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Calculus II for Engineering Spring, 2009
QUIZ 1 – SOLUTION
ID No: Solution Name: Solution
Date: Tuesday, February 17, 2009 Score: 10/10
1/2 (5 points) Find 2a+3b, ‖a‖ and ‖a−b‖ for a= 4i+j and b= ⟨ 1,−2 ⟩. Here, i and j arethe standard basis vectors of the V2 space.
Answer. We observe a = 4i+j = ⟨ 4,1 ⟩ and b = i−2j = ⟨ 1,−2 ⟩. One can use either theform of standard basis vectors or the component form.
2a+3b= 2(4i+j)+3(i−2j)= 11i−4j = ⟨ 11,−4 ⟩
‖a‖ = ‖⟨ 4,1 ⟩‖ =√
42+12 =p
17
a−b= ⟨ 4,1 ⟩−⟨ 1,−2 ⟩ = ⟨ 3,3 ⟩ = 3⟨ 1,1 ⟩
‖a−b‖ = ‖3⟨ 1,1 ⟩‖ = 3‖⟨ 1,1 ⟩‖ = 3√
12+12 = 3p
2. ä
2/2 (5 points) The thrust of an airplane’s engines produces a speed of 300 mph in still air.The wind velocity is given by ⟨ 50,0 ⟩. In what direction should the airplane head to flydue north?
Answer. Let v = ⟨ x, y ⟩ be the direction of the plane and w = ⟨ 50,0 ⟩ represent the windvelocity. We want v+w = ⟨ 0, c ⟩, where c > 0 so that the plane is traveling due north. Wehave
⟨ 0, c ⟩ = v+w = ⟨ x, y ⟩+⟨ 50,0 ⟩ = ⟨ x+50, y ⟩ , i.e., x+50= 0, and y= c.
That is, x =−50 and y= c and so v = ⟨ −50, c ⟩. We have ‖v‖ = 300, which implies
300= ‖v‖ = ‖⟨ −50, c ⟩‖ =√
(−50)2+ c2,
i.e., (−50)2+ c2 = 3002, c2 = 3002− (−50)2 = 87500, c = 50p
35,
where we take the positive square root to have the plane moving north (c > 0). Therefore,we conclude that the plane should fly in the direction:
v = ⟨ −50,50p
35 ⟩ = 50⟨ −1,p
35 ⟩ . ä
Page 1 of 1
Calculus II for Engineering Spring, 2009
QUIZ 2 – SOLUTION
ID No: Solution Name: Solution
Date: Tuesday, February 24, 2009 Score: 10/10
1/2 (5 points) Find the displacement vectors−−→PQ and
−−→QR and determine whether the points
P(2,3,1), Q(0,4,2) and R(4,1,4) are colinear (i.e., on the same line).
Answer.
−−→PQ = ⟨ 0−2,4−3,2−1 ⟩ = ⟨ −2,1,1 ⟩ ,
−−→PR = ⟨ 4−2,1−3,4−1 ⟩ = ⟨ 2,−2,3 ⟩ .
We observe that there does not exist a scalar s such that
−−→PQ = ⟨ −2,1,1 ⟩ = s ⟨ 2,−2,3 ⟩ = s
−−→PR,
i.e., any scalar s does not satisfy all the equations at the same time:
−2= 2s, 1=−2s, 1= 3s.
It implies that the vectors,−−→PQ and
−−→PR, are not parallel and thus the points are not
colinear. ä
2/2 (5 points) A car makes a turn on a banked road. When the road is banked at 15◦, thevector parallel to the road is ⟨ cos15◦,sin15◦ ⟩. If the car has weight 2500 pounds, find thecomponent of the weight vector along the road vector.
Answer. The vector b= ⟨ cos15◦,sin15◦ ⟩ represents the direction of the banked road. Thevector deduced by the weight of the car is w = ⟨ 0,−2500 ⟩. The component of the weightvector in the direction of the bank is
Compbw = w ·b‖b‖ =−2500sin15◦ ≈−647.0 lbs
toward the inside of the curve. ä
Page 1 of 1
Calculus II for Engineering Spring, 2009
QUIZ 3 – SOLUTION
ID No: Solution Name: Solution
Date: Tuesday, March 10, 2009 Score: 10/10
1/2 (5 points) Use the cross product to determine the angle between the vectors a= ⟨ 2,2,1 ⟩,b= ⟨ 0,0,2 ⟩, assuming that 0≤ θ ≤ π
2.
