6.1 Areas between Curves 曲線間的面積

(1)

§

6.1 Areas between Curves 曲線間的面積

[Ex1] Find the area of the region bounded above by , bounded below by , and bounded on sides by

[Sol]:

Formula 1:

(

( ) )

b

a g x

A =

∫

f x − dx

1 2

0

1 0

1 3

( )

2 2

x x

A =

∫

e − xdx = e − x = −e y = ex

0 and 1.

x = x =

y = x

(2)

[Ex2] Find the area of the region enclosed ^圍住 by the parabolas [Sol]:

[Ex5] Find the area of the region bounded by the curves

[Sol]:

and x π2

=

2 2

and 2

y = x y = x − x

4 2

0 4

s

sin sin cos tan 1. (0 )

cos 2 4

( ) ( in ) (sin ) ( )

2 2 c

2

os x cos

y x

x x x x x

y x

A x dx x x dx

π π

π

π π

⎧ =

⇒ = ⇒ = ≤ ≤ ⇒ =

⎨ =⎩

⇒ = − + −

= ⋅⋅⋅⋅

= −

∫ ∫

sin , cos , 0 y = x y = x x =

1 2

0

2 1

( ) ( )

2 3

A =

∫

x − x − x dx = ⋅⋅⋅⋅ =

(3)

[Ex6] Find the area enclosed by the line and the parabola [Sol]:

2 2 6

y = x+

2 2

2

1 2( 1) 6 2 8 0

2 6

4 5

( 4)( 2) 0 {

2 1

y x

y y y y

y x

y y

y x

⎧ = −

⇒ = + + ⇒ − − =

⎨ = +

⎩

= ⇒ =

⇒ − + = ⇒

= − ⇒ = −

1 5

3 1

1 1

1 5

2 2

3 1

3 3

2 1 2 2 5

3 1

( ) ( ) ( ) ( )

(2 6) 2 1 (2 6) 2 1 2

2 1 2 1

(2 6

2 6 2

) (2 6)

3

2 6

2

1

3 2

=18

6

A dx dx

x dx x x dx

x x x x

x x

x

− x

− −

−

− −

−

− −

= − + −

= + ⋅ + + ⋅ − +

⎡ ⎤

= + + ⋅ + −

+ −

⎢ + ⎥

⎣ ⎦

+ + −

∫ ∫

1 y = −x

(4)

Formula 2:

(

( ) )

d

c g y

A =

∫

f y − dy

[Ex] Use formula 2 to solve Ex6 again.

[Sol]:

2

1 2, 4

2 6 y x

y x y

⎧ = −

⇒ ⋅⋅⋅⋅⋅ ⇒ = −

⎨ = +

⎩

4 2

2

3 2 4

2

( ) ( )

1 1

2 3

( 4 )

6 2

18

A y dy

y y

y

−

= −

= − + +

+

=

∫

−

(5)

§

6.2 Volumes 體積

Let’s divide S (the solid ^立體) into n “slabs” ^厚片 of equal width by using the planes ^平面 to slice ^切 the solid ^立體.

If we choose sample points we can approximate the ith slab by a cylinder ^柱狀體 with base ^底 area and height

*

in [ 1, ],

i i i

x x ₋ x

1, 2 , 3 , , 1

x x x xn

P P P P ₋

b a

x n

Δ = −

( _i*)

A x Δx.

