ATH 2 Z Z 3 1 ATH 2 Z Z 3 : SA MP L ET E ST2

MATH 2ZZ3: SAMPLE TEST #2

1. Let D be a bounded region in thex, y plane whose boundary is a smooth simple closed curve C, which is positively oriented (with respect to D). Then, the area of D can be computed as the line integral:

(A)

I

C 1 3y

3_dx− 2 3x 3_dy (B) I C −1 2y

2_dx− 1 2x 2_dy → (C) I C −2

3y dx+ 1 3x dy

(D)

I

C 1

3y dx− 2 3x dy

(E)

I

C 1

3y dx+ 2 3x dy

(Hint: Use Green’s theorem.)

Solution. By Green’s Theorem, if ∂Q_∂x − ∂P

∂y = 1, we have

I

C

P dx+Q dy =

Z Z

D ∂Q

∂x − ∂P

∂y dA=

Z Z

D

1dA= Area(A).

If P(x, y) =−2

3y and Q(x, y) = 1

3x, we have ∂Q

∂x − ∂P

∂y = 1 3 −(−

2

3) = 1 so the choice (C) will work while this last equation is not satisfied for the other choices.

2. Evaluate the integral

I =

I

C

x3dx−(x−1)2y2dy,

where C is the triangle with vertices at (0,0),(1,0) and (0,1), oriented positively.

→(A) I = 2

15

(B) I = 2 9

(C) I =− 3

16

(D) I =− 7

10

(E) I = 6 5

Solution. By Green’s theorem,

I =

I

C

x3dx−(x−1)2y2dy=

Z Z

R

∂(−(x−1)2y2) ∂x −

∂(x3) ∂y dA =

Z Z

R

−2 (x−1)y2dA,

where R is the region bounded byC and described as the type I region

R ={(x, y), 0≤x≤1, 0≤y≤1−x}.

Thus,

I =

Z 1

0

Z 1−x

0

−2 (x−1)y2dy dx=

Z 1

0

−2 (x−1)

y3 3

1−x

0

dx= 2 3

Z 1

0

(x−1)4dx

= 2 3

(x−1)5 5

1

0 = 2

Alternatively, letting C1 denote the line segment from (0,0) to (1,0) C2 denote the line segment from (1,0) to (0,1) and C3 denote the line segment from (0,1) to (0,0), we have

I = I C1 + I C2 + I C3

x3dx−(x−1)2y2dy.

For C1, we let x=t, dx=dt, y= 0, dy = 0, so

I

C1

x3dx−(x−1)2y2dy=

Z 1

0

t3dt = 1 4.

For C2, we let x= 1−t, dx=−dt, y=t, dy=dt, so

I

C2

x3dx−(x−1)2y2dy=

Z 1

0

−(1−t)3 −t4dt=

(1−t)4

4 −

t5 5

1

0 =−1

4− 1 5.

For C3, we let x= 0, dx= 0, y= 1−t,dy =−dt, so

I

C3

x3dx−(x−1)2y2dy=

Z 1

0

(t−1)2dt =

(t−1)3 3

1

0 = 1

3.

Thus, I = 1₃ −1 5 =

2 15.

3. LetDbe the region bounded by the positively oriented, simple closed curveCparametrized by

x=ecost, y=esint, 0≤t ≤2π.

Use Green’s theorem to compute the double integral

I =

Z Z

D 1 x ydA.

(Hint: LetP(x, y) = 0 and Q(x, y) = lnx y .)

(A) I = 3π 4

(B) I =−2π

(C) I = π 4

(D) I =−π

2

→(E) I =π

Solution. Letting P(x, y) = 0 and Q(x, y) = ln_yx, we have ∂Q_∂x − ∂P ∂y =

1

x y. Thus, using Green’s theorem

Z Z

R 1

x ydA=

Z Z

R ∂Q

∂x − ∂P

∂y dA=

I

C

P dx+Q dy=

I

C lnx

y dy.

Letting x=ecost and y =esint, we have dy = cost esintdt and

I

C lnx

y dy=

Z 2π

0

cost

esint cost e sint

dt =

Z 2π

0

cos2t dt= 1 2

Z 2π

0

1 + cos(2t)dt

= 1 2

t+sin(2t) 2

2π

4. Let f(x, y, z) = x2_+y3_{+z and let C be the path parametrized by the vector-function}

r(t) = 2 +t, t2, 6 2 +t

, 0≤t≤1.

The value of the path integral

I =

Z

C

∇f·dr

is

(A) I = 2

(B) I =−2

→(C) I = 5

(D) I = 0

(E) I = 3

Solution. Since r(0) = h2, 0, 3i and r(1) = h3, 1, 2i, the fundamental theorem of line integrals shows that

Z

C

∇f·dr=f(3,1,2)−f(2,0,3) = 12−7 = 5.

5. Let D be the region in the first quadrant (x, y ≥ 0) bounded by the line y = 1−x and the circle x2_+y2 _{= 1. Use polar coordinates to compute the double integral}

I =

Z Z

D

x+y x2_+y2 dA.

