P463-464. #3, 5, 6, 8, 9, 11, 14, 17, 22, 24, 27, 29, 31, 34 (see. 33 if problems), 62
, 63a-c look at Example 6. 3. Disk. 2. V. r h π. = ( ). 4. 2. 1. 4. 1. V x dx x dx π π.
Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
7.1 Area of a Region Between Two Curves
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Apply & Practice 7.1 Set ?: P452- 453 #2, 5, 6, 7, 8, 10, 11, 13, 16, 20, 23, 25, 29, 32 #37, 39, 50, 51, 52 part b no calculator, 44, 46, 53, 55, 68
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
S. Stirling 2011-2012
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
S. Stirling 2011-2012
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
S. Stirling 2011-2012
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
S. Stirling 2011-2012
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
S. Stirling 2011-2012
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
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7.2 Volume: The Disk Method
Apply & Practice 7.2: P463-464 #3, 5, 6, 8, 9, 11, 14, 17, 22, 24, 27, 29, 31, 34 (see 33 if problems), 62, 63a-c look at Example 6
r= x
R r
3. Disk V = π r h
R
r
2
4
V = ∫π
( )
(
5. Washer V = π R − r
2
x dx
2
2
R = x 2 and r = x3
1
1
1
4
⎡1 ⎤ V = π ⎢ x2 ⎥ ⎣ 2 ⎦1 ⎡1 2 1 2 ⎤ = π ⎢ ( 4 ) − (1) ⎥ 2 ⎣2 ⎦ ⎡ 1 ⎤ 15 = π ⎢8 − ⎥ = π ⎣ 2⎦ 2
(
6. Washer V = π R − r 2
2
)h
x2 R = 4− and r = 2 4
4
= π ∫ x dx
)h
2 2 V = ∫ π ⎡⎢( x2 ) − ( x3 ) ⎤⎥ dx ⎣ ⎦ 0 1
= π ∫ ⎡⎣ x 4 − x6 ⎤⎦ dx 0
1
1 ⎤ ⎡1 V = π ⎢ x5 − x 7 ⎥ 7 ⎦0 ⎣5 ⎡1 5 1 7 ⎤ = π ⎢ (1) − (1) − 0⎥ 7 ⎣5 ⎦ ⎡1 1 ⎤ 2 =π ⎢ − ⎥ = π ⎣ 5 7 ⎦ 35
Intersections:
x2 2 → 8= x → x=± 8 4 2 8 ⎡⎛ ⎤ x2 ⎞ 2 V = ∫ π ⎢⎜ 4 − ⎟ − ( 2 ) ⎥ dx 4⎠ ⎢⎣⎝ ⎥⎦ − 8 2 = 4−
=π
⎡ x4 ⎤ 2 x dx − + 12 2 ∫⎢ 16 ⎥⎦ − 8⎣
= 2π
8
⎡ x4 ⎤ 2 − + 6 x ⎥ dx ∫ ⎢⎣ 32 ⎦ − 8 8
Function is even so… 8 ⎡ x4 ⎤ V = 4π ∫ ⎢6 − x2 + ⎥ dx 32 ⎦ 0 ⎣ 8
8. Disk V = π r h 2
16 − y 2 = x 2 , x = ± 16 − y 2 4
V = ∫π 0
(
16 − y 2
) dy 2
r
4
= π ∫ (16 − y 2 ) dy 0
⎡⎛ 1 3⎞ ⎤ 1 ⎤ ⎡ V = π ⎢16 y − y 3 ⎥ = π ⎢⎜16 ( 4 ) − ( 4 ) ⎟ − 0⎥ = 3 ⎦0 3 ⎣ ⎠ ⎦ ⎣⎝ ⎡192 64 ⎤ 128 − ⎥= π⎢ π 3⎦ 3 ⎣ 3 4