Si

*

* 1

1

So , ( ) ( ) and then

we define lim ( ) ( )

i i

n

i i

n n i

i

V S A x x

V A x x

→∞ =

=

≈ Δ

= Δ

∑

(6)

Definition of Volume

體積的定義

Let S be a solid that lies between . If the cross-sectional

橫截面area of S in the plane , through x and perpendicular ^垂直 to the x-axis, is , then the volume of S is

* 1

lim ( ) ( )

n b

i a

n i

V A x x A x dx

→∞ =

=

∑

Δ =

∫

Px

and

x = a x = b

( ) A x

(7)

[Ex1] Show that the volume of a sphere ^球體 of radius ^半徑r is [Sol]:

4 3

V = 3 π r

2 2

2 0

( ) ( )

r r r

r r

V A x dx π r x dx π r x dx

− −

=

∫

=

∫

− =

∫

−

2 2

(r x )

π −

( is even)∵

2 3 3 3

0

1 1

2 ( ) 2 ( )

3 3

r x x r r r

π π

= − = −

4 3

4.18879 3πr

= ≈

2 2 2

( ) ( )

A x π y π r x

∴ = = −

2 2

y = r − x

∵

(8)

[Ex7] A solid with a circular base of radius 1 is

shown on the right. Parallel cross-sections ^{平行橫切面}

perpendicular to the base are equilateral triangles ^{等邊三角形}. Find the volume of the solid.

[Sol]:

2

1 1

2

2 1

2 3

1 1 0

1 0

( ) 3 (1 )

( )

the area of 1 (2 ) 3 3 2

1 4 3

2 3 1 2 3 ( )

3 (1 ) 3

3

A x x

ABC y y y

V dx dx x dx x x

− −

= Δ = ⋅ ⋅ = =

∴ = = − = =

−

− − =

∫ ∫ ∫

(

^∵⁽¹⁻ ^x²^{) is even}

)

(9)

[Ex8] Find the volume of a pyramid ^角錐體 whose base is a square with side L and whose height is h.

[Sol]:

2 2 2

2

2 2 2

2

0

0 2

3 2

2

( )

1

3

h

A x

A x L x

S x x

S L

L h h

x L

S L x

h h

V dx

dx

L h

h L h

h

= ⇒ =

∴ = = =

⇒ =

=

= ⋅

=

∫

∵

(10)

[Ex9] A wedge is cut out of a circular cylinder ^圓柱體 of radius 4 by two planes.

One plane is perpendicular to the axis ^軸 of the cylinder. The other intersects ^相

交the first at an angle of 30° along a diameter ^直徑of the cylinder.

Find the volume of the wedge.

[Sol]:

2

4 4 4 4

4 2

0

3

2

4 0

tan 30 1

3

1 1 1

2 3 2 3

1 16 3

1 1

(16 ) 3 3

128 ( ) 16

2 3 ( )

1

6 2

3

BC y y

y y y

V dx

dx x dx

x x

A x x

−

= =

∴ = ⋅ ⋅ = =

⇒ =

=

= −

=

∫

∵

(11)

[Ex2] Find the volume of the solid obtained by rotating ^旋轉 about the x-axis the region under the curve from 0 to 1.

[Sol]:

y = x

[Ex3] Find the volume of the solid obtained by rotating the region bounded by , y = 8, and x = 0 about the y-axis.

[Sol]:

y = x3

2 2

1 1

2

0 0

1 0

( )

1 ( )

( ) 2 2

y x

V d

A x x

A x x xdx x

π π

= = =

∴ =

∫

=

∫

= ⋅ =

∵

2 3 2

8 8 5

3

0 0

2 3

8 0

( )

3 96

5 5

x y

V d

A y y

A y y y dy y

π π

= = =

∴ =

∫

=

∫

= ⋅ =

∵

(12)

[Ex4] The region R enclosed by the curve is rotated about the x-axis. Find the volume of the resulting solid.

