微積分甲

982微積分甲^01-05班期末考解答與評分標準

1. (10%) Let f (x, y) = xⁿ+ yⁿ

2 , x > 0, y > 0, n > 2, n∈ N.

(a) Apply the method of Lagrange multipliers to find the extreme values of the function f (x, y) on the line x + y = C, where C > 0. (No credits if the method is not used.) (b) Use part (a) to prove the inequality xⁿ+ yⁿ

2 ≥ (x + y

2 )ⁿ, x > 0, y > 0, n∈ N.

Sol:

(a) Apply lagrange multiplier:

nxⁿ⁻¹

2 = λ, nyⁿ⁻¹ 2 = λ x + y = C

Slove it to get (x, y) = (c 2,c

2)

Because x + y = C, x> 0, y > 0 is bounded and closed line segment and f (x, y)is differentiable on this line segment.

It must contain maximum and minimun value.

Compare (c 2, c

2) with end points (c, 0), (0, c) f (c

2,c 2) = (c

2)ⁿ is minimum on x + y = C, x> 0, y > 0 f (0, c) = f (c, 0) = cⁿ

2 is maximum on x + y = C, x> 0, y > 0 thus f (c

2, c 2) = (c

2)ⁿ is minimum on x + y = C, x > 0 , y > 0 (b) n = 1, x + y

2 > x + y 2 n = 2, x²+ y²

2 − (x + y

2 )² = (x− y)²

4 > 0, so x²+ y²

2 > (x + y 2 )² n > 2, when x + y = C , C > 0 . By (a), xⁿ+ yⁿ

2 > (c

2)ⁿ = (x + y 2 )ⁿ thus xⁿ+ yⁿ

2 > (x + y

2 )ⁿ, x > 0, y > 0, n∈ N 評分標準^:

Use lagrange method to find possible extreme values. (5 pts) Verify f (c

2,c

2) is the extreme value(minimum). (3 pts) prove inequality xⁿ+ yⁿ

2 > (x + y

2 )ⁿ. (2 pts) 2. (8%) Evaluate

Z ₁

0

Z _y

y 2

y³e^x5dxdy + Z ₂

1

Z ₁

y 2

y³e^x5dxdy.

Sol:

Change the integral order (2 pts)

the domain of integral:x≤ y ≤ 2x, 0 ≤ x ≤ 1 Z 1

0

Z 2x x

y³e^x5dydx (2 pts)

= Z ₁

0

1

4y⁴e^x5¯¯¯^y=2x

y=x

dx

= Z 1

0

15

4 x⁴e^x5dx (2 pts)

= 3

4e^x5¯¯¯^x=1

x=0

= 3

4(e− 1) (2 pts)

3. (8%) Evaluate ZZ

D

x²

px²+ y² dA, where D ={(x, y) ∈ R²|1 ≤ x²+ y² ≤ 4, y ≥ x}.

Sol:

Using polar coordinate (1 pt) D ={(r, θ)|1 ≤ r ≤ 2,π

4 ≤ θ ≤ 5π

4 } (2 pts) Z 5π/4

π/4

Z 2 1

r²cos²θ

r rdrdθ (1 pt)

= Z 5π/4

π/4

1 3r³¯¯¯^r=2

r=1cos²θdθ (1 pt)

= 7 3(θ

2 +sin 2θ

4 )¯¯¯^θ=5π/4

θ=π/4

(2 pts)

= 7π

6 (1 pt) 4. (14%) (a) Evaluate I₁ =

ZZ

R1

e^{−(x2+xy+y2)}dA, where R₁ ={(x, y)|x²+ xy + y² ≤ 1}.

(b) Evaluate I₂ = ZZ

R2

x²y²dA, where R₂ is the region bounded by xy = 1, xy = 2, y = x, y = 4x, and x > 0, y > 0.

Sol:

(a) Method1:

x²+ xy + y² = (x + 1

2y)²+ 3 4y² Let u = x +1

2y, v =

√3y

2 , x = u− v

√3, y = 2v

√3 (1 pt) Preimage of R₁ is R⁰₁ ={(u, v)|u²+ v² ≤ 1} (1 pt)

J =

¯¯¯¯

¯¯¯¯

1 −1√ 3

0 2

√3

¯¯¯¯

¯¯¯¯= 2

√3 (2 pts)

I₁ = Z Z

R1

e^{−(x2+xy+y2)}dA = Z Z

R⁰₁

e^−(u2+v2) 2

√3dudv

= 2

√3 Z _2π

0

Z ₁

0

e^−r2r drdθ

= 2

√3 Z 2π

0

1 2(1− 1

e) dθ

= 2

√3π(1−1

e) (3 pts) Method2:

Let u = (x + y)

