (0, 0, 1) dS =∬

S1

dS (1 point)

=Area (S₁) =6π − π = 5π. (1 point) The flux of F across S₂ is

∬S2

F ⋅ n dS =∬

S2

( x

x²+y², y

x²+y², 1) ⋅ (0, 0, −1) dS = −∬

S2

dS (1 point)

= −Area (S₂) = −5π. (1 point)

(b) Method 1: Note that the outward normal of S3 is given by n = (x, y, 0) . (2 points) So the flux of F across S3is

∬ F ⋅ n dS =∬ (

x , y

, 1) ⋅ (x, y, 0) dS =∬ dS (2 points)

=Area (S3) =2π. (2 points)

Method 2: The surface S₃ can be parametrized as

(x, y, z) = (cos θ, sin θ, z) , 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 1, (1 point), so

n dS = (− sin θ, cos θ, 0) × (0, 0, 1) dθ dz = (cos θ, sin θ, 0) dθ dz. (2 points) Then the flux of F across S3 is

∬S₃

F ⋅ n dS =∫

1 0 ∫

2π 0

(cos θ, sin θ, 1) ⋅ (cos θ, sin θ, 0) dθ dz (2 points)

= ∫

1 0 ∫

2π

0 dθ dz = 2π. (1 point)

(c) Let E be the solid region enclosed by the surfaces S₁, S₂, S₃, and S₄. Using the divergence theorem we get

∭E

∇ ⋅F dV =

∬S1

F ⋅ n dS +∬

S2

F ⋅ n dS −∬

S3

F ⋅ n dS +∬

S4

F ⋅ n dS (3 points) Since

∇ ⋅F = ∂

∂x( x x²+y²) +

∂

∂y( y x²+y²)

=

(x²+y²) −x ⋅ 2x (x²+y²)²

+

(x²+y²) −y ⋅ 2y (x²+y²)²

=0 in E, (2 points) we obtain

5π − 5π − 2π +∬

S4

F ⋅ n dS = 0

⇒ ∬S₄

F ⋅ n dS = 2π. (1 point)

5. (a) (8 pts) Determine whether

∞

∑

n=1

( 2n 2n + 1−

2n − 1

2n ) is convergent or divergent. Compute the sum if it converges.

(b) (12 pts) Find the interval of convergence of the power series

∞

∑

n=3

(

ln(n + 1) − ln n ln(n) ) xⁿ. Solution:

(a) 此題的最後級數數值為1 − ln 2。這部分佔1分。其餘的7分配分比例如下：

(7分) 有正確地說明級數收斂。例如採用下列兩個方法：

● 觀察到0 ⩽_2n+1²ⁿ −²ⁿ⁻¹_2n = _(2n)(2n+1)¹ ⩽ _4n¹₂並使用 comparison test。

● （以下將原級數的前n項部份和記為sn。）正確說明交錯級數¹₂−¹₃+¹₄+ ⋯收斂，並（以某種形式考慮）sn與 上述交錯級數的比較，藉此說明原級數也收斂（到同一個極限）。如果只從交錯級數收斂就做出原級數收斂的

結論不能算對（參看下一條）。

(4分) 正確說明交錯級數¹₂−¹₃+¹₄+ ⋯收斂，但直接宣稱原級數收斂。

(1-3分) The answer might conclude divergence by applying tests with correct reasons to wrongly obtained expres- sions. If this is the case, the grader may give 1 to 3 points depending the situations.

(b) Let an=ln(n+1)−ln n

ln n . One see that lim_a^an

n+1 =1 and hence the radius of convergence is 1. On the other hand, by the mean value theorem we have ln(n + 1) − ln n >_n+1¹ , and hence

ln(n + 1) − ln n

ln n ⩾

1 (n + 1) ln n. Since _{(x+1) ln x}¹ is decreasing and nonnegative and∫

∞ 3

1

(x+1) ln xdx >∫

∞ 3

1

(x+1) ln(x+1)dx diverges, we see that x = 1 does not lie in the interval of convergence. As for x = −1, the series

∞

∑

n=3

(−1)ⁿln(n + 1) − ln n

ln n =

∞

∑

n=3

(−1)ⁿ(

ln(n + 1) ln n −1)

is alternating. We examine if its terms get smaller and smaller as n ↗: the function ^ln(x+1)_{ln x} −1 has derivative x ln x − (x + 1) ln(x + 1)

x(x + 1)(ln x)² <0.

Finally,

ln(n + 1) − ln n

ln n =

ln(n + 1)

ln n −1 ↘ 0

as can be seen by L’Hospital. According Leibniz’s test, −1 lies in the interval of convergence.

此題滿分12分，分成兩部分：

(i) 收斂半徑（5分）：

(5分) 正確求得收斂半徑。以下是兩種可接受的寫法：

● 將冪級數係數記為an，並正確求得

nÐ→∞lim ∣ an

an+1

∣ =1 或 lim

nÐ→∞

1

∣an∣

1 n

=1。

● 令冪級數的第n項為bn並正確算出

nÐ→∞lim ∣bn+1

b_n ∣ = ∣x∣ 或 lim

nÐ→∞∣bn∣

1 n= ∣x∣，

接著「直接」宣稱由ratio test或root test可知收斂半徑為1，或更詳細地說明∣x∣ < 1時級數收斂且∣x∣ > 1時級

數發散。這裡的重點是如果學生只寫其中一邊（多半是∣x∣ < 1這邊）是不夠的。如果只是考慮x = 1的情況就

說收斂半徑是1也是不夠的。參看下一條。

(1分) 正確說明冪級數在某端點收斂，但直接宣稱這保證收斂半徑是1。此時在收斂半徑的部分只給1分。（但是他會在端

點處理的部分的分，見下面(ii)項。）

(1-3分) 如果是想採用收斂半徑公式、ratio test或root test來求得半徑，但是極限的數值計算錯誤，可以給1至3分，由助

教自行判斷。

（7分）： 右端點x = 1佔4分、左端點x = −1佔3分，有做出適當的嘗試可得部分分，由助教自行判斷。

6. Let f (x) =∫

x 0

√

4 + t³dt.