Answer. The cross product of a and b is
a×b= ⟨ 4,−4,0 ⟩ , and ‖a×b‖ = 4p
2, ‖a‖ = 3, ‖b‖ = 2.
We recall the formula:
‖a×b‖ = ‖a‖‖b‖sinθ, i.e., 4p
2= 3(2)sinθ,
sinθ = 2p
23
, i.e., θ = sin−1
(2p
23
)≈ 1.23096 radian. ä
2/2 (5 points) Use the parallelepiped volume formula to determine whether the vectors a =⟨ 1,−3,1 ⟩, b= ⟨ 2,−1,0 ⟩ and c= ⟨ 0,−5,2 ⟩ are coplanar.
Answer. The cross product of a and b is a×b= ⟨ 1,2,5 ⟩.The volume formula implies
V = |c · (a×b)| = |⟨ 0,−5,2 ⟩ · ⟨ 1,2,5 ⟩| = 0.
Since the parallelepiped has the volume 0, we conclude that those three vectors are copla-nar. ä
Page 1 of 1
Calculus II for Engineering Spring, 2009
Quiz 4 { SOLUTION
ID No: Solution Name: Solution
Date: Tuesday, March 24, 2009 Score: 10/10
1/2 (5 points) Evaluate the integral:Z D
cos(3t); sin t; e4tEdt+
Z 2
0
*4
t+ 1 ; et�2; t2
+dt.
Answer.Z Dcos(3t); sin t; e4t
Edt+
Z 2
0
*4
t+ 1 ; et�2; t2
+dt
=� Z
cos(3t) dt;Z
sin t dt;Ze4t dt
�+* Z 2
0
4t+ 1 dt;
Z 2
0et�2 dt;
Z 2
0t2 dt
+=*
sin(3t)3 + c1; � cos t+ c2;
e4t
4 + c3
++*
[4 ln(t+ 1)]20 ; [et�2]20 ;"t3
3
#2
0
+=*
sin(3t)3 ; � cos t; e
4t
4
++ c +
*4 ln 3; 1� e�2;
83
+;
where c = h c1; c2; c3 i is the constant of the integration. ¤
2/2 (5 points) For the vector{valued function r(t) = h t2; t; t2 � 5 i, �nd all values of tsuch that r(t) and r0(t) are perpendicular.
Answer.r0(t) =
D(t2)0; (t)0; (t2 � 5)0
E= h 2t; 1; 2t i :
We recall the theorem that a and b ar perpendicular if and only if a � b = 0. Using thetheorem, we get
0 = r(t) � r0(t) =Dt2; t; t2 � 5
E � h 2t; 1; 2t i= 2t3 + t+ 2t(t2 � 5) = t(4t2 � 9) = t(2t� 3)(2t+ 3);
i:e:; t = 0; t = �32 : ¤
Page 1 of 1
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Calculus II for Engineering Spring, 2009
QUIZ 7 – SOLUTIONSection 13.3 Double Integrals in Polar Coordinates
Section 13.5 Triple Integrals
ID No: Solution Name: Solution
Date: Tuesday, May 19, 2009 Score: 10/10
1/2 (5 points) (1) Sketch the region R and (2) evaluate the given integral by changing to polar coordinates:∫∫R
(x + y) dA, where R is the region that lies to the left of the y–axis between the circles x2 + y2 = 1
and x2 + y2 = 4.
Answer. The region R is represented by
R ={
(x, y) | 1 ≤ x2 + y2 ≤ 4, x ≤ 0}
.
Using the polar coordinates, the region R can be expressed by
R =
{(r, θ) | π
2≤ θ ≤ 3π
2, 1 ≤ r ≤ 2
}.