[Sol]:

[Ex5] Find the volume of the solid obtained by rotating the region in Ex4 about the line

[Sol]:

2.

y =

and 2

y = x y = x

2 2

2

1 0

3 5

2 4

4

1 0

( )

1 1 2

( )

3 5 5

)

1 (

( ) x x

x A x

V dx

x

x x

x

π π π

π

π π

= − = −

∴ = −

= − =

∫

∵

1 0

1 4 2

2 2

0

5 2

2

3 1

0

( )

5 4

1 5 8

( 2 )

5

2 2

( )

(

) )

3 (

15 A x

V dx

x x x

x

dx

x x x

x x

x π

π π

π

π π

π

= −

∴ = −

= − +

= − + =

−

∫

−

∫

∵

(13)

[Ex6] Find the volume of the solid obtained by rotating the region in Ex4 about the line

[Sol]:

The solid in Example 2~6 are called solid of revolution ^旋轉體. 1.

x = −

2

1 0

1 2

2

3 2

0 3

2 2 1

0

( )

(1 ) (1 )

2

4 1 1

( )

3 2 3

1 )

2

( (1 )

A y

V dy

y y y dy y y

y

y y

y π

π π

π π +

= −

∴ = −

=

+

− −

−

=

+ +

= −

∫

∵

(14)

§

6.3 Volume by Cylindrical Shells 圓柱殼法求體積

1

lim 2 (

(

)

2 )

n

i i

n n

i i

i

V x f x

V x f x x

π x π

=

= →∞ Δ

≈

∑

Δ

∑

(15)

[Ex1] Find the volume of the solid obtained by rotating about the y-axis the region between

[Sol]:

The volume of the solid obtained by rotating about the y-axis the region under the curve from a to b, is y = f x( )

* *

1

lim 2 ( ) 2 ( ) where 0

n b

i i a

n i

V π x f x x π xf x dx a b

→∞ =

=

∑

Δ =

∫

≤ <

2 3

2 0

2 3 4

0

4 5 2

0

2 2

1 1

2 ( )

2 5

2 (8 32) 5 16

2 (2 )

5

x

V dx

x x dx x

x x

x π

π π

π

−

=

= −

=

∫

2 3

2 and 0.

y = x − x y =

(16)

[Ex2] Find the volume of the solid obtained by rotating about the y-axis the region between

[Sol 1] By method of cylindrical shells:

[Sol 2] By washer method:

and .2

y = x y = x

1 0

1 2 3

0 3

2

4 1

0

2

1 1

2 ( )

3 4

(

6

2 )

V x dx

x x dx x

x x

π x π π π

=

−

=

∫

1 0

1 2

2

0

2

3

2

2 1

0

( )

1 1

(

) (

2 3 )

)

6 A y

V dy

y y dy y

y y

y

y y

π π π

π π

= −

∴ = −

= −

=

∫

∵

(17)

[Ex3] Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 1. y = x

[Sol 1] By cylindrical shells method:

[Sol 2] By disk method ^圓盤法:

1 2

0

1 3

0

2 4 1

0

2

1 1

2 ( )

2 4

1

2 ( )

2

V y dy

y y dy

y y

y π

π π

π

=

−

=

∫

1

2

1 2 1

0 0 0

2 1

0

( ) ( )

( ) 1

)

2

(

2

V dx dx

A x x

A x xdx

x

x π

π π π

π

=

∴ = = =

= ⋅ =

∫ ∫ ∫

∵

(18)

[Ex4] Find the volume of the solid obtained by rotating the region bounded

1 0

1 3 2

2

0

4 3 2 1

0

2 3 2

1

2 (2 )(

2 ( )

4

)

2

x

V dx

x x x dx

x x

x

x x

π π π

π

=

= − +

− −

=

∫

2

2 0 ( 1) 0 0, 1

0 y x x

x x x x x x

y

⎧ = −

⇒ − = ⇒ − = ⇒ = =

⎨ =

⎩

by y = −x x2 and y = 0 about the line x = 2.

[Sol]:

(19)

§

6.4 Work

If an object moves along a straight line with position function , then the force^力 F on the object (in the same direction) is defined by Newton’s Second Law of Motion ^{牛頓運動定律} as

( ) s t

2 2

F m d s

= dt

m: mass (kg)

s: displacement (m) t: time (s)

F: force or pound (lb) (N=kg m 2)

⋅ s

(1) In the case of constant acceleration ^{加速度是常數} , the force and the work done is defined to be

W = ⋅F d

is also constant F

If F is measured in newtons and d in meters, then the unit for W is a newton-meter ^{牛頓-公尺}, which is called a joule ^焦耳(J).