(a) (6 pts) Write down the Maclaurin series (the Taylor series at 0) for√

4 + t³and find its radius of convergence.

(b) (6 pts) Write down the Maclaurin series for f (x) and find its radius of convergence.

(c) (8 pts) Write down T4(x), the fourth degree Taylor polynomial of f (x) at 0.

Estimate ∣f (¹₂) −T4(¹₂)∣. (You need to justify your answer.)

Solution:

(a)

√

4 + t³=2

√ 1 +t³

4 =2 (

∞

∑

n=0

Cn¹²( t³

4)

n

) =

∞

∑

n=0

Cn¹²⋅ 1

2²ⁿ⁻¹t³ⁿ. (3pts)

The radius of convergence of the binomial series is 1. Thus the series converges absolutely if ∣^t₄³∣ <1 and it diverges if ∣^t₄³∣ >1 (2 pts). Hence the radius of convergence of the series is √³

4 = 2²³. (1 pt) ( Or by the Ratio Test,

lim

n→∞∣ C

1 2

n+1⋅_22n+1¹ t³ⁿ⁺³ C

1

n2 ⋅₂2n−1¹ t³ⁿ

∣ = lim

n→∞

n −¹₂ n + 1

∣t∣³ 4 =

∣t∣³

4 . (2pts)

Hence, the Maclaurin series converges if ∣^t₄³∣ < 1 and the series diverges if ∣^t₄³∣ > 1. Hence the radius of convergence is √³

4 = 2²³. (1 pt) )

(b)

f (x) =∫

x 0

√

4 + t³dt =∫

x 0 (

∞

∑

n=0

C

1

n2 ⋅ 1

2²ⁿ⁻¹t³ⁿ)dt =

∞

∑

n=0

C

1

n2

1 2²ⁿ⁻¹ ⋅

1

3n + 1x³ⁿ⁺¹.(3pts)

The radius of convergence is still 2²³ because a power series and its integral have the same radius of conver- gence. (3 pts)

( Or you can apply the Ratio Test,

n→∞lim∣

C_n+1¹² ⋅_22n+1¹ _3n+4¹ x³ⁿ⁺⁴ C

1

n2 ⋅_22n−1¹ _3n+1¹ x³ⁿ⁺¹

∣ = lim

n→∞

n −¹₂ n + 1

3n + 1 3n + 4

∣x∣³ 4 =

∣x∣³

4 . (2pts)

Hence, the series converges if ∣^x₄³∣ <1 and the series diverges if ∣^x₄³∣ >1. Hence the radius of convergence is

√3

4 = 2²³. (1 pt) )

(c) T₄(x) = 2x + 1

16x⁴. (2 pts) f (¹₂) =2 ⋅¹₂+ ∑^∞_n=1C

1

n2 1 2²ⁿ⁻¹

1 3n+1

1

2³ⁿ⁺¹ =1 + ∑^∞n=1C

1

n2 1 2⁵ⁿ

1

3n+1 and the series

∞

∑

n=1

C

1

n2 1 2⁵ⁿ

1

3n+1 is an alternating series (2 pts).

Moreover, ∣C_n¹²₂¹_5n3n+1¹ ∣is decreasing and approaches 0 as n → ∞

(If students CLAIM / MENTION that the sequence is decreasing and tends to 0, then they get 2 points.

The complete argument is as below.

The sequence ∣C

1

n2 1 2⁵ⁿ

1

3n+1∣is decreasing because

∣C_n+1¹² _25n+5¹ _3n+4¹ ∣

∣Cn¹² 1 2⁵ⁿ

1 3n+1∣

= n −¹₂ n + 1

1 2⁵

3n + 1 3n + 4<

1

2⁵ <1 for n ≥ 1.

Also by the above ratio, we know that ∣C

1

n2 1 2⁵ⁿ

1

3n+1∣ ≤ ₂¹5n for n ≥ 1. Thus it approaches 0 as n → ∞.) Hence by the Alternating Series Estimation Theorem,

∣f (1 2) −T4(

1 2)∣ = ∣

∞

∑

n=2

C

1

n2

1 2⁵ⁿ

1

3n + 1∣ ≤ ∣C

1 2

2

1 2¹⁰

1 6 + 1∣ =

1

2¹³7 (2pts)

Students may use Taylor’s inequality to estimate the error. Writing down the inequality ∣f (1 2) −T₄(

1 2)∣ ≤ M

5!( 1

2)⁵, where M is an upper bound for ∣f⁽⁵⁾(x)∣ on the interval [0,¹₂], students can get 2 points.

It is complicated to estimate ∣f⁽⁵⁾(x)∣. If students have some progress in computing ∣f⁽⁵⁾(x)∣, they can get 1 more point.