Applications of Definite Integral
Applications of Definite Integral
• Area between Curve…..………...…………….2
• Volume of Solid of Revolution……………..12
• Volume by Cylindrical Shells……………....21
• The Length of Curve……………………..….25
Area between Curves
A. Let D = {( x, y) | a ≤ x ≤ b, 0 ≤ y ≤ f ( x)}
f(x)
Area of region D = ?
D
Steps :
a
∆x
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 ∆x → 0 , the area of D is
b
A = f ( x)dx
a
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2
Example :
2
y
=
x
,
Find the area of region that is bounded by parabola
x axis , and x = 2.
Area of pieces:
∆A ≈ x 2 ∆x
y = x2
x
2
Area of region:
2
∆x 2
1 3
A = x dx = x
3
0
2
2
8
=
3
0
Applications of Definite Integral
B. Let D = {( x, y) | a ≤ x ≤ b, g ( x) ≤ y ≤ h( x)}
Area of region D = ?
Steps:
chapter
7
h(x)
D
h(x)-g(x)
1. Divided D into n pieces, the area
g(x)
of each pieces is approximated
by area of rectangular with
∆x
a
b
height h(x) - f(x) and length of
base x ; ∆A ≈ (h( x) − g ( x))∆x
2. The area of D is approximated by sum of area rectangular.
If ∆x → 0 , the area of D is
b
(h( x) − g ( x))dx
a
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Applications of Definite Integral
chapter
Example:
Find the area of region that is bounded by y = x+4
and parabola y = x2 – 2.
The straight line and
parabola intersects at
7
x + 4 = x2 − 2
x2 − x − 6 = 0
( x + 4) − ( x 2 − 2)
y = x2 − 2
y=x+4
-2
∆x
3
( x − 3)(x + 2) = 0
x = -2, x = 3
Area of pieces:
∆A ≈ ((x + 4) − ( x 2 − 2))∆x
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Applications of Definite Integral
chapter
7
Area of region :
3
3
A = (( x + 4) − ( x 2 − 2))dx = (− x 2 + x + 6)dx
−2
−2
1 3 1 2
= − x + x + 6x
3
2
3
−2
125
=
6
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6
Applications of Definite Integral
chapter
7
Example: Find the area of region that is bounded by x axis,
y = x2 and y = -x + 2.
Answer: Intersection points
x2 = −x + 2
y = x2
y=-x+2
∆x
1
∆x
x2 + x − 2 = 0
( x + 2)(x − 1) = 0
x = -2, x = 1
If pieces is vertical , then region
must be divided into 2 sub region.
Area of pieces I
∆A1 ≈ x 2 ∆x
2
Area of pieces II
∆A2 ≈ (−x + 2)∆x
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Applications of Definite Integral
chapter
The area of region I
7
1
1
A1 = x dx = x | =
3
0
2
1
3
3 1
0
The area of region II
2
A2 = − x + 2 dx = − 12 x 2 + 2x |12
1
= (−2 + 4) − (− 12 + 2) =
1
2
The total area is
1 1 5
A = A1 + A2 = + =
3 2 6
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Applications of Definite Integral
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
area of rectangular with height
h(y) - f(y) and length of base y ;
chapter
7
!
∆∆
yy
h(y)-g(y)
∆A ≈ (h( y) − g( y))∆y
2. The area of D is approximated by sum area of rectangular.
If ∆y → 0 , the area of D is
d
A=
(h( y ) − g ( y )) dy
c
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Applications of Definite Integral
chapter
7
Example: Find the area of region that is bounded by x = 3 − y 2
and y = x − 1.
Answer :
The intersection point between
parabola and straight line are
y = x −1
y +1 = 3 − y 2
y2 + y − 2 = 0
1
( y + 2)( y − 1) = 0
∆y
(3 − y2 ) − ( y +1)
x = 3 − y2
-2
y = -2 and y = 1
Area of pieces:
∆A ≈ ((3 − y 2 ) − ( y + 1)) ∆y
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Applications of Definite Integral
chapter
7
The area of region is :
1
1
L = ((3 − y 2 ) − ( y + 1))dy = (− y 2 − y + 2)dy
−2
−2
1 3 1 2
= − y − y + 2y
3
2
1
−2
9
= .
2
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11
Volume of a Solid
of Revolution
Volume by Disk Method
• The Region D = {( x, y) | a ≤ x ≤ b , 0 ≤ y ≤ f ( x)}
is revolved around the x-axis.
#
!
"
Region D
Solid of revolution
volume of a solid of revolution?
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f(x)
D
a
x
b
f(x)
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).
Thus,
∆V ≈ π f 2 ( x) ∆x
b
x
V = π f 2 ( x) dx
a
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
and height x.