(A) I = 3− 2π

3

(B) I = 4 + π 6

(C) I = 1− 3π

2

→(D) I = 2− π

2

(E) I = 3− 2π

3

Solution. The liney = 1−x has equation r cosθ+r sinθ = 1 or r= 1

cosθ+sinθ in polar coordinates while the circle x2_+y2 _{= 1 has equation r = 1. The region D is expressed} as the region

D∗ ={(r, θ), 0≤θ ≤ π

2,

1

cosθ+ sinθ ≤r ≤1}. We have thus

I =

Z Z

D∗

r cosθ+r sinθ

r2 r dr dθ =

Z π/2

0

Z 1

1 cosθ+sinθ

cosθ+ sinθ dr dθ

=

Z π/2

0

(cosθ+ sinθ)

1− 1

cosθ+ sinθ

dθ

=

Z π/2

0

cosθ+ sinθ−1dθ =sinθ−cosθ−θπ/2₀ = 2−π

6. LetDbe the region in thex, y plane bounded by the triangle with vertices at (0,0), (0,3) and (2,3). The integral

I =

Z Z

D

ey2dA

is given by:

(A) I = 0

→(B) I = e

9₋₁

3

(C) I = e 4₋₁

2

(D) I =e3+ 2

(E) I = 3e2_{+ 2e}3

Solution. The region Dcan be expressed as the type II region

D =

(x, y), 0≤y≤3, 0≤x≤ 2y

3

.

Thus,

I =

Z 3

0

Z 2₃y

0

ey2dx dy=

Z 3

0

2y ey2

3 dy= 1 3

h

ey2i 3

0 = e

9₋₁

3 .

7. Let L be the lamina bounded by the curves y=e−x, x = 0,x = 1 and y = 0 with mass density per unit of area given by ρ(x, y) = 1. The x-coordinate of its center of mass is:

(A) x= e

−1₋₂

e−1₋₁

(B) x= e−2 e−1₋₁

(C) x= e−1 e−2

(D) x= e

−1₋₂

e−1

→(E) x= e−2

e−1

Solution. The mass of the lamina is

m=

Z 1

0

Z e−x

0

1dy dx=

Z 1

0

e−xdx=−e−x1₀ = 1−e−1.

We have also

My =

Z 1

0

Z e−x

0

x dy dx=

Z 1

0

x e−xdx=

Z 1

0

x(−e−x)0dx

=x(−e−x)1₀+

Z 1

0

e−xdx=−e−1+ 1−e−1 = 1−2e−1.

Thus,

x= My m =

1−2e−1 1− e−1 =

8. Compute the volume V of the solid region below the graph of the function z = y (1 +x3₎2 and above the triangle with vertices (0,0,0), (1,0,0) and (1,1,0) in the x, y plane.

→(A) V = 1

12

(B) V = 1 3

(C) V = 2 7

(D) V = 3 8

(E) V = 1 8

Solution. IfD denotes the region bounded by the triangle in the x, y plane, we have

V =

Z Z

D y

(1 +x3₎2 dA.

The region D can be expressed as the region of type 1

D={(x, y), 0≤x≤1, 0≤y≤x}.

Therefore, V = Z 1 0 Z x 0 y

(1 +x3₎2 dy dx =

Z 1

0

1 (1 +x3₎2

y2 2

x

0

dx= 1 2

Z 1

0

x2

(1 +x3₎2 dx

= 1 6

− 1

1 +x3

1 0 = 1 6 1 2 = 1 12.

9. When written in polar coordinates, the integral

I =

Z 1

−1

Z √

2−x2

|x|

1

1 +x2 _+y2 dy dx

becomes

(A) I =

Z π₂

0 Z √ 2 0 r

1 +r2 dr dθ

(B) I =

Z 2π₃

π 3

Z 2

0 r

1 +r2 dr dθ

→(C) I =

Z 3π₄

π 4 Z √ 2 0 r

1 +r2 dr dθ

(D) I =

Z 3π₂

0 Z √ 2 0 r

1 +r2 dr dθ

(E) I =

Z 3π₄

π 4

Z 1

0 r

1 +r2 dr dθ

Solution. Note that the curvey=√2−x2 _{is the part of the circle of radius}√_{2 centered} at the origin above the x-axis. Thus,

I =

Z Z

D

1

whereDis the region in the half-planey ≥0, bounded by the linesy =±xand the circle of radius √2 centered at the origin, which is expressed in polar coordinates as the region

D∗ =

(r, θ), π

4 ≤θ ≤ 3π

4 , 0≤r ≤

√

2

.

Hence,

I =

Z 3π₄

π 4

Z √

2

0

r

1 +r2 dr dθ.

10. Find the area A of the region in the x, y plane bounded by the curves y = x2−4 and y = 14−x2_.

(A) A= 36

(B) A= 25

(C) A= 4

→(D) A= 72

(E) A= 96

Solution. The curves y =x2−4 and y = 14−x2 intersect when y=x2−4 = 14−x2. This yields x2 _{= 9 or x = ±3, so the points of intersection are (±3,5). The region} bounded by the two curves can thus be written as

D={(x, y), −3≤x≤3, x2−4≤y≤14−x2}.

