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REVIEW FOR THE EXAM 3 MATH 203 F 2014 (6584966)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Question
1.
Question Details
SEssCalc2 12.1.008. [2155836]
-
SEssCalc2 12.1.015. [2165544]
-
SEssCalc2 12.1.027. [2156155]
-
Evaluate the double integral by first identifying it as the volume of a solid.
(5 − x) dA, R = {(x, y) | 0 ≤ x ≤ 5, 0 ≤ y ≤ 3}
R
37.5
Solution or Explanation
Click to View Solution
2.
Question Details
Calculate the iterated integral.
π/2
5
−5 0
(y + y2 cos x) dx dy
Solution or Explanation
5
−5 0
π/2
(y + y2 cos x)dx dy =
=
5
−5
5
−5
=
3.
xy + y2 sin x
x = π/2
dy
x=0
π y + y2 dy =
2
25π
125
+
−
4
3
π y2 + 1 y3
4
3
25π
125
−
4
3
Question Details
5
−5
=
250
3
Sketch the solid whose volume is given by the iterated integral.
1
0
1
0
(6 − x − 4y)dx dy
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Solution or Explanation
z = f(x, y) = 6 − x − 4y ≥ 0 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. So the solid is the region in the first octant which lies below the
plane z = 6 − x − 4y and above [0, 1]
4.
× [0, 1].
Question Details
SEssCalc2 12.1.023. [2165847]
-
Calculate the double integral.
3x sin(x + y) dA, R = 0,
R
π × 0, π
6
3
Solution or Explanation
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Question Details
SEssCalc2 12.1.029. [2165990]
-
Find the volume of the solid that lies under the plane 4x + 10y − 2z + 17 = 0 and above the rectangle
R = {(x, y) | −1 ≤ x ≤ 2, −1 ≤ y ≤ 1}.
Solution or Explanation
The solid lies under the plane 4x + 10y − 2z + 17 = 0 or z = 2x + 5y +
R
1
=
−1
6.
1
2
1
17
17
17 x = 2
dA =
2x + 5y +
dx dy =
x2 + 5xy +
x
dy
2
2
2
x = −1
−1 −1
−1
1
1
15
57
21 + 10y − −
− 5y dy =
+ 15y dy = 57 y + 15 y2
= 36 − (−21) = 57
2
2
2
2
−1
−1
2x + 5y +
V =
17
so
2
Question Details
SEssCalc2 12.1.031. [2166028]
-
Find the volume of the solid that lies under the elliptic paraboloid x2/9 + y2/16 + z = 1 and above the rectangle
R = [−1, 1]
× [−3, 3].
Solution or Explanation
Click to View Solution
7.
Question Details
SEssCalc2 12.1.035. [2166164]
-
Find the volume of the solid enclosed by the paraboloid z = 2 + x2 + (y − 2)2 and the planes z = 1, x = −2, x = 2, y = 0, and
y = 4.
Solution or Explanation
Click to View Solution
8.
Question Details
SEssCalc2 12.1.034. [2165404]
-
Find the volume of the solid in the first octant bounded by the parabolic cylinder z = 16 − x2 and the plane y = 4.
Solution or Explanation
Click to View Solution
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Question Details
SEssCalc2 12.2.001. [2166100]
-
SEssCalc2 12.2.003. [2165216]
-
SEssCalc2 12.2.015.MI. [2560458]
-
Evaluate the iterated integral.
y
4
0
3xy2 dx dy
0
Solution or Explanation
Click to View Solution
10.
Question Details
Evaluate the iterated integral.
1
x
x2
0
(7 + 14y) dy dx
Solution or Explanation
Click to View Solution
11.
Question Details
Evaluate the double integral.
D
3x cos y dA, D is bounded by y = 0, y = x2, x = 7
Solution or Explanation
Click to View Solution
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Question Details
SEssCalc2 12.2.020. [2165671]
-
SEssCalc2 12.2.019. [2166136]
-
SEssCalc2 12.2.033. [2155958]
-
Evaluate the double integral.
D
8xy dA,
D is the triangular region with vertices (0, 0), (1, 2), and (0, 3)
Solution or Explanation
D
8xy dA =
1
0
=
1
3−x
2x
1
4xy2
0
4x (3 − x)2 − (2x)2 dx = 4
0
y=3 − x
dx
y=2x
1
0
= 4 −
13.
