Exam

Name___________________________________

Find the volume of the indicated region by an iterated integration.

1) The solid cut from the first octant by the surface z = 9 - x2 - y 1)

Find the volume by using polar coordinates.

2) The region bounded by the paraboloid z = x2 + y2, the cylinder x2 + y2 = 25, and the

xy-plane

2)

Evaluate the integral.

3)9

0

9

y

sin (x2) dx dy∫∫ 3)

Calculate the surface area of the given surface.

4) The parabolic cylinder with equation z = y2 and lying over the triangle in the xy-plane

with vertices (0, 0), (0, 4), and (4, 4).

4)

Find all local extreme values of the given function and identify each as a local maximum, local minimum, or saddle point.

5) f(x, y) = x3 + y3 - 48x - 147y + 8 5)

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Find div F.

6) F(x, y, z) = ei + ln(2y + z)j + xyezk

A)1

ln(2y + z) + ez B)

2

ln(2y + z) + xyez

C) ei + ln(2y + z)j + xyezk D) e + 2

ln(2y + z) + xyez

6)

Find curl F.

7) F(x, y, z) = ei + ln(2y + z)j + xyezk

A) xez - 1

(2y + z)2 i -yezj + 0k B) xez +

1

2y + z i + yezj + 1k

C) xez - 1

2y + z i - yezj + 1k D) xez -

1

2y + z i - yezj + 0k

7)

Evaluate the line integral.

8)

C

x + y + z

5 ds∫ ; C is the curve x = 3t, y = (5 cos

4

5t), z = (5 sin

4

5t), 0 ≤ t ≤

5

4π

A)75

32π2 + 25 B)

75

32 +

25

4C)

75

32π D)

75

32π2 +

25

4

8)

1

9)

C

y2 dx + x dy∫ ; C is the curve x = t2 - 1, y = 6t, 0 ≤ t ≤ 1

A) 72 B) 14 C)5

6D) 26

9)

10)

C

(2x - 4y + 9z) dx + (3x - 6y - 3z) dy + (5x +y - z) dz∫ ; C is the line segment path from (0, 0, 0) to

(1, 0, 0) to (1, 2, 0) to (1, 2, 4)

A) 40 B) 39 C) 15 D) 16

10)

Determine whether F is conservative.

11) F(x,y,z) = 7i + -ezj - yezk

A) conservative B) not conservative

11)

Find f so that F = ∇f. If the fuction is not conservative state so.

12) F(x,y) = (x - 3y + 7)i + (-3x + 3y + 8)j

A) f(x, y) = x2

2 - 3xy +

3y2

2 + C B) f(x, y) =

x2

2 + 3xy - 7x +

3y2

2 + C

C) f(x, y) = x2

2 - 3xy + 7x +

3y2

2 + 8y + C D) not conservative

12)

Evaluate the integral.

13)(4, 2)

(-2, 3)

(3x2y - y2) dx + (x3 - 2xy) dy∫A) 154 B) 346 C) 118 D) 246

13)

14)(3, π/2)

(-4, 0)

ex sin y dx + ex cos y dy∫

A) 0 B) e-4 C) e3 D) e3 - e-4

14)

Apply Green's Theorem to evaluate the integral.

15) C

(6y + x) dx + (y + 2x) dy

C: The circle (x - 2)2 + (y - 3)2 = 9

A) -120 B) -24 C) -36π D) 36π

15)

16) C

(y2 + 5) dx + (x2 + 6) dy; C: The triangle bounded by x = 0, x + y = 2, y = 0

A) 0 B)16

3C) -24 D) 32

16)

2

Using Green's Theorem, calculate the area of the indicated region.

17) The area bounded above by y = 4 and below by y = 4

9x2

A) 32 B) 16 C) 0 D) 8

17)

Evaluate

G

g(x, y, z) dS∫∫ .

18) g(x, y, z) = xz2

27; G is the cap cut from the sphere x2 + y2 + z2 = 9 by the cone z = x2 + y2

A) 4π B) 0 C) 12π D) 6π

18)

19) g(x, y, z) = z; G: y2 + z2 = 49, z ≥ 0, 2 ≤ x ≤ 3

A) 49 B) 98 C) 490 D) 14

19)

Use Gauss's Divergence Theorem to calculate

dS

F · n dS∫∫ .

20) F(x, y, z) = zi + xyj + zyk; S: the solid cube cut by the coordinate planes and the planes x = 2, y = 2,

and z = 2

A) 16 B) 4 C) 32 D) 2

20)

Use Stokes's Theorem to calculate

S

(curl F) · n dS∫∫ .

21) F = -3x2yi + 3xy2j + z5k; S is the portion of the paraboloid 1 - x2 - y2 = z that lies above the x-y

plane

A) 6 B) 1π C)3

2π D) 6π

21)

22) F = (x-y)i + (x-z)j + (y-z)k; S is the portion of the cone z = 5 x2 + y2 below the plane z = 1

A)2

25π B) -

2

25π C) -

4

25π D)

4

25

22)

Evaluate the given double integral by changing it to an iterated integral.