Thus,
y = x2
x2
∆x
∆V ≈ π ( x 2 ) 2 ∆x = π x 4 ∆x
$
x2
∆x
volume of a solid of revolution:
2
π 5 2 32
4
V = π x dx = x |0 = π
5
5
0
• The region D = {( x, y) | c ≤ y ≤ d , 0 ≤ x ≤ g ( y)}
is revolved around y-axis.
#%
!
Region D
Solid of revolution
volume of a solid of revolution?
&
∆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) .
Thus,
x = g(y)
!
g (y)
∆V ≈ π g 2 ( y) ∆y
∆y
d
V = π g 2 ( y) dy
c
'
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
y
y = x2
⇔x= y
y
If the slice is revolved around y-axis
we get a circular disk with radius y
and height y.
Thus,
∆V = π ( y ) 2 ∆y = π y ∆y
∆y
volume of a solid of revolution:
4
π 2 4
V = π ydy = y |0 = 8π
0
2
Applications of Definite Integral
chapter
Volume by Washer Method
7
• The Region D = {( x, y) | a ≤ x ≤ b , g ( x) ≤ y ≤ h( x)}
is revolved around x-axis.
h(x)
D
g(x)
a
b
Region D
Solid of revolution
volume of a solid of revolution?
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Applications of Definite Integral
h(x)
b
∆x
∆V ≈ π (h 2 ( x) − g 2 ( x))∆x
g(x)
h(x)
∆x
7
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),
and height x, thus
D
a
chapter
g(x)
b
V = π (h 2 ( x) − g 2 ( x))dx
a
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Applications of Definite Integral
chapter
7
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
radius 1 and outer radius 1 + x2 , thus
∆V ≈ π (( x2 +1)2 −12 )∆x
y = x2
1+ x2
!
∆x
≈ π ( x4 + 2x2 +1 −1)∆x
$
%
≈ π ( x4 + 2x2 )∆x
2
V = π x 4 + 2x 2 dx = π ( 15 x 5 + 23 x3 |02 ) = π ( 325 + 163 ) = 186
15 π
0
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Volume by Cylindrical
Shells
The region D = {( x, y ) | a ≤ x ≤ b , 0 ≤ y ≤ f ( x)}
is revolved around y-axis
#
!
"
Region D
Solid of revolution
volume of a solid of revolution ?
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21
Applications of Definite Integral
7
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
shell of radius x, height f(x), and
thickness x. Thus, its volume is
#
!
∆x
chapter
"
#
∆x
∆V ≈ 2π x f ( x) ∆x
#
#
b
V = 2π xf ( x)dx
a
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Applications of Definite Integral
chapter
7
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
radius x, height x2 , and thickness x.
Thus,
y = x2
x2
!
#
∆x
$
∆V ≈ 2π x x2∆x ≈ 2π x3 ∆x
volume of a solid of revolution:
2
π 4 2
3
V = 2π x dx = x |0 = 8π
2
0
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Applications of Definite Integral
chapter
7
Remark :
- The method of disks and washer:
The slices are perpendicular to rotating axis
- The method of Cylindrical Shells:
The slices are parallel to rotating axis
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The Length of a Curve
A parameter form of curve on R2
x = f(t) ,
y = g(t)
a
t
b
(*)
Point A(f(a), g(a)) is called original point and B(f(b), g(b))
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)
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Applications of Definite Integral
chapter
7
Let a curve on parameter form (*), we will find length
of that curve.
Steps:
1. Divided interval [a,b] into n subintervals
a = t o < t1 < t 2 < ... < t n = b
Qi −1
t1
ti−1 ti
t n−1 "
A partition on [a,b]
(
Qi
Qn
(
Qo
(
Q1(
(
A partition on curve
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Applications of Definite Integral
chapter
2. Approximate length of curve
∆si length of arc
∆si
∆wi
Qi−1
∆xi
7
Qi−1Qi
length of line segment
Qi
∆wi
∆yi
The length of arc is approximated by
length of line segment
Qi −1Qi
2
2
∆si ≈ ∆wi = (∆xi ) + (∆yi )
= [ f (ti ) − f (ti−1 )]2 + [ g(ti ) − g(ti−1 )]2
We apply mean value theorem to function f and g.
