(b) (3 pts) Find an equation of the normal plane for Viviani’s curve at the point

1092 !D06-12í ®M3 ãTŒU–

1. The intersection of the sphere x²+ y²+ z² = 1 and the cylinder x²+ y² = x is a space curve known as the Viviani’s curve. One way to express this parametric curve is

r(t) = ⟨sin²t, sin t cos t, cos t⟩ where 0 ≤ t ≤ 2π.

(a) (5 pts) Find the unit tangent vector T(t).

(b) (3 pts) Find an equation of the normal plane for Viviani’s curve at the point (³4,−^√4³,−¹2).

(c) (4 pts) Find its curvature κ(π) when t = π.

Figure : Viviani’s curve.

Solution:

(a) r^′(t) = ⟨sin 2t, cos 2t, − sin t⟩ or ⟨2 sin t cos t, cos²t− sin²t,− sin t⟩ (1 pt)

∥r^′(t)∥ =√

1+ sin²t (1 pt)

T(t) = _∥r^r′′(t)_(t)∥ = ^√_1+sin¹ 2t⟨sin 2t, cos 2t, − sin t⟩ or ^√_1+sin¹ 2t⟨2 sin t cos t, cos²t− sin²t,− sin t⟩ (3 pt: correct formula for T 2 pt, answer 1 pt))

Marking: (2nd line) if student use the second expression of r’(t) and simplify until ∥r’(t)∥ =

√sin²t+ cos⁴t+ sin⁴t+ 2 sin²t cos²t, still consider correct. (1 pt)

(b) Solve (sin²t, sin t cos t, cos t) = (³4,−^√4³,−¹2) and obtain t = ^2π3 . (1 pt) Vector normal to the normal plane at t=^2π₃ is T(^2π₃ ) = ⟨−√

3

7,−^√1₇,−√

3

7⟩. (1 pt) Equation of normal plane: √

3x+ y +√

3z= 0 (1 pt)

Marking: (2nd line) any direction parallel to T(2π/3) is acceptable (1 pt).

Solve the wrong t but plug in the correct equation of normal plane (2 pt) (c) r^′′(π) = ⟨2, 0, 1⟩ (1 pt)

r^′(π) × r^′′(π) = ⟨0, 1, 0⟩ × ⟨2, 0, 1⟩ = ⟨1, 0, −2⟩ (1 pt),

∥r^′(π)∥ = 1

κ(π) = ^{∣r′(π)×r}∣r^′(π)∣^{′′3(π)∣} =√

5 (2 pt: Correct formula 1 pt, answer 1 pt)

2. Consider the function

f(x, y) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩

x²sin(x)

x²− xy + y² if (x, y) ≠ (0, 0) 0 if (x, y) = (0, 0). (a) (5 pts) Show that f is continuous at (0, 0).

(b) (5 pts) Let u= ⟨a, b⟩ be a unit vector. Find the directional derivative D^uf(0, 0) in terms of a and b.

(c) (4 pts) By appealing to your answer in (b), prove that f(x, y) is not differentiable at (0, 0).

Solution:

(a)

Marking scheme.

(1M) *Attempt to compute the limit lim

(x,y)→(0,0)

x²sin(x) x²− xy + y² (1M) Use of the fact lim

x→0

sin x x = 1

(2M) **Correct method to evaluate lim

x→0

x³

x²− xy + y² (e.g. by Polar coordinates &

Squeeze Theorem)

(1M) Correct limit by using a correct method Remarks.

1. If a student attempts by taking paths, he/she can get at most 1M overall from *.

(Note it’s clear from the question that the relevant limit exists) 2. If a student writes lim

x→0

x³

x²− xy + y² = 0 (or any equivalent limits) without any appropriate explanations, no marks will be awarded from **.

