Applications of Definite Integral

Applications of Definite Integral

• Area between Curve…..………...……….2

• Volume of Solid of Revolution………..12

• Volume by Cylindrical Shells………....21

a _b f(x)

D

x

∆

A. Let D =

{

(x, y)|a ≤ x ≤b,0≤ y ≤ f (x)

}

Area of region D = ?

Steps :

1. Divided D into n pieces, the area

Area between Curves

a _b

1. Divided D into n pieces, the area

of each pieces is approximated by area of rectangular with height f(x) and length of base x ; .∆A ≈ f (x)∆x

2. The area of D is approximated by sum area of rectangular. If , the area of D is

( )

b

a

A

=

f x dx

0

x

Example :

Find the area of region that is bounded by parabola

x axis , and x = 2.

, 2

x y =

Area of pieces:

3 8 3

1 2

0 3 2

0 2

= =

= x dx x

A 2

x y =

2

x

∆

2

x

x x

A ≈ ∆

∆ 2

h(x)

g(x)

D h(x)-g(x)

Area of region D = ? Steps:

1. Divided D into n pieces, the area of each pieces is approximated

{

(x, y)| a x b,g(x) y h(x)

}

D = ≤ ≤ ≤ ≤ B. Let

chapter

7

Applications of Definite Integral

a ∆x _b

x

x

g

x

h

A

≈

−

∆

∆

(

(

)

(

))

of each pieces is approximated by area of rectangular with

height h(x) - f(x) and length of base x ;

2. The area of D is approximated by sum of area rectangular. If , the area of D is

−

b

a

dx x

g x

h( ) ( )) (

0

x

Example:

Find the area of region that is bounded by y = x+4 and parabola y = x2 – 2.

2

The straight line and parabola intersects at

chapter

7

Applications of Definite Integral

2

Area of pieces:

− −

Area of region :

chapter

7

Applications of Definite Integral

Example: Find the area of region that is bounded by x axis,

y = x2 and y = -x + 2. Answer: Intersection points

2

2

+ − = x

x x2 + x− 2 = 0 (x + 2)(x −1) = 0

x = -2, x = 1

chapter

7

Applications of Definite Integral

2

x y=

x

x

A

≈

∆

∆

₁ 2

2

x = -2, x = 1

y=-x+2

1

If pieces is vertical , then region must be divided into 2 sub region.

x

∆

x

∆

x x

A ≈ − + ∆

∆ ₂ ( 2) Area of pieces I

Area of pieces II

1

The area of region I

The area of region II

2

chapter

7

Applications of Definite Integral

y

∆

!

y

∆

h(y)-g(y)

C. Let D =

{

₍x_, y_)| c ≤ y ≤ d_,g₍y₎ ≤ x ≤ h₍y₎

}

Area of region D = ? Steps:

1. Divided D into n pieces, the area of each pieces is approximated by

chapter

7

Applications of Definite Integral

( ( ) ( ))

d

c

A = h y − g y dy

area of rectangular with height

h(y) - f(y) and length of base y ;

y y

g y

h

A ≈ − ∆

∆ ( ( ) ( ))

2. The area of D is approximated by sum area of rectangular. If , the area of D is ∆ →y 0

2

Example: Find the area of region that is bounded by

and . y = x −1

Answer : _{The intersection point between}

parabola and straight line are

1 − = x y

chapter

7

Applications of Definite Integral

3

Area of pieces:

2

((3 ) ( 1))

A y y y

The area of region is :

Applications of Definite Integral

Volume by Disk Method

{

(x, y) | a x b , 0 y f (x)

}

D = ≤ ≤ ≤ ≤

• The Region

is revolved around the x-axis.

Volume of a Solid

of Revolution

" #

!

Solid of revolution

Region D

If the rectangular slice with height f(x) and length of base x is revolved

around x-axis we will get a circular disk with height x and radius f(x).

f(x)

D

x

x

f

V

≈

∆

∆

π

2

(

)

=

b

a

dx x

f

V

π

2( )

Thus,

f(x)

a x _b

2

x y =

Example: The region D is bounded by y = x2, x-axis, and line x = 2. Find the volume of solid that is generated by rotating region D around the x-axis.

If the slice is revolved around x-axis we get a circular disk with radius x2

x

{

(x, y)|c y d , 0 x g(y)

}

D = ≤ ≤ ≤ ≤

• The region

is revolved around y-axis.

#% #% !