Using this representation with the polar coordinates, the integral becomes∫∫R
(x + y) dA =
∫ θ= 3π2
θ=π2
∫ r=2
r=1
(r cos θ + r sin θ)r drdθ
=
∫ θ= 3π2
θ=π2
∫ r=2
r=1
r2(cos θ + sin θ) drdθ
=
[∫ θ= 3π2
θ=π2
(cos θ + sin θ) dθ
] [∫ r=2
r=1
r2 dr
]=
[sin θ − cos θ
]θ= 3π2
θ=π2
[r3
3
]r=2
r=1
= −14
3. �
-2 -1 0 1 2
-2
-1
0
1
2
X
Y
Region R bounded by x2 + y2 = 1 and x2 + y2 = 4 and x ≤ 0
Page 1 of 3
Calculus II for Engineering Spring, 2009
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
∫∫∫Q
z dV , where Q is bounded
by the cylinder y2 + z2 = 9 and the planes x = 0, y = 3x, and z = 0 in the first octant.
Answer. We use the projected region R of the solid Q onto the yz–plane. Then Q has the followingrepresentation:
Q ={
(x, y, z) | (y, z) ∈ R, 0 ≤ x ≤ y
3
}R =
{(y, z) | 0 ≤ y2 + z2 ≤ 9, 0 ≤ y, 0 ≤ z
}=
{(y, z) | 0 ≤ y ≤ 3, 0 ≤ z ≤
√9 − y2
}(Vertical Cut)
Using this, the integral becomes
∫∫∫Q
z dV =
∫∫R
∫ x= y3
x=0
z dxdA =
∫ y=3
y=0
∫ z=√
9−y2
z=0
∫ x= y3
x=0
z dxdzdy =
∫ y=3
y=0
∫ z=√
9−y2
z=0
yz
3dzdy
=
∫ y=3
y=0
[yz2
6
]z=√
9−y2
z=0
dy =
∫ y=3
y=0
y(9 − y2)
6dy =
27
8. �
0.0
0.5
1.0
X
-2
0
2Y
-2
0
2
Z
Cylinder y2 + z2 = 9
0.0
0.5
1.0
X
0
1
2
3
Y
0
1
2
3
Z
Solid bounded by cylinder y2 + z2 = 9 andthe planes x = 0, x = 1 and z = 0
0.0
0.5
1.0
X
0
1
2
3
Y
0
1
2
3
Z
Solid bounded by cylinder y2 + z2 = 9 and the planes x = 0 and z = 0 and y = 3x
Page 2 of 3
Calculus II for Engineering Spring, 2009
QUIZ 7 – SOLUTION
Section 13.3 Double Integrals in Polar Coordinates
Section 13.5 Triple Integrals
ID No: Solution Name: Solution
Date: Tuesday, May 19, 2009 Score: 10/10
1/2 (5 points) (1) Sketch the region R and (2) evaluate the given integral by changing to polar coordinates:∫∫R
yex dA, where R is the region in the first quadrant enclosed by the circle x2 + y2 = 25.
Answer. The region R is represented by
R ={
(x, y) | x2 + y2 ≤ 25, 0 ≤ x, 0 ≤ y}
.
Using the polar coordinates, the region R can be expressed by
R ={
(r, θ) | 0 ≤ θ ≤ π
2, 0 ≤ r ≤ 5
}.
Using this representation with the polar coordinates, the integral becomes∫∫R
yex dA =
∫ θ=π2
θ=0
∫ r=5
r=0
r sin θer cos θr drdθ
=
∫ θ=π2
θ=0
∫ r=5
r=0
r2 sin θer cos θ drdθ = 4e5 − 23
2. �
-4 -2 0 2 4
-4
-2
0
2
4
X
Y
Region R bounded by x2 + y2 = 25 in the first quadrant
Page 1 of 2
Calculus II for Engineering Spring, 2009
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
∫∫∫Q
x dV , where Q is bounded
by the paraboloid x = 4y2 + 4z2 and the plane x = 4.