3. No deductions for students not explaining why cos³θ

1− cos θ sin θ is bounded. (But students are expected to at least mention it before applying the Squeeze Theorem) 4. L’Hospital’s rule is not applicable. (note that θ can be implicitly dependent on r) Sample Solution 1.

lim

(x,y)→(0,0)

x²sin(x) x²− xy + y²

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

= lim

(x,y)→(0,0)

x³

x²− xy + y² ⋅sin x x

Passing to polar coordinates, we have lim

(x,y)→(0,0)

x³

x²− xy + y² = lim

r→0⁺r⋅ cos³θ

1− cos θ sin θ = lim

r→0⁺r⋅ cos³θ 1−¹₂sin(2θ)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

bounded

= 0 by Squeeze Theorem

(2M)

Hence, lim

(x,y)→(0,0)

x²sin(x)

x²− xy + y² = lim

(x,y)→(0,0)

x³

x²− xy + y² ⋅sin x

x = 0 ⋅ 1®

(1M)

= 0®

(1M)

= f(0, 0)

Therefore, f(x, y) is continuous at (0, 0).

Sample Solution 2.

lim

(x,y)→(0,0)

x²sin(x) x²− xy + y²

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

= lim_r→0+

⎛⎜⎜

⎜⎜⎜

⎝

®r

→0

⋅ cos³θ 1−¹2sin(2θ)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

bounded

⎞⎟⎟

⎟⎟⎟

⎠

⋅sin(r cos θ)

r cos θ = 0

Squeeze(2M)®

⋅ 1®

(1M)

= 0®

(1M)

= f(0, 0).

Therefore, f(x, y) is continuous at (0, 0).

Sample Solution 3.

lim

(x,y)→(0,0)

x²sin(x) x²− xy + y²

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

= lim_r→0+

cos²θ

1−¹2sin(2θ)⋅ sin(r cos θ)

Since lim

r→0⁺sin(r cos θ) = sin( lim_r→0+r cos θ) = 0(1M) and cos²θ

1−¹2sin(2θ) is a bounded function, we have by Squeeze Theorem that

r→0lim⁺

cos²θ

1−¹2sin(2θ) ⋅ sin(r cos θ) = 0 = f(0, 0). (2+1M, Correct argument) Therefore, f(x, y) is continuous at (0, 0).

(b)

Marking scheme.

(1M) Correct definition of D_uf(0, 0)

(2M) Correct simplification of the expression D_uf(0, 0) (2M) Correct evaluation of D_uf(0, 0)

Remarks.

1. No marks are awarded to students who use (incorrectly) that D_uf(0, 0) =

∇f(0, 0) ⋅ u.

2. No deductions for students not separating the case a = 0 and leave a³

1− ab as the final answer.

3. No deductions for students not simplifying a²+ b² = 1.

Sample Solution.

By definition,

D_uf(0, 0) = lim_t→0 f(ta, tb) − f(0, 0)

t ,(1M)

= lim_t→0

t²a²sin(ta) t²(a²−ab+b²)− 0

t = lim_t→0a²sin(ta)

t(1 − ab) ,(2M)

=⎧⎪⎪⎪

⎨⎪⎪⎪⎩

0 if a= 0

limt→0

a³ 1− ab ⋅

sin(ta)

ta = a³

1− ab if a≠ 0 (2M)

(c)

Marking scheme.

(1M+1M) **Correct computation of ∇f(0, 0) = ⟨f^x(0, 0), f^y(0, 0)⟩

(1M) Pick an explicit u= ⟨a, b⟩ and find ∇f(0, 0) ⋅ u

(1M) Establishing that Duf(0, 0) ≠ ∇f(0, 0) ⋅ u for some explicit u Remarks.

1. 3M for students who, without offering a concrete u, write ‘_1−ab^a3 is not of the form ac₁+ bc² for some constants c₁, c₂’ or ‘_1−ab^a3 is not a (bi)linear function in a and b’.

2. Marks for ** can be awarded even if a student wrote them in his/her work in (a) or (b)

3. The last 2M will not be awarded if (b) is incorrect.

4. We accept students for writing ‘a is not identically equal to_1−ab^a3 ’.

5. We also accept proofs of non-differentiability by definition. In this case, the last 1M+1M will be allocated to a correct proof that lim

(x,y)→(0,0)

f(x, y) − L(x, y)√

x²+ y² does not exist/does not vanish.