23)

S

1

ln x dA;∫∫ S is the region bounded by the x-axis, the line x = 9, and the curve y = ln x

A) 1 B) 8 C) 10 D) 9

23)

Find the volume of the indicated region by an iterated integration.

24) The region bounded by the surface z = x2 + y2, the cylinder x2 + y2 = 100, and the xy-plane

A) 5000π B) 2500π C)5000

3π D)

10000

3π

24)

Find the area of the region specified in polar coordinates.

25) The region inside both r = 8 sin θ and r = 8 cos θ

A) 8(π - 2) B) 8(π - 1) C) 16(π - 1) D) 16(π - 2)

25)

3

Find the volume by using polar coordinates.

26) The region bounded by the paraboloid z = x2 + y2, the cylinder x2 + y2 = 25, and the xy-plane

A)625

4π B)

625

3π C)

625

2π D)

625

6π

26)

Write an iterated integral for

S

f(x, y, z) dV.∫∫∫

27) S is the region in the first octant bounded by the cylinder y2 + z2 = 1 and the planes x = 5 and x = 7.

A)7

5

1

0

1 - z2

0

f(x, y, z) dx dz dy∫∫∫ B)7

5

1

0

1 - z2

0

f(x, y, z) dy dz dx∫∫∫

C)7

5

1

0

1 - z2

0

f(x, y, z) dy dz dx∫∫∫ D)7

5

1

0

1 - y2

0

f(x, y, z) dx dy dz∫∫∫

27)

Evaluate the integral.

28)1

0

1

0

1

0

(8x + 7y + 4z) dz dy dx∫∫∫

A) 58 B)19

6C)

19

2D)

19

3

28)

Solve the problem.

29) Find the volume of the region enclosed by the paraboloids z = x2 + y2 - 4 and z = 46 - x2 - y2.

A) 625π B) 1875π C) 1250π D) 2500π

29)

30) Find the volume of the region that lies inside the sphere x2 + y2 + z2 = 64 and outside the cylinder

x2 + y2 = 4

A)2(512 - 603/2)π

3B)

3(512 - 603/2)π

2C)

5(512 - 603/2)π

2D)

4(512 - 603/2)π

3

30)

Use the given transformation to evaluate the integral.

31) u = x + y, v = -2x + y;

R

-5x dx dy,∫∫where R is the parallelogram bounded by the lines y = -x + 1, y = -x + 4, y = 2x + 2, y = 2x + 5

A) -10 B) 10 C) 5 D) -5

31)

Solve the problem.

32) Find the direction in which the function is increasing or decreasing most rapidly at the point p.

f(x, y) = xy2 - yx2, p = (2, -1)

A)-8

89 i +

5

89 j B)

5

89 i +

-8

89 j

C)5

89 i +

8

89 j D) 5 89 i + -8 89 j

32)

4

Find the equation of the tangent plane to the given surface at the indicated point.

33) ex sin(yz) - 5x = 0, (0, 3π, 3)

A) 5x - 3(y - 3π) = -3π(z - 3) B) 5x - 1(y - 3π) - 3(z - 3) = 0

C) -5x + 3(y - 3π) + 3π(z - 3) = 0 D) -5x - 3(y - 3π) - 3π(z - 3) = 0

33)

Find the derivative of the function at the point p in the direction of a.

34) f(x, y) = 4x2 - xy + 5y2, p = (-3, -5), a = i - 2j

A) 1 3 B) 25 3 C)3

55 D) 15 5

34)

Find dw

dt by using the Chain Rule. Express your final answer in

35) w = y2 - x ln y; x = st, y = s

t

A) s2 1 - 2

t3 - ln

s

tB) s 1 -

2s

t3 - ln

s

tC) s 1 +

2s

t3 - ln

s

tD) s 1 -

2

t3 - ln

s

t

35)

Provide an appropriate answer.

36) Find ∂w

∂u when u = -6 and v = -4 if w(x, y, z) =

xy2

z, x =

u

v, y = u + v, and z = u · v.

A)∂w

∂u = -

15

2B)

∂w

∂u = -

5

8C)

∂w

∂u = -

5

4D)

∂w

∂u =

10

27

36)

Find the equation of the tangent plane (or tangent "hyperplane" for a function of three variables) at the given point p.

37) x2 + 6xyz + y2 = -8z2, p = (1, 1, 1).

A) -8x - 8y + 10z = -6 B) x + y + z = -6

C) x + y + z = 1 D) -8x - 8y + 10z = 1

37)

Solve the problem.

38) Write parametric equations for the tangent line to the surface x + y2 + 8z = 10 at the point (1, 1, 1)

whose projection on the xy-plane is parallel to the y-axis.

