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, kak and ka − bk for a = 4i + j and b = 〈 1, −2 〉. Here, i and j are
the 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 the
form of standard basis vectors or the component form.
2a + 3b = 2 (4i + j ) + 3 (i − 2j ) = 11i − 4j = 〈 11, −4 〉
kak = k 〈 4, 1 〉 k =
p
42 + 12 =
p
17
a − b = 〈 4, 1 〉 − 〈 1, −2 〉 = 〈 3, 3 〉 = 3 〈 1, 1 〉
ka − bk = k3 〈 1, 1 〉 k = 3k 〈 1, 1 〉 k = 3
p
p
12 + 12 = 3 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 fly
due north?
Answer. Let v = 〈 x, y 〉 be the direction of the plane and w = 〈 50, 0 〉 represent the wind
velocity. We want v + w = 〈 0, c 〉, where c > 0 so that the plane is traveling due north. We
have
〈 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 kv k = 300, which implies
300 = kv k = k 〈 −50, c 〉 k =
i.e.,
(−50)2 + c2 = 3002 ,
p
(−50)2 + c2 ,
c2 = 3002 − (−50)2 = 87500,
p
c = 50 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:
p
p
v = 〈 −50, 50 35 〉 = 50 〈 −1, 35 〉 .
Page 1 of 1
ä
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◦ , the
vector parallel to the road is 〈 cos 15◦ , sin 15◦ 〉. If the car has weight 2500 pounds, find the
component of the weight vector along the road vector.
Answer. The vector b = 〈 cos 15◦ , sin 15◦ 〉 represents the direction of the banked road. The
vector deduced by the weight of the car is w = 〈 0, −2500 〉. The component of the weight
vector in the direction of the bank is
Compb w =
w·b
= −2500 sin 15◦ ≈ −647.0 lbs
kbk
ä
toward the inside of the curve.
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 〉,
π
b = 〈 0, 0, 2 〉, assuming that 0 ≤ θ ≤ .
2
Answer. The cross product of a and b is
a × b = 〈 4, −4, 0 〉 ,
and
p
ka × bk = 4 2,
kak = 3,
kbk = 2.
We recall the formula:
ka × bk = kak kbk sin θ ,
p
2 2
sin θ =
,
i.e.,
3
p
4 2 = 3(2) sin θ ,
à p !
2 2
θ = sin−1
≈ 1.23096 radian.
3
i.e.,
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 coplanar.
ä
Page 1 of 1
Spring, 2009
Calculus II for Engineering
Quiz 4 { SOLUTION
ID No: Solution
Name: Solution
Date: Tuesday, March 24, 2009
Score:
1/2 (5 points) Evaluate the integral:
Z D
cos(3t); sin t;
e4t
E
dt +
Z 2*
0
10/10
4
; e t 2 ; t2
t+1
+
dt.
Answer.
Z 2*
+
4
cos(3t); sin t; e4t dt +
; et 2 ; t2 dt
t+1
0
* Z
+
Z
Z 2
Z
Z
Z 2
2 4
4
t
t
2
2
t dt
=
e dt;
cos(3t) dt; sin t dt; e dt +
dt;
0
0 t+1
0
*
+
*
" #2 +
sin(3t)
e4t
t3
2
2
t
2
=
+ c1 ; cos t + c2 ;
+ c3 + [4 ln(t + 1)]0 ; [e ]0 ;
3
4
3 0
*
+
*
+
sin(3t)
8
e4t
+ c + 4 ln 3; 1 e 2 ;
;
=
; cos t;
3
4
3
Z D
E
where c = h c1 ; c2 ; c3 i is the constant of the integration.
2/2 (5 points) For the vector{valued function r(t) =
such that r(t) and r0 (t) are perpendicular.
Answer.
D
h t2 ;
¤
t; t2
5 i, nd all values of t
E
(t2 )0 ; (t)0 ; (t2 5)0 = h 2t; 1; 2t i :
We recall the theorem that a and b ar perpendicular if and only if a b = 0. Using the
theorem, we get
r 0 (t) =
D
E
0 = r(t) r0 (t) = t2 ; t; t2 5 h 2t; 1; 2t i
= 2t3 + t + 2t(t2 5) = t(4t2 9) = t(2t 3)(2t + 3);
3
i:e:; t = 0;
t= :
2
Page 1 of 1
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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:
(x + y) dA, where R is the region that lies to the left of the y–axis between the circles x2 + y 2 = 1
R
and x2 + y 2 = 4.