Sample Solution 1. (By using (b)) By (b), we have that f_x(0, 0) = 1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

and f_y(0, 0) = 0

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

.

Then we take, for example, u= ⟨¹2,

√ 3

2 ⟩. We have that

∇f(0, 0) ⋅ u = 1 2⋅ 1 +

√3 2 ⋅ 0 = 1

2 (1M) but

D_uf(0, 0) = (1/2)³ 1− (1/2)(√

3/2) = 1 8− 2√

3

Therefore, D_uf(0, 0) ≠ ∇f(0, 0) ⋅ u (1M) and hence the function is not differentiable at (0, 0).

Sample Solution 2. (By definition of differentiability) By (b), we have that f_x(0, 0) = 1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

and f_y(0, 0) = 0

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M)

.

Therefore, the linearisation of f at (0, 0) is given by L(x, y) = x. Now consider the limit

lim

(x,y)→(0,0)

f(x, y) − L(x, y)√

x²+ y² = lim

(x,y)→(0,0)

x²sin(x) x²−xy+y² − x

√x²+ y²

Along the path (0, t), we have

x²sin(x) x²−xy+y² − x

√x²+ y² = 0 → 0 , as t → 0

Along the path (t, 2t) with t > 0, we have

x²sin(x) x²−xy+y² − x

√x²+ y² = ¹³sin t√ − t

5t → 1

3√ 5− 1

√5 , as t→ 0

Since the limit along two paths do not agree(2M), we conclude that lim

(x,y)→(0,0)

f(x, y) − L(x, y)√ x²+ y² does not exist and hence the function is not differentiable at (0, 0).

3. Consider the level surface defined by the equation z⁵+ (sin x)z³+ yz = 4.

(a) (5 pts) Find an equation of the tangent plane at (0, 3, 1).

(b) The equation of the level surface defines z= z(x, y) implicitly near (0, 3, 1) as a differentiable function in x and y.

i. (4 pts) Find ∂z

∂x ∣

(x,y)=(0,3)

and ∂z

∂y ∣

(x,y)=(0,3)

.

ii. (4 pts) Use the linear approximation of z(x, y) at (0, 3) to approximate the real root of the quintic polynomial P(z) = z⁵+ 2.9z − 4.

Solution:

(a)

Marking scheme.

(2M) Find ∇F(x, y, z) (2M) Find ∇F(0, 3, 1)

(1M) Equation of tangent plane Remarks.

1. At most 4M can be awarded to students who made calculation error to at most one of the coordinates of ∇F.

Sample solution.

Let F(x, y, z) = z⁵+ (sin x)z³+ yz. Then

∇F = ⟨(cos x)z³, z, 5z⁴+ 3(sin x)z²+ y⟩. (2M) Therefore,

∇F(0, 3, 1) = ⟨1, 1, 8⟩. (2M) Hence,

⟨1, 1, 8⟩ ⋅ ⟨x, y − 3, z − 1⟩ = 0 (1M) or x+ y + 8z = 11 gives the equation of the tangent plane.

(b) (i)

Marking scheme.

(2M) Use of Implicit Function Theorem or Chain rule (correctly) (1M) Correct value for z_x(0, 3)

(1M) Correct value for zy(0, 3) Remarks.

1. At most 1M will be taken away if a student has forgotten to evaluate these partial derivatives at (0, 3)

Sample solution.

By implicit function theorem, z_x(0, 3) = −Fx(0, 3, 1)

F_z(0, 3, 1)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(2M)

= −1

°8

(1M)

and z_y(0, 3) = −F_y(0, 3, 1) F_z(0, 3, 1) = −

1

°8

(1M)

.

(ii)

Marking scheme.

(2M) Writing down the linear approximation of z(x, y)

(1M+1M) Association with the substitution (x, y) = (0, 2.9) and correct answer Remarks.

1. 1M can be awarded to students who demonstrate some understandings of the concept of ‘linear approximation in multivariables’.