There are tˆi , ti ∈ (ti−1, ti ) , such that
f (ti ) − f (ti−1 ) = f ′(ti )∆t ,
g (ti ) − g (ti−1 ) = g′(tˆi )∆t
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Applications of Definite Integral
chapter
where ∆ti = ti − ti−1
so
7
∆wi = [ f ' (ti )∆ti ]2 + [ g' (tˆi )∆ti ]2
= [ f ' (ti )]2 + [ g ' (tˆi )]2 ∆ti
The length of curve is approximated by length
of polygonal arc,
n
[ f ' (ti )]2 + [ g ' (tˆi )]2 ∆ti
L≈
i =1
If ||P||
0, the length of curve is
b
[ f ' (t )]2 + [ g ' (t )]2 dt
L=
a
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Applications of Definite Integral
chapter
Remark:
7
For curve y = f(x), a ≤ x ≤ b
b
b
2
L=
2
[ f ' (t )] + [ g' (t )] dt
=
[
a
a
2
b
dx 2 dy 2
] + [ ] dt
dt
dt
b
dx 2
dy
dy
( ) (1+
)dt = 1+
dt
dx
dx
a
=
a
2
dx
For curve x = g(y), c ≤ y ≤ d
d
d
[ f ' (t )]2 + [ g ' (t )]2 dt =
L=
c
d
=
c
[
c
dy
dx
( )2 1 +
dt
dy
2
d
dt =
c
dx 2 dy 2
] + [ ] dt
dt
dt
dx
1+
dy
2
dy
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29
Applications of Definite Integral
chapter
7
Example: Find length of curve
1. x = t 3 , y = t 2 ; 0 ≤ t ≤ 4
x′(t ) = 3t 2 , y' (t ) = 2t
The length of curve
4
4
(3t 2 ) 2 + (2t ) 2 dt =
L=
4
t 2 (9t 2 + 4)dt
9t 4 + 4t 2 dt =
0
0
4
0
4
2
d
t
(
9
+ 4)
2
2
1/ 2
= t 9t + 4 dt = t (9t + 4)
18t
0
0
= 181 23 (9t 2 + 4) 3 / 2 |04
= 271 (40 40 − 8) = 271 (80 10 − 8)
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Applications of Definite Integral
chapter
7
1
2. y = 2x , ≤ x ≤ 7
3
3/ 2
Answer :
dy
= 3x1 / 2
dx
7
L=
(
1 + 3x
1/ 3
) dx =
1/ 2 2
7
7
1 + 9xdx = 19 (1 + 9x)1/ 2 d (1 + 9x)
1/ 3
1/ 3
= 272 (1 + 9x) 3 / 2 |17/ 3 = 272 (512 − 8) = 37 13
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Problem Set 1
For problem 1 – 5, sketch and find area of region that
is bounded by
1. y = x2 and y = x + 2
2. y = x3 , y = −x, and y = 8
3. y = x , y = 4x , and y = -x +2
4. y = sin x, y = cos x, x = 0 , and x = 2π.
5. x − 2 y + 7 = 0 and y 2 − 6 y − x = 0
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Problem Set 2
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. y = x3 , y = 0, and x = 2
2
y
=
9
−
x
7.
and y = 0
2
8. y = x and y = 4x
9. y = sin x, y = cos x, x = 0 , and x = π/4
3
10. y = x and y = x, in the first kuadrant.
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Problem Set 3
11. Region D is bounded by y = x and x = 2y. Find the
volume of solid that is generated by rotating region
D around
(a) x-axis
(d) y-axis
(b) line x = -1
(e) line y = -2
(c) line y = 4
(f) line x = 4
12. Region D is bounded by parabola y = 4x − x 2 and line
x+ y = 4. Find the volume of solid that is generated
by rotating region D around
(a) x-axis
(b) line x = 6
(c) y-axis
(d) line y = -1
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Problem Set 4
For problem 13 – 18, find length of curves.
13. x = 4 sin t, y = 4 cost − 5; 0 ≤ t ≤ π
14. x = 3t 2 + 2, y = 2t 3 − 1/ 2; 1 ≤ t ≤ 4
1 2
y
=
( x + 2)3 / 2 , 0 ≤ x ≤ 1
15.
3
x 2 ln x
16. y = − , 2 ≤ x ≤ 4
2
4
17. y = ln(1 − x 2 ), 0 ≤ x ≤ 1/ 2
18.
x=
1
y ( y − 3), 0 ≤ y ≤ 9
3
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35
• Area between Curve…..………...…………….2
• Volume of Solid of Revolution……………..12
• Volume by Cylindrical Shells……………....21
• The Length of Curve……………………..….25
Area between Curves
A. Let D = {( x, y) | a ≤ x ≤ b, 0 ≤ y ≤ f ( x)}
f(x)
Area of region D = ?
D
Steps :
a
∆x
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 ∆x → 0 , the area of D is
b
A = f ( x)dx
a
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
2
Example :
2
y
=
x
,
Find the area of region that is bounded by parabola
x axis , and x = 2.
Area of pieces:
∆A ≈ x 2 ∆x
y = x2
x
2
Area of region:
2
∆x 2
1 3
A = x dx = x
3
0
2
2
8
=
3
0
Applications of Definite Integral
B. Let D = {( x, y) | a ≤ x ≤ b, g ( x) ≤ y ≤ h( x)}
Area of region D = ?