8xy dydx =
(−3x3 − 6x2 + 9x)dx
1
3 4
9
3
9
x − 2x3 + x2 = 4 −
−2+
4
2
4
2
0
=7
Question Details
Evaluate the double integral.
D
(3x − 7y) dA,
D is bounded by the circle with center the origin and radius 1
Solution or Explanation
Click to View Solution
14.
Question Details
Sketch the solid whose volume is given by the iterated integral.
7
0
7−x
0
(7 − x − y)dy dx
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Solution or Explanation
The solid lies below the plane z = 7 − x − y or x + y + z = 7 and above the region
D = {(x, y) | 0 ≤ x ≤ 7, 0 ≤ y ≤ 7 − x} in the xy-plane. The solid is a tetrahedron.
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Question Details
SEssCalc2 12.2.505.XP.MI. [2560095]
-
Evaluate the iterated integral.
π/2
0
cos θ
0
3esin θ dr dθ
Solution or Explanation
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Question Details
SEssCalc2 12.2.509.XP. [2166295]
-
SEssCalc2 12.3.018. [2165609]
-
Evaluate the double integral.
(6x + 3y) dA, D is bounded by y =
D
x and y = x2
Solution or Explanation
x
1
0
6x + 3y dy dx =
x2
1
6xy +
0
1
=
6x3/2 +
0
=
17.
x
3 2
y
dx
2
x2
3
3 4
x − 6x3 −
x dx
2
2
12 5/2
3 2
3 4
3 5 1
27
x
+
x −
x −
x
=
5
4
2
10
20
0
Question Details
Use polar coordinates to find the volume of the given solid.
Bounded by the paraboloids z = 8x2 + 8y2 and z = 9 − x2 − y2
Solution or Explanation
The two paraboloids intersect when 8x2 + 8y2 = 9 − x2 − y2 or x2 + y2 = 1. So
2π
(9 − x2 − y2) − 8(x2 + y2) dA =
V =
x2 + y2 ≤ 1
2π
1
=
0
dθ
0
(9r − 9r3)dr = θ
0
2π
0
1
0
9(1 − r2) r dr dθ
9 2
9 4 1
9
r −
r
=
π
2
4
2
0
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Question Details
SEssCalc2 12.3.019. [2165959]
-
Use polar coordinates to find the volume of the given solid.
Inside both the cylinder x2 + y2 = 4 and the ellipsoid 4x2 + 4y2 + z2 = 64
Solution or Explanation
The given solid is the region inside the cylinder x2 + y2 = 4 between the surfaces z =
64 − 4x2 − 4y2 . So
z=−
V =
= 4
x2 + y2 ≤ 4
2π
2
0
0
64 − 4x2 − 4y2 − −
16 − r2 r dr dθ = 4
64 − 4x2 − 4y2
2π
0
dθ
1
8π
(123/2 − 163/2) =
(64 − 24
= 8π −
3
3
19.
64 − 4x2 − 4y2 and
2
r
0
dA =
16 − r2 dr = 4 θ
x2 + y2 ≤ 4
2π
0
−
2
64 − 4x2 − 4y2 dA
2
1
(16 − r2)3/2
3
0
3)
Question Details
SEssCalc2 12.3.502.XP. [2165118]
-
Evaluate the given integral by changing to polar coordinates.
D
x dA , where D is the region in the first quadrant that lies between the circles x2 + y2 = 4 and x2 + y2 = 2x
Solution or Explanation
Click to View Solution
20.
Question Details
SEssCalc2 12.3.504.XP. [2166168]
-
Use a double integral to find the area of the region.
The region inside the cardioid r = 1 + cos θ and outside the circle r = 3 cos θ
Solution or Explanation
Click to View Solution
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Question Details
SEssCalc2 12.4.010. [2165018]
-
Find the mass and center of mass of the lamina that occupies the region D and has the given density function ρ.
D is bounded by the parabolas y = x2 and x = y2;
ρ(x, y) = 11
x
m =
(x, y) =
Solution or Explanation
Click to View Solution
22.