Answer. We use the projected region R of the solid Q onto the yz–plane. Then Q has the followingrepresentation:
Q ={
(x, y, z) | (y, z) ∈ R, 4y2 + 4z2 ≤ x ≤ 4}
R ={
(y, z) | 0 ≤ y2 + z2 ≤ 1}
={
(y, z) | 0 ≤ y ≤ 1, 0 ≤ z ≤√
1 − y2}
(Vertical Cut)
Using this, the integral becomes∫∫∫Q
x dV =
∫∫R
∫ x=4
x=4y2+4z2
x dxdA
=
∫ y=1
y=0
∫ z=√
1−y2
z=0
∫ x=4
x=4y2+4z2
x dxdzdy
=
∫ y=1
y=0
∫ z=√
1−y2
z=0
42 − (4y2 + 4z2)2
2dzdy =
4π
3. �
0
1
2
3
4
X
-1.0
-0.5
0.0
0.5
1.0
Y
-1.0
-0.5
0.0
0.5
1.0
Z
Solid Q bounded by the paraboloid x = 4y2 + 4z2 and the plane x = 4
Page 2 of 2
Calculus II for Engineering Spring, 2009
QUIZ 7 – SOLUTION
Section 13.3 Double Integrals in Polar Coordinates
Section 13.5 Triple Integrals
ID No: Solution Name: Solution
Date: Tuesday, May 19, 2009 Score: 10/10
1/2 (5 points) (1) Sketch the region R and (2) evaluate the given integral by changing to polar coordinates:∫∫R
tan−1(y/x) dA, where R ={
(x, y) | 1 ≤ x2 + y2 ≤ 4, 0 ≤ y ≤ x}.
Answer. Using the polar coordinates, the region R can be expressed by
R ={
(r, θ) | 0 ≤ θ ≤ π
4, 1 ≤ r ≤ 2
}.
Using this representation with the polar coordinates, the integral becomes∫∫R
yex dA =
∫ θ=π4
θ=0
∫ r=2
r=1
θr drdθ
=
[∫ θ=π4
θ=0
θ dθ
] [∫ r=2
r=1
r dr
]=
3π2
64. �
-2 -1 0 1 2
-2
-1
0
1
2
X
Y
Region R ={
(x, y) | 1 ≤ x2 + y2 ≤ 4, 0 ≤ y ≤ x}
Page 1 of 2
Calculus II for Engineering Spring, 2009
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
∫∫∫Q
xy dV , where Q is bounded
by the parabolic cylinders y = x2 and x = y2 the planes z = 0 and z = x + y.
Answer. We use the projected region R of the solid Q onto the xy–plane. Then Q has the followingrepresentation:
Q = { (x, y, z) | (x, y) ∈ R, 0 ≤ z ≤ x + y }R =
{(x, y) | 0 ≤ x ≤ 1, x2 ≤ y ≤
√x
}(Vertical Cut)
Using this, the integral becomes∫∫∫Q
xy dV =
∫∫R
∫ z=x+y
z=0
xy dzdA =
∫ x=1
x=0
∫ y=√
x
y=x2
∫ z=x+y
z=0
xy dzdydx
=
∫ x=1
x=0
∫ y=√
x
y=x2
xy(x + y) dydx =3
28. �
0.0
0.5
1.0
X
0.0
0.5
1.0
Y
0.0
0.5
1.0
1.5
2.0
Z
Parabolic cylinders x = y2 and y = x2
0.0
0.5
1.0
X
0.0
0.5
1.0
Y
0.0
0.5
1.0
1.5
2.0
Z
Solid bounded by cylinders x = y2 and y = x2 and the planes z = 0 and z = x + y
Page 2 of 2
Calculus II for Engineering Spring, 2009
QUIZ 7 – SOLUTION
Section 13.3 Double Integrals in Polar Coordinates
Section 13.5 Triple Integrals
ID No: Solution Name: Solution
Date: Tuesday, May 19, 2009 Score: 10/10
1/2 (5 points) (1) Sketch the region R and (2) evaluate the given integral by changing to polar coordinates:∫∫R
x dA, where R is the region in the first quadrant that lies between the circles x2 + y2 = 4 and
x2 + y2 = 2x.