2. At most 3M can be given to students who inherited an incorrect answer from (b) Sample solution.

The required linear approximation is L(x, y) = 1 − 1

8x−1

8(y − 3). (2M)

The root of the quintic polynomial P(X) = X⁵+ 2.9X − 4 equals to z(0, 2.9) (1M) which can be approximated by L(0, 2.9) = 81

80 (1M).

4. Let f(x, y) = ln(x + y + 1) + x²− y.

(a) (7 pts) Find and classify the critical point(s) of f .

(b) (6 pts) Find the absolute maxima and minima of f on D= {(x, y) ∈ R²∣ 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

Solution:

(a) f_x= 1

x+ y + 1 +2x, f_y = 1

x+ y + 1 −1. (1%)

To solve f_x = 0 and fy = 0, we have x = −1/2 and y = 1/2. So the critical point of f is (−1/2, 1/2). (2%)

f_xx= −1

(x + y + 1)² + 2, f^xy = −1

(x + y + 1)² and f_yy= −1

(x + y + 1)². (1%) Then

D(−1 2,1

2) = f^xx(−1 2,1

2)f^yy(−1 2,1

2) − [f^xy(−1 2,1

2)]²= −2 < 0.(1%) So the critical point (−1/2, 1/2) is a saddle point.(2%)

(b) Since (−1/2, 1/2) is not in D, the extreme value of f does not occur in the interior of D.

(1%)

The boundary of D consists of the four line segments L₁ ∶ x = 0, 0 ≤ y ≤ 1, L² ∶ y = 0, 0 ≤ x≤ 1, L³∶ x = 1, 0 ≤ y ≤ 1 and L⁴ ∶ y = 1, 0 ≤ x ≤ 1. On L¹, we have

f₁(y) ∶= f(0, y) = ln(y + 1) − y, 0 ≤ y ≤ 1.

Since f₁^′(y) = −y

y+ 1 ≤0 for 0≤ y ≤ 1, we have f1 is decreasing in y. So its maximum value is f(0, 0) = 0 and its minimum value is f(0, 1) = ln 2 − 1. (1%)

On L₂, we have

f₂(x) ∶= f(x, 0) = ln(x + 1) + x², 0≤ x ≤ 1.

Since f₂^′(x) = 1

x+ 1 +2x≥ 0 for 0 ≤ x ≤ 1, we have f² is increasing in x. So its maximum value is f(1, 0) = ln 2 + 1 and its minimum value is f(0, 0) = 0. (1%)

On L₃, we have

f₃(y) ∶= f(1, y) = ln(y + 2) + 1 − y, 0 ≤ y ≤ 1.

Since f₃^′(y) = −y− 1

y+ 2 ≤0 for 0≤ y ≤ 1, we have f³ is decreasing in y. So its maximum value is f(1, 0) = ln 2 + 1 and its minimum value is f(1, 1) = ln 3. (1%)

On L₄, we have

f4(x) ∶= f(x, 1) = ln(x + 2) + x²− 1, 0 ≤ x ≤ 1.

Since f₄^′(x) = 1

x+ 2 +2x≥ 0 for 0 ≤ x ≤ 1, we have f4 is increasing in x. So its maximum value is f(1, 1) = ln 3 and its minimum value is f(0, 1) = ln 2 − 1. (1%)

Therefore, the absolute maximum value of f on D is ln 2+ 1 and the absolute minimum value of f on D is ln 2− 1. (1%)

5. (10 pts) Use the method of Lagrange multipliers to find the absolute maximum and minimum values (if exists) of f(x, y, z) = x²+ 3z² subject to the constraints x+ y + z = 1 and x − y + 2z = 2.