Region D _{Solid of revolution}

volume of a solid of revolution?

x = g(y)

!

y

∆

If the rectangular slice with height

g(y) and length of base y is

revolved around y-axis we will get a circular disk with height y and radius g(y) .

y

∆

) (y g

Thus,

y y g

V ≈ ∆

∆ π 2( )

=

d

c

dy y

g

y ∆

Example: The region D is bounded by y = x2_{, y-axis, and}

line y = 4. Find the volume of solid that is generated by rotating region D around the y-axis.

y If the slice is revolved around y-axis we get a circular disk with radius

{

(x, y)| a x b , g(x) y h(x)

}

D = ≤ ≤ ≤ ≤

• The Region

is revolved around x-axis.

h(x)

Volume by Washer Method

chapter

7

Applications of Definite Integral

g(x)

a _b

D

Region D Solid of revolution

h(x)

g(x) D

If the rectangular slice with height

h(x) - g(x) and length of base x

is revolved around x-axis we will get an annular ring. The ring has

inner radius g(x), outer radius h(x),

chapter

7

Applications of Definite Integral

a ∆x _b

x

∆ ∆V ≈

π

(h2(x) − g2(x))∆x

− =

b

a

dx x

g x

h

V

π

( 2( ) 2( ))

h(x)

g(x)

inner radius g(x), outer radius h(x), and height x, thus

Example: The region D is bounded by y = x2, x-axis, and line x = 2. Find the volume of solid that is generated by rotating region D around y = -1.

If the slice is revolved around y = -1 we get an annular ring with inner

chapter

7

Applications of Definite Integral

{

(x, y) | a x b , 0 y f (x)

}

D = ≤ ≤ ≤ ≤

The region

#

is revolved around y-axis

Volume by Cylindrical

Shells

" !

Region D

Solid of revolution

volume of a solid of revolution ?

#

!

Let the slice are rectangular with height f(x), length of base is x, and its distance from y axis is x. If this slice is revolved around

y-axis we will get a thin cylindrical

chapter

7

Applications of Definite Integral

"

x

∆

x

∆

y-axis we will get a thin cylindrical shell of radius x, height f(x), and thickness x. Thus, its volume is

#

#

#

x

x

f

x

V

≈

∆

∆

2

π

(

)

=

b

a

dx

x

xf

Example: The region D is bounded by y = x2, x-axis, and line x = 2. Find the volume of solid that is generated by rotating region D around y-axis.

If the slice is revolved around y-axis we get a thin cylindrical shell of

chapter

7

Applications of Definite Integral

2

x y =

x

∆ _$

2

x

!

#

2 3

2 2

V π x x x π x x

∆ ≈ ∆ ≈ ∆

= =

=

2

0

2 0 4 3

8 |

2

2π x dx π x π V

we get a thin cylindrical shell of

radius x, height x2 , and thickness x. Thus,

volume of a solid of revolution:

Remark :

- The method of disks and washer:

The slices are perpendicular to rotating axis

- The method of Cylindrical Shells:

chapter

7

Applications of Definite Integral

- The method of Cylindrical Shells:

A parameter form of curve on R2

x = f(t) , a t b y = g(t)

Point A(f(a), g(a)) is called original point and B(f(b), g(b)) is called terminal point of curve.

(*)

The Length of a Curve

is called terminal point of curve.

Definition : A curve is called smooth if

(i) f ′ and g′ _{are continuous on [a,b]}

(ii) f ′_and g′ _{is not zero at the same time on (a,b)}

Let a curve on parameter form (*), we will find length of that curve.

Steps:

1. Divided interval [a,b] into n subintervals

chapter

7

Applications of Definite Integral

b t

t t

t

a = _o < ₁ < ₂ < ...< _n =

"

1

t ti−1 t_i tn−1

A partition on [a,b]

A partition on curve

(

1

Q (

( (

(

o Q

1

−

i

Q Q_i

n

2. Approximate length of curve

The length of arc is approximated by length of line segment

y

∆

chapter

7

Applications of Definite Integral

1

length of line segment

where ∆ti =ti −ti−₁

Applications of Definite Integral

The length of curve is approximated by length of polygonal arc,

Remark:

Applications of Definite Integral

Example: Find length of curve

The length of curve

chapter

7

Applications of Definite Integral

dt

The length of curve

2.

2

3/ 2

,

1

7

3

y

=

x

≤ ≤

x

Answer :

2

Applications of Definite Integral

For problem 1 – 5, sketch and find area of region that is bounded by

2

2

y

=

x

y

= +

x

1. and

3

,

,

8

y

=

x y

= −

x

y

=

2. and

Problem Set 1

3

,

,

8

y

=

x y

= −

x

y

=

2. and

3. y = x , y = 4x , and y = -x +2

4. y = sin x, y = cos x, x = 0 , and x = 2π.

2

2

7 0

6

0

x

−

y

+ =

y

−

y

− =

x

For problem 6 – 10, find the volume of solid that is generated by rotating region that is bounded by curves below around x-axis .

6. and

_y

=

_{x y}

3

_,

=

_0,

_x

=

₂

Problem Set 2

,

0,

2

y

=

x y

=

x

=

7. and

y

= −

9

x

2

y

=

0

8. and

y

=

x

2

y

=

4

x

9. y = sin x, y = cos x, x = 0 , and x = π/4

3

,

y

=

x

y

=

x

10. and in the first kuadrant.

11. Region D is bounded by and x = 2y. Find the volume of solid that is generated by rotating region D around

x

y =

(a) x-axis (d) y-axis

(b) line x = -1 (e) line y = -2

Problem Set 3

(b) line = -1 (e) line = -2 (c) line y = 4 (f) line x = 4

12. Region D is bounded by parabola and line

x+ y = 4. Find the volume of solid that is generated by rotating region D around

2

4x x

y = −

(a) x-axis (c) y-axis