QUIZ 1 – SOLUTION - faculty.uaeu.ac.aefaculty.uaeu.ac.ae/jaelee/class/1120/09Sp_QZS.pdf · Calculus II for Engineering Spring, 2009 QUIZ 2 – SOLUTION ID No: Solution Name: Solution

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


QUIZ 2 – 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


QUIZ 3 – 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


Quiz 4 { 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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QUIZ 7 – SOLUTIONSection 13.3 Double Integrals in Polar Coordinates

Section 13.5 Triple Integrals


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


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


QUIZ 7 – SOLUTION

Section 13.3 Double Integrals in Polar Coordinates





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

}.


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



∫∫∫Q

x dV , where Q is bounded

by the paraboloid x = 4y2 + 4z2 and the plane x = 4.


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


QUIZ 7 – SOLUTION






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

}.


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



∫∫∫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)


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


QUIZ 7 – SOLUTION






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

}.


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



∫∫∫Q

x2ey dV , where Q is bounded

by the parabolic cylinder z = 1 − y2 and the planes z = 0, x = 1 and x = −1.


Q = { (x, y, z) | (y, z) ∈ R,−1 ≤ x ≤ 1 }R =

{(y, z) | −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y2

}(Vertical Cut)


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


QUIZ 7 – SOLUTION






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

}.


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


-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)


∫∫∫Q

y dV , where Q is bounded

by the planes x = 0, y = 0, z = 0 and 2x + 2y + z = 4.


Q = { (x, y, z) | (x, y) ∈ R, 0 ≤ z ≤ 4 − 2x − 2y }R = { (x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x } (Vertical Cut)


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


QUIZ 7 – SOLUTION






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

}.


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


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.


Q = { (x, y, z) | (x, y) ∈ R, 0 ≤ z ≤ 1 + x + y }R =

{(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤

√x

}(Vertical Cut)


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

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QUIZ 1 – SOLUTION - faculty.uaeu.ac.aefaculty.uaeu.ac.ae/jaelee/class/1120/09Sp_QZS.pdf · Calculus II for Engineering Spring, 2009 QUIZ 2 – SOLUTION ID No: Solution Name: Solution