√2 , v = −x + y√

2 (1 pt) x²+ xy + y² = 3

2u²+1

2v² ≤ 1 (1 pt) Z Z

R1

e^{−(x2+xy+y2)}dA = Z Z

3

2u²+¹₂v²≤1

e^{−(32u2+12v2)}dudv

= 1

√3 Z Z

η²+v²≤2

e^−η2+v22 dηdv (2 pts)

= 1

√3 Z 2π

0

Z ^√2 0

e^−r22 r drdθ

= 2

√3π(1− 1

e) (3 pts) (b) Let u = xy, v = y

x x =

ru

v, y =√

uv (1 pt)

Preimage of R₂ is R⁰₂ ={(u, v)|1 ≤ u ≤ 2, 1 ≤ v ≤ 4} (1 pt)

J =

¯¯¯¯

¯¯¯ 1

2u⁻¹² v⁻¹² −1 2 u¹²v⁻³² 1

2u⁻¹² v⁻¹² 1 2u¹²v⁻¹²

¯¯¯¯

¯¯¯

= 1

2v (2 pts)

I₂ = Z Z

R2

x²y²dA = Z Z

R2

u² 1 2v dudv

= 1 2

Z ₄

1

Z ₂

1

u²v⁻¹dudv

= 1 2

Z 4 1

7

3v⁻¹dv = 7

3ln 2 (3 pts) 5. (18%) Let F(x, y) =

D x− y

x²+ y², x + y x²+ y²

E

, (x, y)6= (0, 0), and H = {(x, y)|y > 0} be the upper half plane.

(a) Compute J₁ = Z

C

F· dr, where C is the unit circle with counterclockwise orientation. Is F conservative on R²\ {(0, 0)}?

(b) Show that Z

C

F· dr = 0 for any piecewise-smooth simple closed curve C in H.

(c) Suppose that Γ is a piecewise-smooth simple curve in H with initial point P₁ = (r₁cos θ₁, r₁sin θ₁) and terminal point P₂ = (r₂cos θ₂, r₂sin θ₂), r_j > 0, 0 < θ_j < π, j = 1, 2. Evaluate J₂ =

Z

Γ

F· dr in terms of r1, r₂, θ₁, and θ₂. (Hint. Try a path with constant r in one piece and constant θ in another piece.)

Sol:

(a) Let r(θ) = (cos θ, sin θ), 0≤ θ < 2π, be the parametric equation for C.

J₁ = Z

C

F· dr = Z _2π

0

F· r⁰(θ)dθ = Z _2π

0

(cos²θ + sin²θ)dθ = 2π 6= 0

Therefore, F is NOT conservative onR²− {(0, 0)}.

(b) LetC be a piecewise-smooth simple closed curve in H and D be the region bounded by C.

GiveC a counterclockwise orientation so that ∂D = C. By Green’s theorem, since D ⊆ H, on which F is well-defined, and H is simply connected,

Z

C

F· dr = ZZ

D

· ∂

∂x

µ x + y x² + y²

¶

− ∂

∂y

µ x− y x² + y²

¶¸

dA = ZZ

D

0dA = 0

(c) It follows from (b) that Z

C

F· dr is independent of path in H. Thus, to compute Z

Γ

F· dr, letC = C1∪ C2 be a path connecting P₁ and P₂.

C1 : r₁(t) = (r₁cos t, r₁sin t), t goes from θ₁ to θ₂

C2 : r2(t) = (t cos θ2, t sin θ2), t goes from r1 to r2

Then it’s easy to see that F(r₁(t))· r⁰1(t) = 1 and F(r₂(t))· r⁰2(t) = 1 t. Z

Γ

F· dr = Z

C1∪C2

F· dr = Z θ2

θ1

1dt + Z r2

r1

1

tdt = θ2− θ1+ ln µr₂

r₁

¶

= J2

6. (12%) Evaluate I = Z

C

y²

√R²+ x²dx + h

4x + 2y ln(x +√

R²+ x²) i

dy, where C is the upper semicircle x² + y² = R², y ≥ 0, R > 0, and is traversed from A(−R, 0) to B(R, 0). (Hint.

Apply Green’s Theorem.)

Sol:

We want to find I, where I is I =

Z

C

y²

√R²+ x²dx + [4x + 2y log(x +√

R²+ x²)]dy = Z

C

P dx + Qdy Worth 2 pts:

Before we do anything, we must deal with our curve C. In order to use Green’s theorem we must make sure C satisfies all of the assumptions for the theorem, that is that C is a positively oriented, piecewise smooth, and simple closed curve. However, C does not satisfy these assumptions; hence we must make it so a line segment, call L, runs from B(R, 0) to A(−R, 0). Now we see that CUL is closed and piecewise smooth with clockwise orientation.

Clockwise orientation is not positive orientation, so we will have a negative sign out in front.