A) x = 1, y = 16t + 1, z = -2t + 1 B) x = 1, y = 16t + 1, z = -t + 1

C) x = 1, y = 8t + 1, z = -t + 1 D) x = 1, y = 8t + 1, z = -2t + 1

38)

Find the indicated limit or state that it does not exist.

39) lim(x, y) → (0, 0)

cos x2

x2 + y2

A) 0 B)π

2C) 1 D) No limit

39)

40) lim(x, y) → (0, 1)

y3 sin x

x

A) 0 B) ∞ C) 1 D) No limit

40)

5

Solve the problem.

41) Find the slope of the tangent to the curve of the intersection of the surface 400z = 16x2 + 25y2 and

the plane x = -5 at the point -5, -4, 2

A) - 1

2B) -

5

8C) -

8

25D) -

2

5

41)

Find all the second order partial derivatives of the given function.

42) f(x, y) = xy2 + yex2 + 5

A) fxx(x, y) = 2yex2; fyy(x, y) = 2x; fyx(x, y) = fxy(x, y) = 2y + 2xex2

B) fxx(x, y) = 2yex2; fyy(x, y) = 2x; fyx(x, y) = fxy(x, y) = 2xex2

C) fxx(x, y) = yex2(1 + 2x2); fyy(x, y) = x; fyx(x, y) = fxy(x, y) = y + xex2

D) fxx(x, y) = 2yex2(1 + 2x2); fyy(x, y) = 2x; fyx(x, y) = fxy(x, y) = 2y + 2xex2

42)

Solve the problem.

43) Find symmetric equations for the line through the points P(-1, -1, -3) and Q(2, -5, -5).

A)x + 1

3 =

y - 1

-4 =

z + 3

-2B)

x - 1

3 =

y + 1

-4 =

z + 3

-2

C)x - 1

3 =

y - 1

-4 =

z + 3

-2D)

x + 1

3 =

y + 1

-4 =

z + 3

-2

43)

Find the curvature κ for the given function.

44) r(t) = -3i + (10 + 2t)j + (t2 + 4)k

A) κ = - 1

2(t2 + 1)3/2B) κ =

1

(t2 + 1)3/2

C) κ = 1

2 t2 + 1D) κ =

1

2(t2 + 1)3/2

44)

Find parametric equations for the line described below.

45) The line through the point P(5, -1, -5) parallel to the vector -6i + 5j - 5k

A) x = 6t - 5, y = 5t + 1, z = -5t + 5 B) x = 6t + 5, y = 5t - 1, z = -5t - 5

C) x = -6t + 5, y = 5t - 1, z = -5t - 5 D) x = -6t - 5, y =5t + 1, z = -5t + 5

45)

Find the curvature κ and radius of curvature R for the curve at the given point.

46) y = cos x, π

4,

2

2

A) κ = 2

3, R =

3

2B) κ =

2

3 3, R =

3 3

2

C) k = 3 3

2, R =

2

3 3D) κ =

3

2 3, R =

2 3

3

46)

Find the point of the curve at which the curvature is at a maximum.

47) y = x2 - 6x

A) (-3, 9) B) (3, -9) C) (-9, 3) D) (9, -3)

47)

6

Find T, N, and B for the given space curve.

48) r(t) = (8t sin t + 8cos t)i + (8t cos t - 8 sin t)j - 6k

A) T = (-cos t)i - (sin t)j; N = sin(t)i - (cos t)j; B = 6k

B) T = (-cos t)i - (sin t)j; N = (sin t)i - (cos t)j; B = k

C) T = (-cos t)i + (sin t)j; N = (sin t)i + (cos t)j; B = -k

D) T = (-cos t)i - (sin t)j; N = (sin t)i - (cos t)j; B = -k

48)

For the curve r(t), write the acceleration in the form aTT + aNN.

49) r(t) = (4t sin t + 4 cos t)i + (4t cos t - 4 sin t)j + 9k

A) a = 4T + 4tN B) a = 4tN C) a = 1

4tN D) a = 4T +

1

4tN

49)

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0).

50) f(x, y) = x2y

x4 + y250)

7

Answer KeyTestname: CALII FINAL REVIEW

1)324

5

2)625

2π

3)1

2(1 - cos 81)

4)1

12(653/2 - 1)

5) (4, 7); local minimum; (4, -7); saddle point; (-4, 7); saddle point; (-4, -7); local maximum

6) B

7) D

8) D

9) B

10) C

11) A

12) C

13) C

14) C

15) C

16) A

17) B

18) B

19) B

20) A

21) C

22) B

23) B

24) A

25) A

26) C

27) B

28) C

29) A

30) D

31) C

32) B

33) D

34) D

35) B

36) C

37) A

38) D

39) D

40) C

41) A

42) D

43) D

44) D

45) C

8

Answer KeyTestname: CALII FINAL REVIEW

46) B

47) B

48) B

49) A

50) Answers will vary. One possibility is Path 1: x = t, y = t ; Path 2: x = t, y = t2

9