Answer. The region R is represented by
R=
{
(x, y) | 1 ≤ x2 + y 2 ≤ 4, x ≤ 0
}
.
Using the polar coordinates, the region R can be expressed by
{
R=
3π
π
≤θ≤
, 1≤r≤2
(r, θ) |
2
2
}
.
Using this representation with the polar coordinates, the integral becomes
∫∫
∫
θ= 3π
2
∫
r=2
(x + y) dA =
(r cos θ + r sin θ)r drdθ
θ= π2
R
∫
θ= 3π
2
r=1
∫
r=2
r2 (cos θ + sin θ) drdθ
=
θ= π2
[∫
r=1
] [∫
θ= 3π
2
=
r=2
(cos θ + sin θ) dθ
θ= π2
]
]θ= 3π2 [ r3 ]r=2
14
=− .
r dr = sin θ − cos θ
π
3 r=1
3
θ= 2
[
2
r=1
2
1
Y 0
-1
-2
-2
-1
0
X
1
2
Region R bounded by x2 + y 2 = 1 and x2 + y 2 = 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:
z dV , where Q is bounded
Q
by the cylinder y 2 + z 2 = 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 following
representation:
{
y }
3
{
}
2
2
R = (y, z) | 0 ≤ y + z ≤ 9, 0 ≤ y, 0 ≤ z
}
{
√
= (y, z) | 0 ≤ y ≤ 3, 0 ≤ z ≤ 9 − y 2
(x, y, z) | (y, z) ∈ R, 0 ≤ x ≤
Q=
(Vertical Cut)
Using this, the integral becomes
∫∫∫
∫∫ ∫
z dV =
Q
∫
R
x=0
y=3 [
=
y=0
∫
x= y3
yz 2
6
z dxdA =
] √ 2
z=
y=3
∫
z=
√
9−y 2
∫
∫
x= y3
y=3
∫
z=
z dxdzdy =
y=0
9−y
∫
z=0
y=3
dy =
y=0
z=0
x=0
y=0
z=0
√
9−y 2
yz
dzdy
3
y(9 − y 2 )
27
dy = .
6
8
3
Y
0
2
Y 2
1
-2
0
3
2
2
Z
Z 0
1
-2
0
0.0
0.0
0.5
X
0.5
X
1.0
1.0
2
Solid bounded by cylinder y 2 + z 2 = 9 and
the planes x = 0, x = 1 and z = 0
Cylinder y 2 + z = 9
3
0.0
X
0.5
2
Y
1
1.0
3
0
2
Z
1
0
Solid bounded by cylinder y 2 + z 2 = 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:
yex dA, where R is the region in the first quadrant enclosed by the circle x2 + y 2 = 25.
R
Answer. The region R is represented by
R=
{
(x, y) | x2 + y 2 ≤ 25, 0 ≤ x, 0 ≤ y
}
.
Using the polar coordinates, the region R can be expressed by
{
R=
(r, θ) | 0 ≤ θ ≤
}
π
, 0≤r≤5 .
2
Using this representation with the polar coordinates, the integral becomes
∫∫
∫
θ= π2
∫
θ=0
θ= π2
∫
x
r=5
r sin θer cos θ r drdθ
ye dA =
R
∫
r=0
r=5
r2 sin θer cos θ drdθ = 4e5 −
=
θ=0
r=0
23
.
2
4
2
Y 0
-2
-4
-4
-2
0
X
2
4
Region R bounded by x2 + y 2 = 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:
x dV , where Q is bounded
Q
by the paraboloid x = 4y 2 + 4z 2 and the plane x = 4.
Answer. We use the projected region R of the solid Q onto the yz–plane. Then Q has the following
representation:
{
}
Q = (x, y, z) | (y, z) ∈ R, 4y 2 + 4z 2 ≤ x ≤ 4
{
}
R = (y, z) | 0 ≤ y 2 + z 2 ≤ 1
}
{
√
2
= (y, z) | 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y
(Vertical Cut)
Using this, the integral becomes
∫∫ ∫
∫∫∫
x=4
x dV =
Q
x dxdA
R
∫
y=1
x=4y 2 +4z 2
∫
z=
√
1−y 2
∫
x=4
=
x dxdzdy
∫
y=0
y=1
z=0
∫
z=
√
x=4y 2 +4z 2
1−y 2
=
y=0
z=0
42 − (4y 2 + 4z 2 )2
4π
dzdy =
.
