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
Evaluate the integral.
9 9
3)
sin (x2) dx dy
0 y
∫ ∫
2)
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 + xyez k
1
A)
+ ez
ln(2y + z)
6)
B)
C) ei + ln(2y + z)j + xyez k
2
+ xyez
ln(2y + z)
D) e +
2
+ xyez
ln(2y + z)
Find curl F.
7) F(x, y, z) = ei + ln(2y + z)j + xyez k
1
A) xez i -yez j + 0k
(2y + z)2
C) xez -
7)
1
i - yez j + 1k
2y + z
B) xez +
1
i + yez j + 1k
2y + z
D) xez -
1
i - yez j + 0k
2y + z
Evaluate the line integral.
8)
∫
C
x+y+z
4
4
5
ds ; C is the curve x = 3t, y = (5 cos t), z = (5 sin t), 0 ≤ t ≤ π
5
5
5
4
A)
75 2
π + 25
32
B)
75 25
+
32
4
C)
1
75
π
32
8)
D)
75 2 25
π +
32
4
9)
∫
y2 dx + x dy ; C is the curve x = t2 - 1, y = 6t, 0 ≤ t ≤ 1
9)
C
A) 72
10)
∫
B) 14
C)
5
6
D) 26
(2x - 4y + 9z) dx + (3x - 6y - 3z) dy + (5x +y - z) dz ; C is the line segment path from (0, 0, 0) to
C
(1, 0, 0) to (1, 2, 0) to (1, 2, 4)
A) 40
B) 39
C) 15
D) 16
Determine whether F is conservative.
11) F(x,y,z) = 7i + -ez j - yezk
11)
A) conservative
B) not conservative
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
x2
3y2
x2
3y2
A) f(x, y) =
B) f(x, y) =
- 3xy +
+C
+ 3xy - 7x +
+C
2
2
2
2
C) f(x, y) =
x2
3y2
- 3xy + 7x +
+ 8y + C
2
2
∫
∫
(3, π/2)
12)
D) not conservative
Evaluate the integral.
(4, 2)
13)
(3x2 y - y2) dx + (x3 - 2xy) dy
(-2, 3)
A) 154
B) 346
14)
10)
13)
C) 118
D) 246
ex sin y dx + ex cos y dy
14)
(-4, 0)
A) 0
B) e-4
C) e3
D) e3 - e-4
Apply Green's Theorem to evaluate the integral.
15)
C
(6y + x) dx + (y + 2x) dy
15)
C: The circle (x - 2)2 + (y - 3)2 = 9
A) -120
B) -24
16)
C
C) -36π
D) 36π
(y2 + 5) dx + (x2 + 6) dy; C: The triangle bounded by x = 0, x + y = 2, y = 0
A) 0
B)
16
3
C) -24
2
16)
D) 32
Using Green's Theorem, calculate the area of the indicated region.
4
17) The area bounded above by y = 4 and below by y = x2
9
A) 32
Evaluate
∫∫
B) 16
C) 0
17)
D) 8
g(x, y, z) dS .
G
18) g(x, y, z) =
xz 2
; G is the cap cut from the sphere x2 + y2 + z 2 = 9 by the cone z =
27
A) 4π
B) 0
19) g(x, y, z) = z; G: y2 + z2 = 49, z ≥ 0, 2 ≤ x ≤ 3
A) 49
B) 98
Use Gauss's Divergence Theorem to calculate
∫ ∫
x2 + y2
C) 12π
D) 6π
C) 490
D) 14
19)
F · n dS .
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
Use Stokes's Theorem to calculate
∫∫
18)
20)
(curl F) · n dS .
S
2
2
5
21) F = -3x yi + 3xy j + z k; S is the portion of the paraboloid 1 - x2 - y2 = z that lies above the x-y
plane
3
A) 6
B) 1π
C) π
D) 6π
2
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
2
2
4
4
A)
π
B) π
C) π
D)
25
25
25
25
21)
22)
Evaluate the given double integral by changing it to an iterated integral.
23)
∫∫
S
A) 1
1
dA; S is the region bounded by the x-axis, the line x = 9, and the curve y = ln x
ln x
B) 8
C) 10
D) 9
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
5000
10000
A) 5000π
B) 2500π
C)
π
D)
π
3
3
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)
23)
24)
25)
C) 16(π - 1)
3
D) 16(π - 2)
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
625
625
625
625
A)
π
B)
π
C)
π
D)
π
4
3
2
6
Write an iterated integral for
∫∫∫
26)
f(x, y, z) dV.
S
27) S is the region in the first octant bounded by the cylinder y2 + z2 = 1 and the planes x = 5 and x = 7.