Solution:

Let g₁(x, y, z) = x + y + z and g²(x, y, z) = x − y + 2z. Then we will find Max f(x, y, z), Min f(x, y, z)

subject to the constraints

{ g₁(x, y, z) = x + y + z = 1,

g₂(x, y, z) = x − y + 2z = 2. (2 pts) By Lagrange multipliers, we have

⎧⎪⎪⎪⎨⎪⎪⎪

⎩

∇f = λ ⋅ ∇g1+ µ ⋅ ∇g2

g₁(x, y, z) = x + y + z = 1 g₂(x, y, z) = x − y + 2z = 2

⇒⎧⎪⎪⎪

⎨⎪⎪⎪⎩

(2x, 0, 6z) = λ ⋅ (1, 1, 1) + µ ⋅ (1, −1, 2) x+ y + z = 1

x− y + 2z = 2 ⇒

⎧⎪⎪⎪⎪

⎪⎪⎪⎪⎨⎪⎪⎪

⎪⎪⎪⎪⎪

⎩

2x = λ + µ 0 = λ − µ 6z = λ + 2µ x+ y + z = 1 x− y + 2z = 2

⇒

⎧⎪⎪⎪⎪

⎪⎪⎪⎪⎨⎪⎪⎪

⎪⎪⎪⎪⎪

⎩

2x = λ + µ λ = µ 6z = λ + 2µ x+ y + z = 1 x− y + 2z = 2

⇒⎧⎪⎪⎪

⎨⎪⎪⎪⎩

x= λ = µ = 2z x+ y + z = 1

x− y + 2z = 2 ⇒⎧⎪⎪⎪

⎨⎪⎪⎪⎩

x = λ = µ = ⁶₇ y = ⁻²7

z = ³7.

(5 pts)

Note that the intersection of two planes g₁(x, y, z) = 1 and g²(x, y, z) = 2 is a line L in R³. By Lagrange multiplier, the absolute minimum value of the function f is f(⁶7,⁻²₇ ,³₇) = ⁹7 since there is another point (3, −1, −1) ∈ L with f(3, −1, −1) = 12 > Γf(⁶₇,⁻²₇ ,³₇). (2 pts) There is no absolute maximum value since the x-coordinate (or z-coordinate) of the point on L can be arbitrary large. (1 pt)

Remark 1. (If you use the following method, you can only get at most 6 credits.) Since

{ x+ y + z = 1

x− y + 2z = 2 ⇒2x+ 3z = 3 ⇒ x = 1

2(3 − 3z), the function f(x, y, z) becomes

f(x, y, z) = x²+ 3z²= (3 2 −3

2z)²+ 3z²= 21 4 (z −3

7)²+9 7 ≥ 9

7. (2 pts)

The absolute minimum value of the function f is f(⁶₇,⁻²₇ ,³₇) = ⁹₇. (3 pts) There is no absolute maximum value since the x-coordinate (or z-coordinate) of the point on L can be arbitrary large. (1 pt)

Remark 2. (If you use the following method, you can only get at most 6 credits.)

The intersection of two planes x+ y + z = 1 and x − y + 2z = 2 is a line L in R³ described by (x, y, z) = (3+3t, −1−t, −1−2t) for each point (x, y, z) ∈ L. Then the function f(x, y, z) becomes

f(x, y, z) = (3 + 3t)²+ 3(−1 − 2t)²= 21(t +5 )²+9

≥ 9

. (2 pts)

Under the constraints, the function f(x, y, z) obtains its absolute minimum value ⁹7 when t=⁻⁵7 . which implies (x, y, z) = (⁶7,⁻²₇ ,³₇). (3 pts) There is no absolute maximum value since the x- coordinate (or z-coordinate) of the point on L can be arbitrary large. (1 pt)

Remark 3. (If you use the following method, you can only get at most 9 credits.) Since

{ x+ y + z = 1

x− y + 2z = 2 ⇔ { 2x+ 3z = 3

x+ y + z = 1 (1 pt)

we will use Lagrange multiplier method to find the extreme values of f(x, y, z) ∶= x²+ 3z² with the constraint g(x, y, z) ∶= 2x + 3z = 3. (2 pts) By Lagrange multipliers, we have

{ ∇f = λ ⋅ ∇g

g(x, y, z) = 2x + 3z = 3

⇒ { (2x, 0, 6z) = λ ⋅ (2, 0, 3)