If F is measured in pounds and d in feet, then the unit for W is a foot-pound ^英呎-磅(ft-lb), which is about 1.36J.

(20)

[Ex1] (a) How much work is done in lifting ^舉起 a 1.2-kg book off the floor to put it on a desk that is 0.7m high? Use that fact that the acceleration

due to gravity ^重力 is

(b) How much work is done in lifting a 20-lb weight 6ft off the ground?

[Sol]:

m 2

9.8 s g =

(a) The force exerted is equal and opposite to that exerted by gravity, so (1.2)(9.8) 11.76 N

Therefore, the work done is

(11.76)(0.7) 8.232 J (b) In the ca

F mg

W Fd

= = =

反向

se of 20 lb and 6 ft, the work done is (20)(6) 120 ft-lb

F d

W Fd

= =

= = =

(21)

(2) If the force is variable, let’s suppose that the object moves along the x-axis in

the positive direction ^正方向from x = a to x = b, and at each point x between a and b a force acts on the object, where f is a continuous function. We divide

the interval [a, b] into n subintervals with endpoints

and equal width . Therefore we can approximate the total work by , where ,

We define as

[Ex2] When a particle ^質點 is located a distance x feet from the origin, a force of pounds acts on it. How much work is done in moving it from x = 1 to x = 3?

( ) f x

0 , 1, , _n

x = a x x = b Δx

*

[ 1, ]

i i i

x ∈ x ₋ x

* 1

( )

n

i i

W f x x

=

≈

∑

Δ

the work done in moving the object form a to b

* 1

lim ( ) ( )

n b

i a

n i

W f x x f x dx

→∞ =

=

∑

Δ =

∫

2 2

x + x

3 3

2 3 2

1 1

3 1

1 50

( ) 2 ( ) ft-lb

3 3

W =

∫

f x dx =

∫

x + xdx = x + x = [Sol]:

Since , the work done in moving it from x = 1 to x = 3 is f x( ) = x² + 2x

(22)

By Hook’s Law ^虎克定律

, where k is the spring constant ^彈性係數 ( )

f x = kx

[Ex3] A force of 40N is required to hold a spring that has been stretched^伸展 from its natural length of 10cm to a length of 15cm. How much work is done in stretching the spring from 15cm to 18cm?

[Sol]:

According to Hook’s Law, the force required to hold the spring stretched x meters beyond its natural length is

When the spring is stretched from 10cm to 15cm. 15 10

0.05 m x 100−

= =

(0.05) 40 i.e. 0.05 40 800.

f k k

∴ = = ⇒ =

Thus, and the work done in stretching the spring from 15cm to 18cm is

( ) 800 f x = x

( )

f x = kx

0.08 2 2 2

0.05

0.08

800 400 0.05 400 (0.08) (0.05) 1.56 J

W =

∫

xdx = x = ⎡⎣ − ⎤⎦ =

(23)

[Ex4] A 200-lb cable is 100ft long and hangs ^垂掛 vertically ^垂直from the top of a tall building. How much work is required to lift ^拉the cable to top of the building?

[Sol]: We don’t have a formula for the force function here, so let’s use the initial approach.

First, we divide the cable into small parts of equal length . So the weight of the ith part is

lb since the cable weights pounds per foot. Thus, the work done on the ith part is

x 100 Δ = n

2 xΔ 200

100 = 2

* *

(2Δ ⋅x) x_i = 2x_i Δx

* 1

So, 2

n

i i

W x x

=

≈

∑

Δ

* 100

1 0

2 100

0

lim 2 2

10000 ft-lb

n n i

i

W x x xdx

x

→∞ =

⇒ = Δ =

= =

∑ ∫

(24)

[Ex5] A tank has the shape of an inverted circular cone ^{倒圓錐形水槽} with height 10m and base radius 4m. It is filled with water to a height of 8m. Find the work required to empty the tank by pumping ^抽 all of the water to the top of the tank.