Question Details
SEssCalc2 12.4.011. [2164981]
-
A lamina occupies the part of the disk x2 + y2 ≤ 81 in the first quadrant. Find its center of mass if the density at any point is
proportional to its distance from the x-axis.
(x, y) =
Solution or Explanation
π/2
ρ(x, y) = ky = kr sin θ, m =
0
243 k,
My =
=
0
6561
k,
8
Mx =
πk.
π/2
9
0
π/2
0
9
0
Hence (x, y) =
9
0
kr3 sin θ cos θ dr dθ =
kr3 sin2 θ dr dθ =
kr2 sin θ dr dθ =
1
k
4
0
1
k
4
0
π/2
π/2
1
k
3
0
9
r4 sin2 θ dθ =
0
π/2
9
r3 sin θ dθ = 243 k
r4 sin θ cos θ dθ =
0
9
π/2
0
0
6561
k
4
0
6561
k
4
0
π/2
π/2
sin θ dθ = 243 k −cos θ
sin θ cos θ dθ =
sin2 θ dθ =
π/2
=
0
π/2
6561
k −cos2 θ
8
0
π/2
6561
1
6561
k θ−
sin 2θ
=
8
2
16
0
27 27
,
π .
8
16
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Question Details
SEssCalc2 12.4.024. [2165435]
-
A lamina with constant density ρ(x, y) = ρ occupies the given region. Find the moments of inertia Ix and Iy and the radii of
gyration
and
.
The region under the curve y = 2 sin(x) from x = 0 to x = π.
Ix =
Iy =
=
=
Solution or Explanation
Click to View Solution
24.
Question Details
SEssCalc2 12.4.023. [2166283]
-
A lamina with constant density ρ(x, y) = ρ occupies the given region. Find the moments of inertia Ix and Iy and the radii of
gyration
and
.
The rectangle 0 ≤ x ≤ 4b, 0 ≤ y ≤ 4h.
Ix =
Iy =
=
=
Solution or Explanation
Click to View Solution
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Question Details
SEssCalc2 12.4.018. [2166034]
-
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 + y2 ≤ 36 in the first quadrant if the
density at any point is proportional to the square of its distance from the origin. (Assume that the coefficient of proportionality
is k.)
Ix =
Iy =
I0 =
Solution or Explanation
Click to View Solution
26.
Question Details
SEssCalc2 12.4.012. [2165083]
-
A lamina occupies the part of the disk x2 + y2 ≤ 16 in the first quadrant. Find the center of mass of the lamina if the density
at any point is proportional to the square of its distance from the origin.
(x, y) =
Solution or Explanation
Click to View Solution
27.
Question Details
The boundary of a lamina consists of the semicircles y =
SEssCalc2 12.4.014. [2165522]
1 − x2 and y =
-
36 − x2 together with the portions of the x-
axis that join them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance
from the origin.
(x, y) =
Solution or Explanation
Click to View Solution
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Question Details
SEssCalc2 12.5.017. [2165229]
-
Use a triple integral to find the volume of the given solid.
The tetrahedron enclosed by the coordinate planes and the plane 9x + y + z = 5
Solution or Explanation
The plane 9x + y + z = 5 intersects the xy-plane when 9x + y + 0 = 5 → y = 5 − 9x, so
x, y, z | 0 ≤ x ≤
E =
5/9
V =
0
=
0
=
0
=
0
5/9
5/9
5/9
5 − 9x
0
5
, 0 ≤ y ≤ 5 − 9x, 0 ≤ z ≤ 5 − 9x − y
9
5 − 9x − y
0
5y − 9xy −
dz dy dx =
5/9
0
and
5 − 9x
0
(5 − 9x − y) dy dx
1 2 y=5−9x
y
dx
2
y=0
5(5 − 9x) − 9x(5 − 9x) −
1
(5 − 9x)2 dx
2
5/9
1
1 81 3
90 2
125
(81x2 − 90x + 25) dx =
x −
x + 25x
=
2
2 3
2
54
0
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Question Details
SEssCalc2 12.5.039. [2165065]
-
SEssCalc2 12.5.040. [2165908]
-
Find the mass and center of mass of the solid E with the given density ρ.
E is the cube 0 ≤ x ≤ a, 0 ≤ y ≤ a, 0 ≤ z ≤ a; ρ(x, y, z) = 9x2 + 9y2 + 9z2.
m=
x, y, z
=
Solution or Explanation
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30.