Answer. Using the polar coordinates, x2 + y2 = 4 corresponds to r2 = 4, i.e., r = 2, while x2 + y2 = 2xcorresponds to r2 = 2r cos θ, i.e., r = 2 cos θ. So the region R can be expressed by
R ={
(r, θ) | 0 ≤ θ ≤ π
2, 2 cos θ ≤ r ≤ 2
}.
Using this representation with the polar coordinates, the integral becomes∫∫R
x dA =
∫ θ=π2
θ=0
∫ r=2
r=2 cos θ
(r cos θ) r drdθ
=
∫ θ=π2
θ=0
∫ r=2
r=2 cos θ
r2 cos θ drdθ =8
3− π
2. �
-2 -1 1 2X
-2
-1
1
2
Y
Region R in the first quadrant that lies between the circles x2 + y2 = 4 and x2 + y2 = 2x
Page 1 of 2
Calculus II for Engineering Spring, 2009
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
∫∫∫Q
x2ey dV , where Q is bounded
by the parabolic cylinder z = 1 − y2 and the planes z = 0, x = 1 and x = −1.
Answer. We use the projected region R of the solid Q onto the yz–plane. Then Q has the followingrepresentation:
Q = { (x, y, z) | (y, z) ∈ R,−1 ≤ x ≤ 1 }R =
{(y, z) | −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y2
}(Vertical Cut)
Using this, the integral becomes∫∫∫Q
x2ey dV =
∫∫R
∫ x=1
x=−1
x2ey dxdA
=
∫ y=1
y=−1
∫ z=1−y2
z=0
∫ x=1
x=−1
x2ey dxdzdy
=
∫ y=1
y=−1
∫ z=1−y2
z=0
2ey
3dzdy =
8
3e. �
-1.0
-0.5
0.0
0.5
1.0
X
-1.0
-0.5
0.0
0.5
1.0
Y
0.0
0.5
1.0
Z
Solid Q bounded by the parabolic cylinder z = 1 − y2 and the planes z = 0, x = 1 and x = −1
Page 2 of 2
Calculus II for Engineering Spring, 2009
QUIZ 7 – SOLUTION
Section 13.3 Double Integrals in Polar Coordinates
Section 13.5 Triple Integrals
ID No: Solution Name: Solution
Date: Tuesday, May 19, 2009 Score: 10/10
1/2 (5 points) (1) Sketch the region R and (2) evaluate the given integral by changing to polar coordinates:∫∫R
1 dA, where R is within both of the circles x2 + y2 = x and x2 + y2 = y.
Answer. Using the polar coordinates, x2 + y2 = x corresponds to r2 = r cos θ, i.e., r = cos θ, whilex2 +y2 = y corresponds to r2 = r sin θ, i.e., r = sin θ. We find the intersection points of these two curves:
r cos θ = r sin θ, cos θ = sin θ, tan θ = 1, θ =π
4.
(We observe that the graphs of r = cos θ and r = sin θ are circles within 0 ≤ θ ≤ π.) Moreover, we
observe the region R is bounded by r = sin θ for 0 ≤ θ ≤ π
4and r = cos θ for
π
4≤ θ ≤ π
2. So the region
R can be expressed by
R ={
(r, θ) | 0 ≤ θ ≤ π
4, 0 ≤ r ≤ sin θ
}∪
{(r, θ) | π
4≤ θ ≤ π
2, 0 ≤ r ≤ cos θ
}.
Using this representation with the polar coordinates, the integral becomes∫∫R
1 dA =
∫ θ=π4
θ=0
∫ r=sin θ
r=0
r drdθ +
∫ θ=π2
θ=π4
∫ r=cos θ
r=0
r drdθ =π − 2
16+
π − 2
16=
π − 2
8. �
Another Answer: Using Rectangular Coordinates. When we use the rectangular coordinates, the regionR is represented by
R =
(x, y) | 0 ≤ x ≤ 1
2,
1
2−
√(1
2
)2
− x2 ≤ y ≤
√(1
2
)2
−(
x − 1
2
)2
(Vertical Cut)
Hence, the integral becomes
∫∫R
1 dA =
∫ x= 12
x=0
∫ y=√
( 12)
2−(x− 1
2)2
y= 12−
√( 1
2)2−x2
1 dydx =π − 2
8. �
Page 1 of 3
Calculus II for Engineering Spring, 2009
-0.5 1X
-0.5
1
Y
Region R is within both of the circles x2 + y2 = x,i.e., r = cos θ (upper circle), and x2 + y2 = y, i.e.,
r = sin θ (lower circle)
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
∫∫∫Q
y dV , where Q is bounded
by the planes x = 0, y = 0, z = 0 and 2x + 2y + z = 4.