7 1
7
1
1 - z2
1 - z2
A)
f(x, y, z) dx dz dy
B)
f(x, y, z) dy dz dx
5 0
5
0 0
0
∫ ∫ ∫
7
C)
1
∫ ∫ ∫
5
0
27)
∫ ∫ ∫
1 - z2
7
f(x, y, z) dy dz dx
D)
∫ ∫ ∫
5
0
Evaluate the integral.
1 1
1
28)
(8x + 7y + 4z) dz dy dx
0 0
0
19
A) 58
B)
6
1
0
1 - y2
f(x, y, z) dx dy dz
0
∫ ∫ ∫
28)
C)
19
2
D)
19
3
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 + z 2 = 64 and outside the cylinder
x2 + y2 = 4
30)
A)
2(512 - 603/2)π
3
B)
3(512 - 603/2)π
2
C)
5(512 - 603/2)π
2
D)
4(512 - 603/2)π
3
Use the given transformation to evaluate the integral.
31) u = x + y, v = -2x + y;
31)
∫ ∫ -5x dx dy,
R
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
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
i+
89
5
j
89
B)
C)
5
i+
89
8
j
89
D) 5 89 i + -8 89 j
4
5
i+
89
-8
j
89
32)
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)
C) -5x + 3(y - 3π) + 3π(z - 3) = 0
33)
B) 5x - 1(y - 3π) - 3(z - 3) = 0
D) -5x - 3(y - 3π) - 3π(z - 3) = 0
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
Find
3
C)
5
B) 25 3
34)
5
D) 15 5
dw
by using the Chain Rule. Express your final answer in
dt
s
35) w = y2 - x ln y; x = st, y =
t
2
s
A) s2 1 - ln
3
t
t
35)
B) s 1 -
2s
s
- ln
3
t
t
C) s 1 +
2s
s
- ln
3
t
t
D) s 1 -
2
s
- ln
3
t
t
Provide an appropriate answer.
36) Find
A)
xy2
u
∂w
when u = -6 and v = -4 if w(x, y, z) =
, x = , y = u + v, and z = u · v.
v
z
∂u
15
∂w
=2
∂u
B)
5
∂w
=8
∂u
C)
5
∂w
=4
∂u
D)
36)
∂w 10
=
27
∂u
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 = -8z 2 , p = (1, 1, 1).
37)
A) -8x - 8y + 10z = -6
C) x + y + z = 1
B) x + y + z = -6
D) -8x - 8y + 10z = 1
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
Find the indicated limit or state that it does not exist.
x2
39)
lim
cos
2
x + y2
(x, y) → (0, 0)
A) 0
40)
lim
(x, y) → (0, 1)
A) 0
B)
38)
39)
π
2
C) 1
D) No limit
y3 sin x
x
40)
B) ∞
C) 1
5
D) No limit
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
1
5
8
2
A) B) C) D) 2
8
25
5
Find all the second order partial derivatives of the given function.
42) f(x, y) = xy2 + yex2 + 5
41)
42)
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
Solve the problem.
43) Find symmetric equations for the line through the points P(-1, -1, -3) and Q(2, -5, -5).
x+ 1 y- 1 z+ 3
x- 1 y+1 z+ 3
A)
B)
=
=
=
=
3
3
-4
-2
-4
-2
C)
x- 1 y- 1 z+ 3
=
=
3
-4
-2
D)
x+ 1 y+1 z+ 3
=
=
3
-4
-2
Find the curvature κ for the given function.
44) r(t) = -3i + (10 + 2t)j + (t2 + 4)k
C) κ =
44)
1
A) κ = -
B) κ =
2(t2 + 1)3/2
1
2 t2 + 1
D) κ =
1
(t2 + 1)3/2
1
2(t2 + 1)3/2
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
Find the curvature κ and radius of curvature R for the curve at the given point.
π
2
46) y = cos x,
,
4 2
A) κ =
C) k =
2
3
,R=
2
3
B) κ =
3 3
2
, R=
2
3 3
D) κ =
2
3 3
3
2 3
B) (3, -9)
C) (-9, 3)
6
45)
46)
,R=
3 3
2
,R=
2 3
3
Find the point of the curve at which the curvature is at a maximum.
47) y = x2 - 6x
A) (-3, 9)
43)
47)
D) (9, -3)
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 a TT + a NN.
49) r(t) = (4t sin t + 4 cos t)i + (4t cos t - 4 sin t)j + 9k
A) a = 4T + 4tN
49)
B) a = 4tN
C) a =
1
N
4t
D) a = 4T +
1
N
4t
Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0).
x2 y
50) f(x, y) =
50)
x4 + y2
7
Answer Key
Testname: CALII FINAL REVIEW
1)
324
5
2)
625
π
2
3)
1
(1 - cos 81)
2
4)
1
(653/2 - 1)
12
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 Key
Testname: 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