Question Details
Find the mass and center of mass of the solid E with the given density function ρ.
E is the tetrahedron bounded by the planes x = 0, y = 0, z = 0, x + y + z = 2; ρ(x, y, z) = 3y.
m=
x, y, z
=
Solution or Explanation
Click to View Solution
31.
Question Details
SEssCalc2 12.5.016. [2166067]
-
Evaluate the triple integral.
E
5z dV, where E is bounded by the cylinder y2 + z2 = 9 and the planes x = 0, y = 3x, and z = 0 in the first octant
Solution or Explanation
Click to View Solution
32.
Question Details
SEssCalc2 12.5.025. [2155930]
-
Sketch the solid whose volume is given by the iterated integral.
1
0
1− x
0
3 − 3z
0
dy dz dx
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Solution or Explanation
E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x, 0 ≤ y ≤ 3 − 3z}, the solid bounded by the three coordinate planes and the planes
z = 1 − x, y = 3 − 3z.
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Question Details
SEssCalc2 12.6.003. [2165550]
-
Change from rectangular to cylindrical coordinates. (Let r ≥ 0 and 0 ≤ θ ≤ 2π.)
(a)
(−3, 3, 3)
(b)
(−9, 9
3 , 1)
Solution or Explanation
3
= −1 and the point (−3, 3) is in the
−3
3π
3π
second quadrant of the xy-plane, so θ =
; z = 3. Thus, the set of cylindrical coordinates is 3 2 ,
,3 .
4
4
(a) From this equation we have r2 = (−3)2 + 32 = 18 so r =
(b)
r2 = (−9)2 + (9
the xy-plane, so θ =
34.
3 )2 = 324 so r = 18; tan θ =
9
3
−9
18 ; tan θ =
=−
3 and the point (−9, 9
2π
; z = 1. Thus, the set of cylindrical coordinates is
3
Question Details
18,
3 ) is in the second quadrant of
2π
,1 .
3
SEssCalc2 12.6.017. [2165996]
-
Use cylindrical coordinates.
Evaluate
E
2
2
x2 + y2 dV, where E is the region that lies inside the cylinder x + y = 16 and between the planes
z = −4 and z = 0.
Solution or Explanation
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Question Details
SEssCalc2 12.6.501.XP. [2165139]
-
Find the rectangular coordinates of the point, whose cylindrical coordinates are given.
(a)
(3, π, e)
(b)
(2, 3π/2, 3)
Solution or Explanation
(a)
x = 3 cos π = −3, y = 3 sin π = 0, and z = e, so the point is (−3, 0, e) in rectangular coordinates.
Assume
r = 3
θ = π
z = e
(b)
x = 2 cos
Assume
3π
3π
= 0, y = 2 sin
= −2, z = 3, so the point is (0, −2, 3) in rectangular coordinates.
2
2
r = 2
3π
θ =
2
z = 3
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Question Details
SEssCalc2 12.6.022. [2166075]
-
Use cylindrical coordinates.
Find the volume of the solid that lies within both the cylinder x2 + y2 = 9 and the sphere x2 + y2 + z2 = 64.
Solution or Explanation
Click to View Solution
37.
Question Details
SEssCalc2 12.6.AE.003. [2378286]
-
A solid E lies within the cylinder x2 + y2 = 1, below the
EXAMPLE 3
plane z = 10, and above the paraboloid z = 1 − x2 − y2. (See the
figure.) The density at any point is proportional to its distance from the axis
of the cylinder. Find the mass of E.
SOLUTION
In cylindrical coordinates, the cylinder is
r=
and the paraboloid is
1
z=
, so we can write
E=
Video Example
(r, θ, z) 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 1 − r2 ≤ z ≤ 10 .
Since the density at (x, y, z) is proportional to the distance from the zaxis, the density function is
x2 + y2 = Kr
f(x, y, z) = K
where K is the proportionality constant. Therefore, from this formula, the
mass of E is
x2 + y2 dV
K
m =
E
2π
0
2π
1
Kr2
=
=
=
dr dθ
0
2π
1
(9r2 + r4) dr
dθ
K
0
=
r dz dr dθ
1 − r2
0
0
10
1
=
0
1
2πK
0
.