2
3
1.0
Y 0.5
0.0
-0.5
-1.0
1.0
0.5
0.0
Z
-0.5
-1.00
1
2
X
3
4
Solid Q bounded by the paraboloid x = 4y 2 + 4z 2 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:
{
tan−1 (y/x) dA, where R =
}
(x, y) | 1 ≤ x2 + y 2 ≤ 4, 0 ≤ y ≤ x .
R
Answer. Using the polar coordinates, the region R can be expressed by
{
R=
(r, θ) | 0 ≤ θ ≤
}
π
, 1≤r≤2 .
4
Using this representation with the polar coordinates, the integral becomes
∫∫
∫
x
θ= π4
∫
r=2
ye dA =
θr drdθ
] [∫
r=1
[θ=0
∫ θ= π
4
R
=
r=2
θ dθ
θ=0
]
r dr =
r=1
3π 2
.
64
2
1
Y 0
-1
-2
-2
Region R =
{
-1
0
X
1
2
(x, y) | 1 ≤ x2 + y 2 ≤ 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:
xy dV , where Q is bounded
Q
by the parabolic cylinders y = x2 and x = y 2 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 following
representation:
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
∫∫∫
∫∫ ∫
xy dV =
Q
∫
z=x+y
x=1
∫
√
y= x
∫
z=x+y
xy dzdA =
∫
R
z=0
x=1
∫
√
y= x
=
xy dzdydx
x=0
xy(x + y) dydx =
y=x2
x=0
y=x2
z=0
3
.
28
1.0
Y
0.5
0.0
2.0
1.5
Z
1.0
0.5
0.0
0.0
0.5
X
1.0
Parabolic cylinders x = y 2 and y = x2
1.0
Y
0.5
0.0
2.0
1.5
Z
1.0
0.5
0.0
0.0
0.5
X
1.0
Solid bounded by cylinders x = y 2 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:
x dA, where R is the region in the first quadrant that lies between the circles x2 + y 2 = 4 and
R
x2 + y 2 = 2x.
Answer. Using the polar coordinates, x2 + y 2 = 4 corresponds to r2 = 4, i.e., r = 2, while x2 + y 2 = 2x
corresponds 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 cos θ ≤ r ≤ 2 .
2
Using this representation with the polar coordinates, the integral becomes
∫∫
∫
θ= π2
∫
r=2
x dA =
R
(r cos θ) r drdθ
θ=0
∫ θ= π
2
r=2 cos θ
∫ r=2
r2 cos θ drdθ =
=
θ=0
r=2 cos θ
8 π
− .
3 2
Y
2
1
X
-2
1
-1
2
-1
-2
Region R in the first quadrant that lies between the circles x2 + y 2 = 4 and x2 + y 2 = 2x
Page 1 of 2
Calculus II for Engineering
Spring, 2009
∫∫∫
x2 ey dV , where Q is bounded
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
Q
by the parabolic cylinder z = 1 − y 2 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 following
representation:
Q = { (x, y, z) | (y, z) ∈ R, −1 ≤ x ≤ 1 }
}
{
R = (y, z) | −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2
(Vertical Cut)
Using this, the integral becomes
∫∫∫
∫∫ ∫
x=1
2 y
x2 ey dxdA
x e dV =
Q
∫
R
x=−1
∫
y=1
z=1−y 2
∫
x=1
x2 ey dxdzdy
=
y=−1
∫
y=1
z=0
∫
z=1−y 2
=
y=−1
z=0
x=−1
2ey
8
dzdy = .
3
3e
1.0
Y 0.5
0.0
-0.5
-1.0
1.0
Z
0.5
0.0
-1.0
-0.5
0.0
X
0.5
1.0
Solid Q bounded by the parabolic cylinder z = 1 − y 2 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:
1 dA, where R is within both of the circles x2 + y 2 = x and x2 + y 2 = y.