Worth 4 pts:

Now note that I = I + J J being:

J = Z

L

P dx + Qdy

Note that for J , dy = 0 ⇒=

Z

L

0dx + Q∗ 0 Hence J = 0

Worth 2 pts:

Now we can deal with the domain bounded by CUL, which we call D. By Green’s Theorem:

I = I + J

= Z

CU L

P dx + Qdy

=− ZZ

D

(Qx− P y)dA

Worth 2 pts:

Finding Qx and P y: Qx = 4 + 2y

√R²+ x², P y = 2y

√R²+ x²

Worth 2 pts:

Plugging the above into I gives: I =− ZZ

D

4dA =−2πR²

7. (12%) Evaluate ZZ

S

curlG· dS, where G(x, y, z) = x²yzi + yz²j + z³e^xyk, S is the part of the

sphere x²+ y²+ z² = 5 that lies above the plane z = 1, and S is oriented upward.

Sol:

Let C be the boundary of the surface S.

Then C can be parameterized as (x, y, z) = (2 cos θ, 2 sin θ, 1), 0 ≤ θ ≤ 2π. (4 pts) G(x, y, z) = (8 sin θ, 2 sin θ, e4 sin θ cos θ), dr = (dx, dy, dz) = (−2 sin θ, 2 cos θ, 0).

By Stoke’s theorem, ZZ

S

curlG· dS = Z

C

G· dr (4 pts)

= Z _2π

0

−16 sin²θ cos²θ + 4 sin θ cos θdθ

= Z 2π

0

−4 sin²2θ + 2 sin 2θdθ

=−4π (4 pts)

8. (18%) Let B be the ball centered at the origin with radius ρ₀ > 0, and W be the smaller wedge cut from B by two planes y = 0 and y = √

3x. The boundary of W consists of 3 surfaces S₁, S₂, and S₃: ∂W = S₁∪ S2 ∪ S3, and is given with outward orientation. Here S₁ and S₂ are semidisks on y = 0 and y = √

3x, respectively, and S₃ is on the boundary of B, a sphere of radius ρ₀. See the figure. Let H(x, y, z) = xzi + yj− x²k.

(a) Find a parametric representation for the surface S3. (b) Compute

ZZ

S3

H· dS.

(c) Compute ZZ

S1∪S²

H· dS. (Hint. Use the Divergence Theorem.)

Sol:

(a) Note that the surface S₃ is a piece of the sphere {(x, y, z) : x² + y² + z² = ρ²₀}. The spherical coordinate works. So let x = ρ₀sin φ cos θ, y = ρ₀sin φ sin θ, and z = ρ₀cos φ.

Notice that θ ∈ [0, π/3] and φ ∈ [0, π]. (2 pts) The parametric representation is given by

r(θ, φ) = (ρ0sin φ cos θ, ρ0sin φ sin θ, ρ0cos φ), θ ∈ [0, π/3], φ ∈ [0, π]. (2 pts)

(b) Since S₃ is a piece of the sphere, the outward normal of S₃ is given by

n(x, y, z) = (x, y, z)

ρ₀ , (x, y, z)∈ S3. And H(x, y, z)· n(x, y, z) = y²

ρ0

. So the integral is given by ZZ

S3

H· dS = ZZ

S3

H· ndS = ZZ

R

y²

ρ₀ρ²₀sin φdθdφ = Z _π/3

0

Z _π

0

ρ³₀sin³φ sin²θdφdθ (4 pts)

Here R = [0, π/3]× [0, π]. Note that Z π/3

0

Z π 0

ρ³₀sin³φ sin²θdφdθ = ρ³₀ Z π/3

0

sin²θdθ Z π

0

sin³φdφ

For the first integral, consider sin²θ = (1− cos 2θ)/2.

Z _π/3

0

sin²θdθ = Z _π/3

0

1− cos 2θ

2 dθ =

µθ 2 −1

4sin 2θ¶¯¯

¯¯^π/3

0

= π 6 −

√3 8 For the later one, we have

Z π 0

sin³φdφ =− Z π

0

(1− cos²φ)d cos φ = µ1

3cos³φ− cos φ¶¯¯

¯¯^π

0

= 4 3

So ZZ

S3

H· dS = 4 3ρ³₀

à π 6 −

√3 8

!

(3 pts)

(c) ∇ · H = z + 1. So by divergence theorem, ZZ

S1∪S²

H· dS + ZZ

S3

H· dS = ZZZ

E

∇ · HdV = ZZZ

E

(z + 1)dV, (4 pts)

where E is the solid bounded by S₁ ∪ S2 ∪ S3. Firstly, note that E is symmetric with respect to xy-plane. Therefore,

ZZZ

E

zdV = 0

So ZZZ

E

(z + 1)dV = ZZZ

E

dV = 4

3πρ³₀π/3 2π = 2

9πρ³₀ (3 pts)

since the volume of E is equal to the volume of the sphere times π/3

2π = 1/6. Now ZZ

S1∪S2

H· dS = ZZZ

E

∇ · HdV − ZZ

S3

H· dS = 2

9πρ³₀− 4 3ρ³₀

à π 6 −

√3 8

!

=

√3 6 ρ³₀