Hence,

A=

Z Z

D

1dA=

Z 3

−3

Z 14−x2

x2₋₄

1dy dx =

Z 3

−3

18−2x2dx= 2

Z 3

0

18−2x2dx

= 2

18x−2x

3

3

3

0

= 2 (36) = 72.

11. Let C be the semicircle x2 _+y2 _{= 4, x ≥ 0, oriented anticlockwise. Compute the path} integral I =R_C(x+y)2ds.

(Hint: You can parametrizeC using cosine and sine functions.)

(A) I = 4π2

(B) I = 4

(C) I = 2

→(D) I = 8π

(E) I = 2π

Solution. Letting x = 2 cost, y = 2 sint for −π

2 ≤ t ≤ π

2, we have x

y0(t) = 2 cost, ds=p(x0_(t))2_{+ (y}0_(t))2_{dt =}p_{4 (sin}2_{t+ cos}2_{t)dt = 2dt. Thus,}

I =

Z π₂

−π 2

(2 cost+ 2 sint)22dt = 8

Z π₂

−π 2

cos2t+ sin2t+ 2 sint cost dt

= 8

Z π₂

−π 2

1 + sin(2t)dt = 8

t− cos(2t)

2

π₂

−π₂

= 8π

12. The closed path C consists of three line segments:

from (1,0,0) to (0,0,2), from (0,0,2) to (0,1,0), and from (0,1,0) back to (1,0,0).

The work W done by the force F(x, y, z) =hxy, xz, yzi along C is

→(A) W =−1

2

(B) W = 1₂

(C) W = 0

(D) W = 2₃

(E) W =−2 3

Solution. If we denote the line segments above by C1, C2 and C3, respectively and we let P =x y, Q=x z and R =y z, we note thatP =R = 0 onC1, P =Q= 0 onC2 and Q=R= 0 on C3. Thus,

W =

Z

C1

Q dy+

Z

C2

R dz+

Z

C3

P dx.

Since y = 0 on C1,

R

C1 Q dy = 0. On C2, we can let y=t and z = 2(1−t), dz =−2dt,

for 0≤t≤1. So

Z

C2

y z dz =−

Z 1

0

t(1−t) 4dt =−4

t2 2 −

t3 3

1

0 =−2

3.

On C2, we can let x=t and y = 1−t, dx=dt, for 0 ≤t ≤1. So

Z

C3

x y dz =

Z 1

0

t(1−t)dt=

t2 2 −

t3 3

1

0 = 1

6.

Thus,

W = 0− 2

3 + 1 6 =−

13. Leta >1. The path integralI =

Z

C

x dy+y dx

x2_+y2 , where C denotes the line segment from (1,1) to (a, a), is

(A) I =aπ 4

→(B) I = lna

(C) I = 1 2 lna

(D) I = (a−1)π 4

(E) undefined because the path passes through (0,0).

Solution.We can parametrizeCby lettingx=y=t, for 1≤t≤aso thatdx=dy=dt. Hence,

I =

Z a

0 2t 2t2 dt=

Z a

1 1

t dt= lna.

14. For what value of c is the vector field F(x, y, z) = h2y + 2z,2x −3z,−3y +c xi a gradient (or conservative) vector field?

→(A) c= 2

(B) c=−2

(C) c=−3

(D) c= 3

(E) no such c exists

Solution. F is conservative iff ∇ ×F=0. We have

∇ ×F=

i j k

∂ ∂x

∂ ∂y

∂ ∂z 2y+ 2z 2x−3z −3y+c x

= 0i+ (2−c)j+ 0k=0

if and only if c= 2. Note that forc= 2, a potential function isg(x, y, z) = 2x y+ 2x z−

3y z.

15. Let C be the path from (1,0,1) to (2,3,2) parametrized by

x= 1

2(3−cos(πt)), y= 3t

3_{, z = 1 +t, 0≤t≤1,}

and consider the functions P = 2x y, Q = x2 _+z2 _{and R = 2y z. Compute the path} integral I =R_C P dx+Q dy+R dz.

(Hint: Does the fieldhP, Q, Ri have a potential function? )

(A) I =−24

(B) I = 0

→(C) I = 24

(D) I = 12

(E) I =−12

Solution. By inspection, a potential function is g(x, y, z) =x2_{y+y z}2_{. Thus,}

16. The vector field F(x, y) =h−1/y, x/y2_{i is conservative in the region}

(A) x >0

→ (B) y >0

(C) x2+y2 <1

(D) x+y >0

(E) x−y >0

Solution. Since the vector field F is undefined when y = 0, the region cannot intersect the line y = 0. The only possible region where a potential function can exist is thus the half-plane y > 0. In fact, a potential function is g(x, y) = −x

y which is well-defined on that region.