R
Answer. Using the polar coordinates, x2 + y 2 = x corresponds to r2 = r cos θ, i.e., r = cos θ, while
x2 + y 2 = 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
π
π
π
observe the region R is bounded by r = sin θ for 0 ≤ θ ≤ and r = cos θ for ≤ θ ≤ . So the region
4
4
2
R can be expressed by
{
R=
(r, θ) | 0 ≤ θ ≤
} {
}
π
π
π
, 0 ≤ r ≤ sin θ ∪ (r, θ) |
≤ θ ≤ , 0 ≤ r ≤ cos θ .
4
4
2
Using this representation with the polar coordinates, the integral becomes
∫∫
∫
θ= π4
∫
θ= π2
∫
r=cos θ
r drdθ +
1 dA =
R
∫
r=sin θ
θ=0
r drdθ =
θ= π4
r=0
r=0
π−2
π−2 π−2
+
=
.
16
16
8
Another Answer: Using Rectangular Coordinates. When we use the rectangular coordinates, the region
R is represented by



√( )
√( )
(
)2 
2
2
1 1
1
1
1
R = (x, y) | 0 ≤ x ≤ , −
− x2 ≤ y ≤
− x−


2 2
2
2
2
(Vertical Cut)
Hence, the integral becomes
∫∫
∫
x= 21
∫
√
y=
1 dA =
R
x=0
y= 21 −
2
2
( 12 ) −(x− 21 )
√
2
( 21 )
Page 1 of 3
1 dydx =
−x2
π−2
.
8
Calculus II for Engineering
Spring, 2009
Y
1
X
1
-0.5
-0.5
Region R is within both of the circles x2 + y 2 = x,
i.e., r = cos θ (upper circle), and x2 + y 2 = y, i.e.,
r = sin θ (lower circle)
∫∫∫
2/2 (5 points) (1) Sketch the region and (2) evaluate the triple integral:
y dV , where Q is bounded
Q
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 following
representation:
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
∫∫∫
∫∫ ∫
y dV =
Q
∫
z=4−2x−2y
x=2
∫
y=2−x
∫
z=4−2x−2y
y dzdA =
∫
R z=0
x=2 ∫ y=2−x
=
x=0
y=0
y dzdydx
x=0
y=0
z=0
4
y(4 − 2x − 2y) dydx = .
3
2.0
Y 1.5
1.0
0.5
0.0
4
3
Z
2
1
0
0.0
0.5
1.0
X
1.5
2.0
Solid Q bounded by the planes x = 0, y = 0, z = 0
and 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:
e−x
2 −y 2
dA, where R is the region bounded by the semicircle x =
√
4 − y 2 and the y–axis.
R
√
Answer. Using the polar coordinates, x = 4 − y 2 corresponds to x2 = 4 − y 2 , x2 + y 2 = 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
π
≤θ≤ , 0≤r≤2 .
2
2
Using this representation with the polar coordinates, the integral becomes
∫∫
−x2 −y 2
e
∫
θ= π2
∫
r=2
dA =
θ=− π2
R
[∫
r=0
θ= π2
=
e−r r drdθ
2
] [∫
1 dθ
θ=− π2
]
r=2
re
−r2
dr =
r=0
(e4 − 1)π
.
2e4
Y
2
1
X
-2
1
-1
2
-1
-2
Region R bounded by the semicircle x =
and the y–axis
Page 1 of 2
√
4 − y2
Calculus II for Engineering
Spring, 2009
∫∫∫
2/2 (5 points) (1) Sketch the region R and (2) evaluate the triple integral:
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 = 0
and x = 1.
Q
Answer. We use the projected region R of the solid Q onto the xy–plane. Then Q has the following
representation:
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
∫∫∫
∫∫ ∫
6xy dV =
Q
∫
z=1+x+y
x=1
∫
√
y= x
∫
z=1+x+y
6xy dzdydx
6xy dzdA =
∫
R
x=1
z=0
∫
x=0
√
y= x
=
y=0
6xy(1 + x + y) dydx =
x=0
y=0
z=0
65
.
28
1.0
1.0
Y
0.5
Y
0.5
0.0
0.0
3
3
2
2
Z
Z
1
1
0
0.0
0
0.0
0.5
X
0.5
X
1.0
Solid Q bounded by the curves y =
y = 0, x = 1, z = 0 and z = 1
√
1.0
x,
Page 2 of 2
√
Solid Q bounded by the curves y = x,
y = 0, x = 1, z = 0 and z = 1 + x + y