(The density ^密度of water is )1000kg ₃ m [Sol]:

We divide the interval [2,10] into n subintervals of equal width with endpoints

and choose in the ith subinterval.

The ith layer is approximated by a circular cylinder with radius and height . Δx

0 2, 1, , _n 10

x = x ⋅⋅⋅⋅ x =

*

xi

ri

Δx

*

4 2

(10 )

10 10 5

i

i i

i

r r x

x = ⇒ = −

−

The volume of the ith layer

2 4 * 2

(10 )

i i 25 i

V ≈ π r Δ =x π − x Δx

(25)

The work done to raise the ith layer to the top is

* * * 2

1568 (10 )

i i i i i

W ≈ F x ≈ πx − x Δx

1

* * 2

1

10 2

2

6

lim

lim 1568 (10 ) 1568 (10 )

1568 2048 3.36 10 J 3

n n i

i n

i i

n i

W W

x x x

x x dx

π π

π

→∞ =

=

= − Δ

= −

=

= ⋅ ≈ ×

∑

∫

So its mass ^質量 4 ^{* 2} ^{* 2}

1000 (10 ) 160 (10 )

i 25 i i

m ≈ ⋅ π − x Δ =x π − x Δx The force required to raise the ith layer is

* 2 * 2

160 (10 ) (9.8) 1568 (10 )

i i i i

F = m g ≈ π − x Δ ⋅x = π − x Δx

Therefore, the total work done is

(26)

§

6.5 Average Value of a Function 函數的平均值

The average value of f on the interval [a, b] is defined as

1 ^b ( )

ave a

f f x dx

b a

= −

∫

[Ex1] Find the average value of the function on the interval [-1,2].f x( ) = +1 x² [Sol]:

2 2 3

1

2 1

1 1 1 1

( ) (1 ) ( ) 2

2 ( 1) 3 3

b

ave a

f f x dx x dx x x

b a ⁻ −

= = + = + =

−

∫

− −

∫

The Mean Value Theorem for Integrals If f is continuous on [a, b], then there exists a number c in [a, b] such that

( ) ( )( )

b

a f x dx = f c b−a

∫

or f_ave = f c( )

(27)

[Ex2] is continuous on . Find all number c that satisfy the conclusion of the Mean Value Theorem for Integrals.

( ) 1 2

f x = + x

According to the Mean Value Theorem for Integrals, there is a number c in such that.

( )

2 2

11 x dx f c( ) 2 ( 1) 6 f c( ) 3

− + = − − ⇒ = ⋅

∫

[Sol]:

( ) 2 _ave

f c f

⇒ = =

[ 1, 2]−

There happen to be two numbers 1 and 1 in [ 1, 2] that work in the Theorem.

c c

=

= − −

i.e. 1+ c2 = ⇒ = ± ∈ −2 c 1 [ 1, 2]

(28)

[Ex3] Show that the average velocity ^速度 of a car over a time interval is the same as the average of its velocities during the trip.

1 2

[ , ]t t

[Sol]:

If is the displacement of the car at time t, then the average velocity of the car over is

( ) s t

1 2

[ , ]t t ( )² ( )¹

2 1

s t s t s

t t t

− Δ =

Δ −

On the other hand, the average value of the velocity function on is

1 2

[ , ]t t

[ ]

2 2

1 1

2 1 2 1

2 1

1 1

( ) ( )

1 ( ) ( ) average velocity

t t

ave t t

v v t dt s t dt

t t t t

s t s t t t

= = ′

− −

= − =

−

∫ ∫

(by the Net Change Theorem) ^淨值定理