Ch 07 Homework Complete Solutions

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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−



⎡ 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

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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

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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

Name _________________________________

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

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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

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Ch 07 Homework Complete Solutions: S. Stirling Calculus: Early Transcendental Functions, 4e Larson

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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 ⎣ ⎦

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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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