Answer. We use the projected region R of the solid Q onto the xy–plane. Then Q has the followingrepresentation:
Q = { (x, y, z) | (x, y) ∈ R, 0 ≤ z ≤ 4 − 2x − 2y }R = { (x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x } (Vertical Cut)
Using this, the integral becomes∫∫∫Q
y dV =
∫∫R
∫ z=4−2x−2y
z=0
y dzdA =
∫ x=2
x=0
∫ y=2−x
y=0
∫ z=4−2x−2y
z=0
y dzdydx
=
∫ x=2
x=0
∫ y=2−x
y=0
y(4 − 2x − 2y) dydx =4
3. �
0.0
0.5
1.0
1.5
2.0
X
0.0
0.5
1.0
1.5
2.0
Y
0
1
2
3
4
Z
Solid Q bounded by the planes x = 0, y = 0, z = 0and 2x + 2y + z = 4
Page 2 of 3
Calculus II for Engineering Spring, 2009
QUIZ 7 – SOLUTION
Section 13.3 Double Integrals in Polar Coordinates
Section 13.5 Triple Integrals
ID No: Solution Name: Solution
Date: Tuesday, May 19, 2009 Score: 10/10
1/2 (5 points) (1) Sketch the region R and (2) evaluate the given integral by changing to polar coordinates:∫∫R
e−x2−y2
dA, where R is the region bounded by the semicircle x =√
4 − y2 and the y–axis.
Answer. Using the polar coordinates, x =√
4 − y2 corresponds to x2 = 4 − y2, x2 + y2 = 4, i.e., r = 2,while the y–axis corresponds to θ = π/2 and θ = −π/2. So the region R can be expressed by
R ={
(r, θ) | −π
2≤ θ ≤ π
2, 0 ≤ r ≤ 2
}.
Using this representation with the polar coordinates, the integral becomes∫∫R
e−x2−y2
dA =
∫ θ=π2
θ=−π2
∫ r=2
r=0
e−r2
r drdθ
=
[∫ θ=π2
θ=−π2
1 dθ
] [∫ r=2
r=0
re−r2
dr
]=
(e4 − 1)π
2e4. �
-2 -1 1 2X
-2
-1
1
2
Y
Region R bounded by the semicircle x =√
4 − y2
and the y–axis
Page 1 of 2
Calculus II for Engineering Spring, 2009
2/2 (5 points) (1) Sketch the region R and (2) evaluate the triple integral:
∫∫∫Q
6xy dV , where Q lies under
the plane z = 1 + x + y and above the region R in the xy–plane bounded by the curves y =√
x, y = 0and x = 1.
Answer. We use the projected region R of the solid Q onto the xy–plane. Then Q has the followingrepresentation:
Q = { (x, y, z) | (x, y) ∈ R, 0 ≤ z ≤ 1 + x + y }R =
{(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤
√x
}(Vertical Cut)
Using this, the integral becomes∫∫∫Q
6xy dV =
∫∫R
∫ z=1+x+y
z=0
6xy dzdA =
∫ x=1
x=0
∫ y=√
x
y=0
∫ z=1+x+y
z=0
6xy dzdydx
=
∫ x=1
x=0
∫ y=√
x
y=0
6xy(1 + x + y) dydx =65
28. �
0.0
0.5
1.0
X
0.0
0.5
1.0
Y
0
1
2
3
Z
Solid Q bounded by the curves y =√
x,y = 0, x = 1, z = 0 and z = 1
0.0
0.5
1.0
X
0.0
0.5
1.0
Y
0
1
2
3
Z
Solid Q bounded by the curves y =√
x,y = 0, x = 1, z = 0 and z = 1 + x + y
Page 2 of 2