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Question Details
SEssCalc2 12.6.504.XP. [2165930]
-
Use cylindrical coordinates.
Evaluate
E
5(x3 + xy2) dV, where E is the solid in the first octant that lies beneath the paraboloid z = 1 − x2 − y2.
Solution or Explanation
Click to View Solution
39.
Question Details
SEssCalc2 12.7.003. [2165234]
Change from rectangular to spherical coordinates. (Let ρ ≥ 0, 0 ≤ θ ≤ 2π, and 0 ≤
(a)
-
≤ π.)
(0, −9, 0)
(ρ, θ, ) =
(b)
(−1, 1, −
2)
(ρ, θ, ) =
Solution or Explanation
(a) From Equation 1 and Equation 2,
ρ=
=
x2 + y2 + z2 =
02 + (−9)2 + 02 = 9, cos φ =
3π
2
[since y < 0]. Thus spherical coordinates are
(b)
ρ=
1 + 1 + 2 = 2, cos φ =
z
ρ
=
−
9,
z
ρ
=
0
=0
9
φ = π , and cos θ =
4
2
x
ρ sin φ
=
0
=0
9 sin(π/2)
θ
3π π
,
.
2
2
φ = 3π , and cos θ =
2
2
−1
x
−1
=
=
=−
2(
ρ sin φ
2 sin(3π/4)
2/2 )
1
2
θ = 3π
4
[since y > 0]. Thus spherical coordinates are
40.
Question Details
2,
3π 3π
,
.
4
4
SEssCalc2 12.7.001. [2166135]
-
Plot the point whose spherical coordinates are given. Then find the rectangular coordinates of the point.
(a)
(7, π/3, π/6)
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(x, y, z) =
(b)
(9, π/2, 3π/4)
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(x, y, z) =
Solution or Explanation
(a) From the equations,
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x = ρ sin φ cos θ = 7 sin
π cos π = 7 · 1 · 1 = 7 ,
6
3
2
2
π sin π = 7 · 1 ·
y = ρ sin φ sin θ = 7 sin
6
and z = ρ cos φ = 7 cos
so the point is
(b)
7 7
,
4 4
3
π =7·
6
3,
7
2
3
x = 9 sin
3π
π =9·
cos
4
2
y = 9 sin
3π
π =9·
sin
4
2
z = 9 cos
3π
=9 −
4
2
2
2
3
=
2
4
3
2
7
2
=
7
4
3,
3,
in rectangular coordinates.
2
· 0 = 0,
2
2
2
·1= 9
=−
2
9
2
2 , and
2 , so the point is 0,
9
2
2, −
9
2
2
in rectangular coordinates.
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41.
11/16/14, 5:29 PM
Question Details
SEssCalc2 12.7.022. [2165173]
-
SEssCalc2 12.7.023. [2166210]
-
Use spherical coordinates.
Evaluate
H
(1 − x2 − y2) dV, where H is the solid hemisphere x2 + y2 + z2 ≤ 36, z ≥ 0.
Solution or Explanation
Click to View Solution
42.
Question Details
Use spherical coordinates.
Evaluate
E
2
2
2
2
2
2
(x2 + y2) dV , where E lies between the spheres x + y + z = 9 and x + y + z = 25.
Solution or Explanation
In spherical coordinates, E is represented by {(ρ, θ, φ) | 3 ≤ ρ ≤ 5, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π} and
x2 + y2 = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ2 sin2 φ(cos2 θ + sin2 θ) = ρ2 sin2 φ. Thus
E
(x2 + y2)dV =
π
0
π
=
2π
0
5
3
(ρ2 sin2 φ)ρ2 sin φ dρ dθ dφ =
(1 − cos2 φ)sin φ dφ θ
0
=
1−
2π 1
5
ρ5 =
5
0
3
π
0
sin3 φ dφ
−cos φ +
2π
0
dθ
5
3
ρ4 dρ
π
1
1
cos3 φ (2π) · (3125 − 243)
3
5
0
1
1
2882
23056
+1−
(2π)
=
π.
3
3
5
15
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Name (AID): REVIEW FOR THE EXAM 3 MATH 203 F 2014 (6584966)
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