2x+ 3z = 3 ⇒⎧⎪⎪⎪

⎨⎪⎪⎪⎩

2x = 2λ 6z = 3λ 2x+ 3z = 3

⇒ { x= λ = 2z 2x+ 3z = 3

⇒ (x, z, λ) = (6 7,3

7,6

7) ⇒ (x, y, z) = (6 7,−2

7 ,3

7). (2 pts) Under the constraints { 2x+ 3z = 1

x+ y + z = 1, we get f(x, y, z) = f(1

2(1 − 3z), 1 − x − z, z) = (3 2−3

2z)²+ 3z²= 21 4 (z −3

7)²+9 7 ≥9

7 = f(6 7,−2

7 ,3 7).

The absolute minimum value of the function f(x, y, z) = x²+ 3z² is f(⁶7,⁻²₇ ,³₇) = ⁹7.(3 pts) There is no absolute maximum value since the x-coordinate (or z-coordinate) the constraint 2x+ 3z = 3 can be arbitrary large. (1 pt)

6. Evaluate the following integrals.

(a) (8 pts)

∫₁²∫^√x_xsin(πx

2y) dydx + ∫₂⁴∫^√2_xsin(πx

2y) dydx.

(b) (8 pts)

∭

E

√z dV,

where E is the solid lying below the cone z² = 4x²+ 4y², within the cylinder x²+ y² = 1 and above the plane z= 0.

Solution:

(a) (Method I)

∫₁²∫^√x_xsinπx

2ydydx+ ∫₂⁴∫^√2_xsinπx 2ydydx

= ∫₁²∫ ^y

2

y

sinπx

2ydxdy (4 pts)

= ∫₁²[−2y

π ⋅ cosπx 2y ∣

x=y²

x=y

]dy

= ∫₁²[−2y

π ⋅ (cosπy

2 − cosπ

2)]dy (1 pt)

= −2 π ∫

2 1

y⋅ cosπy

2 dy= −4

π² ∫₁²y d sinπy 2

= −4

π²[y ⋅ sinπy 2 ∣

2

1 − ∫₁²sinπy

2 dy] (2 pts)

= −4

π²[−1 + 2

πcosπy 2 ∣

2

1

] = 4

π³(2 + π). (1 pt) (Method II)

∫₁²∫^√x_xsinπx

2ydydx+ ∫₂⁴∫^√2_xsinπx 2ydydx

= ∫₁²∫_y²sinπx

2ydxdy+ ∫₁²∫ ^y

2

2 sinπx

2ydxdy (4 pts)

= ∫₁²[−2y

π ⋅ cosπx 2y ∣

x=2

x=y]dy + ∫₁²[−2y

π ⋅ cosπx 2y ∣

x=y²

x=2

]dy

= ∫₁²[−2y

π ⋅ (cos2π

2y − cosπ

2)]dy + ∫₁²[−2y

π ⋅ (cosπy

2 − cos2π

2y)]dy (1 pt)

= ∫₁²[−2y

π ⋅ (cosπy

2 − cosπ 2)]dy

= −2 π ∫

2 1

y⋅ cosπy

2 dy= −4

π² ∫₁²y d sinπy 2

= −4

π²[y ⋅ sinπy 2 ∣

2

1 − ∫₁²sinπy

2 dy] (2 pts)

= −4

[−1 + 2

cosπy

∣

2

] = 4

(2 + π). (1 pt)

(b) (Method I) The solid E can be described as

E= {(x, y, z) ∣ x²+ y²≤ 1, 0 ≤ z ≤√

4(x²+ y²),}

= {(r, θ, z) ∣ 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 0 ≤ z ≤ 2r}. (3 pts) Using cylinder coordinate, we have

∭

E

√z dV

= ∫₀^2π∫₀¹∫₀^2r√

z⋅r dz dr dθ (2 pts)

= ∫₀^2π∫₀¹[ 2 3z³² ] ∣

z=2r

z=0

⋅r dr dθ

=2 3 ∫

2π

0 ∫₀¹2√

2⋅ r⁵²dr dθ (2 pts)

