Sol: Method 1 (i) Find the four vertices of the region

1. (10%) Evaluate the iterated integral Z a

0

Z a x

sin (y²) dy dx, a > 0.

Sol:

Z a 0

Z a x

sin y²dy dx = Z a

0

Z y 0

sin y²dx dy

= Z a

0

y sin y²dy

= − cos y² 2

y=a y=0

= 1 − cos a² 2

2. (12%) Compute the area of the domain in the first quadrant bounded by the four curves xy = 1, xy = 4, y

x² = 1, and y x² = 2.

Sol:

Method 1

(i) Find the four vertices of the region:

(2⁻¹³, 2¹³), (1, 1), (2¹³, 2⁵³), (2²³, 2⁴³).

(From left to right.) (ii)

Area = Z 1

2⁻

1 3

(2x²− 1 x) dx +

Z 2¹³ 1

(2x²− x²) dx

+ Z 2²³

2¹³

(4

x − x²) dx

= ln 2

or

Area = Z 2¹³

1

(y¹² − y⁻¹) dy

+ Z 2⁴³

2¹³

(y¹² − (y 2)¹²) dy +

Z 2⁵³ 2⁴³

(4 y − (y

2)¹²) dy

= ln 2.

Method 2

(i) Make the change of variables:







u = xy v = ^y_x²

⇒







x = (^u_v)¹³ y = u²³v¹³ (ii)

Area = Z 2

1

Z 4

1 |∂(x, y)

∂(u, v)|dudv = ln 2.

3. (12%) Find the region E ⊂ R³ for which the triple integral Z Z Z

E

(4 − x²− 4y² − 9z²) dV is a maximum, and compute this maximum value.

Sol:

First we note that the region E which maximize the integralR 4 −x²− 4y²− 9z² is the ellipsoid (x, y, z) : 4 − x²− 4y²− 9z² > 0

Here we use the ”change of coordinate”, that is x = r sin φ cos θ, y = 1

2r sin φ sin θ, z = 1 3r cos φ and the corresponding Jacobian is

∂(x, y, z)

∂(r, φ, θ) = 1

6r²sin φ

With the above transformation, the original integral R

x²+4y²+9z²<44 − x²− 4y²− 9z²dxdydz is now changed to

Z 2π 0

Z π 0

Z 2

0 (4 − r²)1

6r²sin φ drdφdθ = 128 45π 4. (14%) Let F(x, y, z) = yzi +

xz + y

y²+ z²+ 1

 j+

xy + z

y²+ z²+ 1 + cos z k.

(a) Find a function f such that F = ∇f.

(b) Compute the line integral Z

C

F· dr, where C is the curve starting from the origin given by r(θ) = hθ cos θ, θ sin θ, θi, 0 ≤ θ ≤ π.

Sol:

(a) ∂f

∂x = yz ⇒ f = xyz + g(y, z)

∂f

∂x = xz + y

y²+ z²+ 1 ⇒ f = xyz +1

2ln(y²+ z²+ 1) + h(z)

∂f

∂x = xy + z

y²+ z² + 1 + cos y ⇒ f = xyz + 1

2ln(y²+ z²+ 1) + sin z (b)

Z

CF · dr = f(x, y, z) 

r(π)=(−π,0,π) r(0)=(0,0,0) = 1

2ln(π²+ 1) 5. (14%) Compute the line integral

I

C

(3xy + 1) dx + (x²+ x) dy, where the closed curve C is the cardioid given by the polar equation r = 1 + sin θ, and is oriented counterclockwise (see figure).

Sol:

Since 3xy + 1, x²+ x are smooth on C and smooth in D ,where D is the region enclosed by C.

Hence using Green’s theorem we have I

c

(3xy + 1)dx + (x²+ x)dy = Z Z

D

∂(x²+ x)

∂x − ∂(3xy + 1)

∂y dA =

Z Z

D(1 − x)dA

by using polar coordinates we have Z Z

D(1 − x)dA = Z 2π

0

Z 1+sin θ

0 (1 − r cos θ)rdrdθ Z 2π

0

Z 1+sin θ

0 (1 − r cos θ)rdrdθ = Z 2π

0

[(1 + sin θ)²

2 −1

3cos θ(1 + sin θ)³]dθ

= Z 2π

0

(1

2 + sin θ +1

2sin²θ)dθ

= π + π 2 = 3π

2 6. (10%) Evaluate

Z Z

S

p1 + x²+ y²dS, where S is the helicoid parametrized as r(u, v) = u cos v i+

u sin v j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ π.

Sol:

r(u, v) = (u cos v, u sin v, v)

∂r

∂u = (cos v, sin v, v)

∂r

∂v = (−u sin v, u cos v, 1)

∂r

∂u × ∂r

∂v = (sin v, − cos v, u) 

∂r

∂u × ∂r

∂v   =

√1 + u²

Z Z

S

p1 + x²+ y²dS = Z 1

0

Z π 0

√1 + u²√

1 + u²dvdu = (u + 1 3u³)

1 0(v)

π 0 = 4

3π

7. (14%) Let closed curve C be the intersection of the plane x + y + z = 1 and the cylinder x²+ y² = 9, oriented counterclockwise as viewed from above, and F(x, y, z) = x²zi + xy²j+ z²k.

Find I

C

F· dr.

Sol:

Compute the curl of vector field F. Find curl(F) = x²j+ y²k. Using Stoke’s Theorem, we have

I

C

F· dr = Z Z

S

curl(F)dS

where S is the surface with boundary C which is parametrized with counterclockwise orienta- tion. And C is the intersection of x²+ y²= 9 and x + y + z = 1. Now observe that the normal

vector of S is (1, 1, 1)/√

3 and that dS =√

3 dA. So Z Z

S

curl(F) = Z Z

A

(x²+ y²)dA

where A is the projection of S on xy plane., that is, x² + y² = 9. Hence by using polar coordinates. We have

Z Z

A

(x²+ y²)dA = Z 3

0

Z 2π 0

r²rdrdθ = 81 2 π

8. (14%) Let F(x, y, z) =

xy²+py²+ z⁴ i+

tan⁻¹x + x²y

j+z³

3 − e^x2+y2 k.

(a) Find div F.

(b) Find Z Z

S

F· dS, where the surface S is the top half of the sphere x²+ y²+ z² = 1 with the unit normal vectors pointing away from the origin. Warning. S is not a closed surface!

Sol:

(a) divF = x²+ y²+ z²

(b) 1^◦ By divergence theorem, we have Z Z

S

F· dS + Z Z

E

F· dS = Z Z Z

x²+ y²+ z²dxdydz

where E is the disk with radius 1 on xy-plane and with the unit normal vectors (0, 0, −1).

2^◦

Z Z

E

F· dS = Z Z

x²+y²≤1

e^x2+y2 −

0 z³

3dxdy = Z 2π

0

Z 1 0

e^r2 · rdrdθ = π(e − 1) 3^◦

Z Z Z

x²+ y²+ z²dxdydz = Z 2π

0

Z π/2 0

Z 1 0

ρ² · ρ²sin φdρdφdθ = 2π 5 4^◦ Thus

Z Z

S

F· dS = 2π

5 − π(e − 1)