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⎡ 1 x5 ⎤ V = 4π ⎢6 x − x 3 + ⎥ 3 160 ⎦ 0 ⎣ ⎡⎛ 16 2 128 2 ⎞ ⎤ + 4π ⎢⎜⎜12 2 − ⎟ − 0⎥ 3 160 ⎟⎠ ⎦⎥ ⎣⎢⎝
⎡⎛ 180 2 80 2 12 2 ⎞ ⎤ − + 4π ⎢⎜⎜ ⎟⎥ 15 15 ⎟⎠ ⎥⎦ ⎢⎣⎝ 15
⎡112 2 ⎤ 448 2 π 4π ⎢ ⎥= 15 ⎣ 15 ⎦ ≈ 132.69
Page 7 of 13
Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
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9. Disk V = π r h 2
y
3
2
=x
( ) dy
1
V = ∫π y 0
3
2
r
2
11a. Disk V = π r h 2
1
= π ∫ ( y 3 ) dy
4
V = ∫π
0
0
1
⎡1 4 ⎤ 1 ⎡1 ⎤ V = π ⎢ y 4 ⎥ = π ⎢ (1) − 0⎥ = π ⎣ 4 ⎦0 ⎣4 ⎦ 4
(
11b. Washer V = π R − r 2
2
4
)h
2 11c. Washer V = π r h
r
2
2 2 V = ∫ π ⎡⎢( 4 ) − ( y 2 ) ⎤⎥ dy ⎣ ⎦ 0
r
0
⎡1 ⎤ ⎡1 2 ⎤ V = π ⎢ x 2 ⎥ = π ⎢ ( 4 ) − 0⎥ = 8π ⎣ 2 ⎦0 ⎣2 ⎦
R
R = 4 and r = y 2
( x ) dx = π ∫ x dx 4
2
r = 4 − y2
r
2
V = ∫ π ⎡⎣ 4 − y ⎤⎦ dy 2 2
0
2
= π ∫ (16 − 8 y 2 + y 4 ) dy
2
= π ∫ (16 − y 4 ) dy
0
0
1 ⎤ ⎡⎛ 1 5⎞ ⎤ ⎡ V = π ⎢16 y − x5 ⎥ = π ⎢⎜16 ( 2 ) − ( 2 ) ⎟ − 0⎥ 5 ⎦0 5 ⎣ ⎠ ⎦ ⎣⎝ 32 ⎤ 128 ⎡ π = π ⎢32 − ⎥ = 5⎦ 5 ⎣ 2
2
8 1 ⎤ ⎡ V = π ⎢16 y − y 3 + y5 ⎥ = 3 5 ⎦0 ⎣ ⎡⎛ 8 3 1 5⎞ ⎤ π ⎢⎜16 ( 2 ) − ( 2 ) + ( 2 ) ⎟ − 0⎥ = 3 5 ⎠ ⎦ ⎣⎝ ⎡ ⎣
π ⎢32 −
(
11d. Washer V = π R − r 2
2
64 32 ⎤ ⎡ 480 320 96 ⎤ 256 + ⎥= π⎢ − + = π 3 5⎦ ⎣ 15 15 15 ⎥⎦ 15
)h
R = 6 − y 2 and r = 2
r
2
2 2 V = ∫ π ⎡⎢( 6 − y 2 ) − ( 2 ) ⎤⎥ dy ⎣ ⎦ 0
R
2
= π ∫ ( 32 − 12 y 2 + y 4 ) dy 0
1 ⎤ ⎡⎛ 1 5⎞ ⎤ 3 ⎡ V = π ⎢32 y − 4 y 3 + y 5 ⎥ = π ⎢⎜ 32 ( 2 ) − 4 ( 2 ) + ( 2 ) ⎟ − 0⎥ = 5 ⎦0 5 ⎠ ⎦ ⎣ ⎣⎝ 32 ⎤ ⎡ ⎡160 32 ⎤ 192 + ⎥= π ⎢64 − 32 + ⎥ = π ⎢ π 5⎦ 5⎦ 5 ⎣ ⎣ 5 2
S. Stirling 2011-2012
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
(
14a. Washer V = π R − r 2
2
)h
R = 6 − 2 x − x and r = x + 6 2
∫ ⎣⎢(
−3
) − ( x + 6) ⎤⎦⎥ dx
2 2
2
( 6 − 2x − x )( 6 − 2x − x ) 2
6 − 2x − x2 = x + 6
r
0
Intersection V = π ⎡ 6 − 2 x − x Expand R:
Intersection
R
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2
0 = x 2 + 3x
36 − 12 x − 6 x 2 − 12 x + 4 x2 + 2 x3 − 6 x2 + 2 x3 + x 4
0 = x ( x + 3)
36 − 24 x − 8x 2 + 4 x3 + x4 Expand r:
x = 0, −3
( x + 6)( x + 6) x 2 + 12 x + 36
Now the integrand is:
36 − 24 x − 8x 2 + 4 x3 + x 4 − x 2 − 12 x − 36 −36 x − 9 x 2 + 4 x3 + x 4 0
