Solution to Assignment 4p.423 #6,8,16,20

p.445 #10,16,20,34

Section 6.3

6.

Find the length of the curve,

x = 8 cos(t) + 8t sin(t), y = 8 sin(t)− 8t cos(t), 0 ≤ t ≤ π

2.

Solution

Remembering

ds =

√(dx

dt

)2

+

(dy

dt

)2

dt,

the length of the curve is

s =

∫ π/2

0

√(dx

dt

)2

+

(dy

dt

)2

dt =

∫ π/2

0

√(8t cos(t))2 + (8t sin(t))2dt

= 8

∫ π/2

0

t dt =8

2[t2]

π/20 = 4

(π2

)2

= π2.

8.

Find the length of the curve

y = x3/2 from x = 0 to x = 4.

Solution

Remembering

ds =

√1 + (

dy

dx)2dx,

1

the length of the curve is

s =

∫ 4

0

√1 +

(dy

dx

)2

dx =

∫ 4

0

√1 +

(3

2x1/2

)2

dx =

∫ 4

0

√1 +

9

4x

=4

9

[2

3

(1 +

9

4x

)3/2]4

0

=8

27(103/2 − 1) =

8

27(10√

10− 1).

16.

Find the length of the curve

y =

∫ x

−2

√3t4 − 1dt, −2 ≤ x ≤ −1.

Solution

By the fundamental theorem of calculus,

dy

dx=√

3x4 − 1.

So the length of the curve is

s =

∫ −1

−2

√1 +

(dy

dx

)2

dx =

∫ −1

−2

√1 + (3x4 − 1)dx =

∫ −1

−2

√3x2dx

=

[√3

3x3

]−1

−2

=1√3(−1− (−8))

=7√3.

20.

x =√

1− y2, −1/2 ≤ y ≤ 1/2

Solution

2

s =

∫ 1/2

−1/2

√1 +

(dx

dy

)2

dy

=

∫ 1/2

−1/2

√1 +

(1

2(1− y2)−1/2(−2y)

)2

dy

=

∫ 1/2

−1/2

√1 + y2(1− y2)−1dy =

∫ 1/2

−1/2

√1

1− y2dy

=π

3(by the numerical computation).

Section 6.5

10.

Find the lateral surface area of the cone generated by revolving the line segment y = x/2,

0 ≤ x ≤ 4 about the y-axis. Check your answer with the geometry formula

Lateral surface area=12× base circumference × slant height.

Solution

Remembering s =∫

2πrds, the lateral surface area is

s =

∫ 4

0

2πx

√1 +

(dy

dx

)2

dx =

∫ 4

0

2πx

√1 +

1

4dx =

√5π

[x2

2

]4

0

= 8√

5π.

Lateral surface area=12× 2π 4× slant height,

where slant height is√

42 + 22 = 2√

5. Therefore, lateral surface area is

1

2× 8π × 2

√5 = 8

√5π.

16

Find the area of the surface generated by revolving the curve about the indicated axis.

y =√x+ 1, 1 ≤ x ≤ 5; x-axis

Solution

3

Since s =∫

2πyds,

s =

∫ 5

1

2πy

√1 +

(dy

dx

)2

dx =

∫ 5

1

2π√x+ 1

√1 +

(1

2(x+ 1)−1/2

)2

dx

=

∫ 5

1

2π√x+ 1

√1 +

(x+ 1)−1

4dx =

∫ 5

1

2π

√x+ 1 +

1

4dx =

∫ 5

1

2π

√x+

5

4dx

= 2π

[2

3

(x+

5

4

)3/2]5

1

=4π

3

[(25

4

)3/2

−(

9

4

)3/2]

=4π

3

(125

8− 27

8

)=

49π

3.

20.

Find the area of the surface generated by revolving the curve about the indicated axis.

x =√

2y − 1, 5/8 ≤ y ≤ 1; y − axisSolution

s =

∫ 1

5/8

2πx

√1 +

(dx

dy

)2

dy =

∫ 1

5/8

2π√

2y − 1

√1 +

(1

2(2y − 1)−1/2 2

)2

dy

=

∫ 1

5/8

2π√

2y − 1 + 1 dy = 2√

2π2

3

[y3/2

]15/8

=4√

2π

3(1− (

5

8)3/2)

=π

12(16√

2− 5√

5).

There was some confusion about # 34. It will not graded this time. But it will be the

question next assignment. I will announce this in the class.

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