=4√ 2

3 ∫₀^2π2 7r⁷² ∣

r=1

r=0

dθ

=8√ 2

21 ⋅ ∫₀^2π1 dθ= 8√ 2

21 ⋅ 2π = 16√ 2

21 π. (1 pt) (Method II)

∭

E

√z dV

= ∬

0≤x²+y²≤1

∫

√

4(x²+y²) 0

√z dz dA (3 pts)

= ∬

0≤x²+y²≤1

2 3z³² ∣

√

4(x²+y²)

0

dA

=2

3 ∬

0≤x²+y²≤1

2√

2⋅ (x²+ y²)³⁴ dA (2 pts) (let x = r cos θ, y = r sin θ)

=4√ 2

3 ∫₀^2π∫₀¹r³² ⋅r dr dθ (2 pts)

=4√ 2

3 ⋅ (∫₀¹r⁵²dr) ⋅ (∫₀^2π1dθ)

=4√ 2 3 ⋅2

7⋅ 2π = 16√ 2π

21 . (1 pt)

7. Let E be an upper-half ball which occupies the region

E= {(x, y, z) ∈ R³∣ x²+ y²+ z² ≤ 16 and z ≥ 0}

and whose density function is ρ(x, y, z) =√

x²+ y² for each point (x, y, z) ∈ E.

(a) (7 pts) Find the mass of E.

(b) (5 pts) Let (0, 0, ¯z) be the center of mass of E. Find the value of ¯z.

Solution:

(a) the formula of mass is

M = ∭_Eρ(x, y, z) dV. (1)

Applying spherical coordinates yields M = ∫

π 2

0 ∫₀^2π∫₀⁴r sin(ϕ)r²sin(ϕ) dr dϑ dϕ (2)

= 32π². (3)

(b) the formula of z is

z= ∭^Ezρ(x, y, z) dV

M . (4)

Applying spherical coordinates yields

∫

π 2

0 ∫₀^2π∫₀⁴r cos(ϕ)r sin(ϕ)r²sin(ϕ) dr dϑ dϕ (5)

= 2¹¹π

15 . (6)

Therefore,

z= 64

15π. (7)

Grading Suggestion.

ˆ For (a), get 1/7 by obtaining (1), 3/7 by obtaining (2), and 3/7 by obtaining (3).

ˆ For (b), get 1/5 by by obtaining (4), 2/5 by obtaining (5), 1/5 by obtaining (6), and 1/5 by obtaining (7).

8. (10 pts) Find the volume of the solid lying below the plane z= 9 and above the surface z = (3x + y − 1)²+ (x − y)².

Solution:

Let D∶= {(x, y) ∣ (3x+y −1)²+(x−y)²≤ 9} be the region enclosed by (3x+y −1)²+(x−y)²= 9 in the xy-plane. Then the solid in the question can be described as

E∶= {(x, y, z) ∣ (x, y) ∈ D, (3x + y − 1)²+ (x − y)² ≤ z ≤ 9}.

So we have

Volume of the solid= ∭

E

1 dV

= ∬

D

∫_(3x+y−1)⁹ 2+(x−y)²

1 dz dA (4 pts)

= ∬

D

[9 − (3x + y − 1)²− (x − y)²] dA (1 pt)

( let {. u= 3x + y − 1

v= x − y ⇒ { x=¹4(u + v + 1)

y= ¹4(u − 3v + 1) ⇒ ∣J∣ = ∣ ∣¹⁴1 ¹⁴ 4

−3 4

∣ ∣ = 1 4. )

= ∬

u²+v²≤9

[9 − u²− v²] ⋅1

4 dA (3 pts) (let u = r cos θ, v = r sin θ)

=1 4 ∫

2π

0 ∫₀³[9 − r²] ⋅r dr dθ

=1

4 ⋅ 2π ⋅ [9 2r²−1

4r⁴] ∣

3

0

=1

4⋅ 2π ⋅ (81 2 −81

4 ) =81π

8 . (2 pts)