1 ⎤ ⎡ V = ∫ π ⎡⎣−36 x − 9 x + 4 x + x ⎤⎦ dx = π ⎢ −18x 2 − 3x3 + x 4 + x5 ⎥ = 5 ⎦ −3 ⎣ −3 0
2
⎡
⎛ ⎝
3
4
π ⎢0 − ⎜ −18 ( −3) − 3 ( −3) + ( −3) + ⎣
2
3
4
(
243 243 1 5 ⎤ ( −3) ⎞⎟⎥ = π ⎡⎢162 − 81 − 81 + ⎤⎥ = π 5 5 ⎦ 5 ⎠⎦ ⎣
14b. Washer V = π R − r
R r
R = ( 6 − 2 x − x2 ) − 3
2
2
)h
and r = ( x + 6 ) − 3 Now the integrand is:
9 − 12 x − 2 x 2 + 4 x3 + x4 − x2 − 6 x − 9 −18x − 3x2 + 4 x3 + x4 0
2 2 V = ∫ π ⎡⎢( 3 − 2 x − x2 ) − ( x + 3) ⎤⎥ dx ⎣ ⎦ −3
Expand R:
( 3 − 2x − x2 )( 3 − 2x − x2 )
9 − 6 x − 3 x 2 − 6 x + 4 x 2 + 2 x 3 − 3 x 2 + 2 x3 + x 4 9 − 12 x − 2 x 2 + 4 x3 + x4 Expand r:
( x + 3)( x + 3) x2 + 6 x + 9
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0
V = ∫ π ⎡⎣−18x − 3x2 + 4 x3 + x4 ⎤⎦ dx −3
⎡
1
⎤
0
π ⎢−9 x 2 − x3 + x4 + x5 ⎥ = 5 ⎦ ⎣ −3
⎡
⎛ ⎝
π ⎢( 0 ) − ⎜ −9 ( −3) − ( −3) + ( −3) + 2
3
4
1 5 ⎤ ( −3) ⎞⎟⎥ = 5 ⎠⎦
⎣ 243 ⎤ ⎡ −135 243 ⎤ 108 ⎡ + = π ⎢81 − 27 − 81 + π =π⎢ ⎥ 5 ⎦ 5 ⎥⎦ 5 ⎣ ⎣ 5
Page 9 of 13
Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson 17.
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2 ⎡ 2 ⎛ 1 ⎞ ⎤ V = ∫ π ⎢( 4 ) − ⎜ 4 − ⎟ ⎥ dx ⎝ 1 + x ⎠ ⎦⎥ 0 ⎣⎢ 3
Expand R:
r
1 ⎞⎛ 1 ⎞ ⎛ ⎜4− ⎟⎜ 4 − ⎟ ⎝ 1 + x ⎠⎝ 1 + x ⎠ 8 1 16 − + 1 + x (1 + x )2
R
(
Washer V = π R − r 2
2
R = 4 and r = 4 −
)h
1 1+ x
3
⎡ 1 ⎤ V = π ⎢8ln (1 + x ) + ⎥ = (1 + x ) ⎦0 ⎣ 1 ⎡ ⎤ π ⎢8ln ( 4) + − ( 8ln (1) + 1) ⎥ = 4 ⎣ ⎦ 3⎤ ⎡ π ⎢8ln ( 4 ) − ⎥ ≈ 32.485 4⎦ ⎣
3 ⎡ ⎛ 8 1 ⎞⎤ + V = ∫ π ⎢16 − ⎜16 − ⎟⎥ dx 2 ⎜ ⎟⎥ 1 + x 1 + x ⎢ ( ) 0 ⎝ ⎠⎦ ⎣
disks
disks
R
r
disks & even function
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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
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disks
disks & need trig identity: sin 2 u =
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1 − cos 2u 2
Page 11 of 13
Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
Name _________________________________
perpendicular to x-axis so all dx: x2 + y2 = 4 →
2
base of cross section = top – bottom curve
−2
2
y = ± 4 − x2
s = 4 − x2 − − 4 − x2 =
2 4 − x2 −2
(a) Squares
V = s2h
∫( 2
V=
2 4 − x2
−2
1 3 3 2 A = i si s = s 2 2 4
(b) Equilateral triangles
) dx 2
2
(
3 3 2 2 4 − x2 V= s h, V = ∫ 4 4 −2
2
2 = 4 ∫ 4 − x dx
) dx = 2
2
3 ∫ 4 − x2 dx −2
Volume from –2 to 0 = from 0 to 2:
−2
2
Volume from –2 to 0 = from 0 to 2: 2
1 ⎤ 1 3 ⎤ ⎡ ⎡ 8 ⎢ 4 x − x3 ⎥ = 8 ⎢ 4 ( 2 ) − ( 2 ) − 0 ⎥ = 3 ⎦0 3 ⎣ ⎣ ⎦ ⎡ 24 8 ⎤ 128 8⎢ − ⎥ = ⎣ 3 3⎦ 3
1 ⎤ 1 3 ⎤ ⎡ ⎡ 2 3 ⎢ 4 x − x3 ⎥ = 2 3 ⎢ 4 ( 2 ) − ( 2 ) − 0 ⎥ = 3 ⎦0 3 ⎣ ⎣ ⎦ ⎡ 24 8 ⎤ 32 3 2 3⎢ − ⎥ = 3 ⎣ 3 3⎦
(c) Semicircles radius =
1 i2 4 − x 2 = 2
1 V = π r 2h 2 2 1 V = ∫ π 4 − x2 2 −2
(
=
π
4 − x2 (d) Isosceles Right Triangles
) dx 2
2
2 ∫ 4 − x dx
2 −2
Volume from –2 to 0 = from 0 to 2:
π⎡
2
1 ⎤ 1 3 ⎤ ⎡ 2i ⎢ 4 x − x3 ⎥ = π ⎢ 4 ( 2 ) − ( 2 ) − 0⎥ = 2⎣ 3 ⎦0 3 ⎣ ⎦ ⎡ 24 8 ⎤ 16 π⎢ − ⎥= π ⎣ 3 3⎦ 3
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height =
A=
(
1 i2 4 − x 2 = 2
1 2 4 − x2 2
)(
4 − x2
)
4 − x2 = 4 − x2
2
V = ∫ 4 − x 2 dx −2
Volume from –2 to 0 = from 0 to 2: 2
1 ⎤ ⎡ 2 ⎢ 4 x − x3 ⎥ = 3 ⎦0 ⎣ ⎡ 24 8 ⎤ 32 2⎢ − ⎥ = ⎣ 3 3⎦ 3
1 3 ⎤ ⎡ 2 ⎢ 4 ( 2 ) − ( 2 ) − 0⎥ = 3 ⎣ ⎦
Page 12 of 13
Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson
Name _________________________________
Perpendicular to y-axis so all dy:
y = x3 → x = 3 y Base of a cross section = 1 − 3 y
(a) Squares
(b) Semicircles
V =s h
radius =
2
(
1
)
2
V = ∫ 1 − 3 y dy 0 1
=
∫ (1 − 2 y
1 3
+y
2
3
0
)
1 V = π r 2h 2
) dy
(
1
⎡ 3 4 3 3 53 ⎤ y + y ⎥ 5 ⎣ 2 ⎦0 3 4 3 5 ⎞ ⎛ = ⎜ (1) − (1) 3 + (1) 3 ⎟ − 0 2 5 ⎝ ⎠ 10 15 6 1 − + = = 10 10 10 10
=
2
)
1 ⎛1 ⎞ ∫−2 2 π ⎜⎝ 2 1 − 3 y ⎟⎠ dx 2
V=
= ⎢x −
1− 2 y 8 ∫(
π
2
−2
π⎡
1
3
+y
2
3
) dx
1
3 4 3 5 ⎤ = ⎢x − y 3 + y 3 ⎥ 8⎣ 2 5 ⎦0
π⎛
3 43 3 53 ⎞ ⎜ (1) − (1) + (1) ⎟ − 0 8⎝ 2 5 ⎠ π ⎛ 10 15 6 ⎞ π = ⎜ − + ⎟= 8 ⎝ 10 10 10 ⎠ 80 =
1 3 3 2 A = i si s = s 2 2 4
(b) Equilateral triangles
(
2
(
1 1− 3 y 2
)
2 3 3 2 1 − 3 y dx = V= s h, V = ∫ 4 4 −2 2
(
)
1 2 3 1 − 2 y 3 + y 3 dx ∫ 4 −2 1
3 ⎡ 3 43 3 53 ⎤ x− y + y ⎥ = 4 ⎢⎣ 2 5 ⎦0 =
3⎛ 3 43 3 53 ⎞ ⎜ (1) − (1) + (1) ⎟ − 0 4 ⎝ 2 5 ⎠
=
3 ⎛ 10 15 6 ⎞ 3 ⎜ − + ⎟= 4 ⎝ 10 10 10 ⎠ 40
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