Steps:
chapter
7
h(x)
D
h(x)-g(x)
1. Divided D into n pieces, the area
g(x)
of each pieces is approximated
by area of rectangular with
∆x
a
b
height h(x) - f(x) and length of
base x ; ∆A ≈ (h( x) − g ( x))∆x
2. The area of D is approximated by sum of area rectangular.
If ∆x → 0 , the area of D is
b
(h( x) − g ( x))dx
a
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
4
Applications of Definite Integral
chapter
Example:
Find the area of region that is bounded by y = x+4
and parabola y = x2 – 2.
The straight line and
parabola intersects at
7
x + 4 = x2 − 2
x2 − x − 6 = 0
( x + 4) − ( x 2 − 2)
y = x2 − 2
y=x+4
-2
∆x
3
( x − 3)(x + 2) = 0
x = -2, x = 3
Area of pieces:
∆A ≈ ((x + 4) − ( x 2 − 2))∆x
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
5
Applications of Definite Integral
chapter
7
Area of region :
3
3
A = (( x + 4) − ( x 2 − 2))dx = (− x 2 + x + 6)dx
−2
−2
1 3 1 2
= − x + x + 6x
3
2
3
−2
125
=
6
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
6
Applications of Definite Integral
chapter
7
Example: Find the area of region that is bounded by x axis,
y = x2 and y = -x + 2.
Answer: Intersection points
x2 = −x + 2
y = x2
y=-x+2
∆x
1
∆x
x2 + x − 2 = 0
( x + 2)(x − 1) = 0
x = -2, x = 1
If pieces is vertical , then region
must be divided into 2 sub region.
Area of pieces I
∆A1 ≈ x 2 ∆x
2
Area of pieces II
∆A2 ≈ (−x + 2)∆x
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7
Applications of Definite Integral
chapter
The area of region I
7
1
1
A1 = x dx = x | =
3
0
2
1
3
3 1
0
The area of region II
2
A2 = − x + 2 dx = − 12 x 2 + 2x |12
1
= (−2 + 4) − (− 12 + 2) =
1
2
The total area is
1 1 5
A = A1 + A2 = + =
3 2 6
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
8
Applications of Definite Integral
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
area of rectangular with height
h(y) - f(y) and length of base y ;
chapter
7
!
∆∆
yy
h(y)-g(y)
∆A ≈ (h( y) − g( y))∆y
2. The area of D is approximated by sum area of rectangular.
If ∆y → 0 , the area of D is
d
A=
(h( y ) − g ( y )) dy
c
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
9
Applications of Definite Integral
chapter
7
Example: Find the area of region that is bounded by x = 3 − y 2
and y = x − 1.
Answer :
The intersection point between
parabola and straight line are
y = x −1
y +1 = 3 − y 2
y2 + y − 2 = 0
1
( y + 2)( y − 1) = 0
∆y
(3 − y2 ) − ( y +1)
x = 3 − y2
-2
y = -2 and y = 1
Area of pieces:
∆A ≈ ((3 − y 2 ) − ( y + 1)) ∆y
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10
Applications of Definite Integral
chapter
7
The area of region is :
1
1
L = ((3 − y 2 ) − ( y + 1))dy = (− y 2 − y + 2)dy
−2
−2
1 3 1 2
= − y − y + 2y
3
2
1
−2
9
= .
2
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
11
Volume of a Solid
of Revolution
Volume by Disk Method
• The Region D = {( x, y) | a ≤ x ≤ b , 0 ≤ y ≤ f ( x)}
is revolved around the x-axis.
#
!
"
Region D
Solid of revolution
volume of a solid of revolution?
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
12
f(x)
D
a
x
b
f(x)
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).
Thus,
∆V ≈ π f 2 ( x) ∆x
b
x
V = π f 2 ( x) dx
a
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
and height x.
Thus,
y = x2
x2
∆x
∆V ≈ π ( x 2 ) 2 ∆x = π x 4 ∆x
$
x2
∆x
volume of a solid of revolution:
2
π 5 2 32
4
V = π x dx = x |0 = π
5
5
0
• The region D = {( x, y) | c ≤ y ≤ d , 0 ≤ x ≤ g ( y)}
is revolved around y-axis.
#%
!
Region D
Solid of revolution
volume of a solid of revolution?
&
∆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) .
Thus,
x = g(y)
!
g (y)
∆V ≈ π g 2 ( y) ∆y
∆y
d
V = π g 2 ( y) dy
c
'
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
y
y = x2
⇔x= y
y
If the slice is revolved around y-axis
we get a circular disk with radius y
and height y.
Thus,
∆V = π ( y ) 2 ∆y = π y ∆y
∆y
volume of a solid of revolution:
4
π 2 4
V = π ydy = y |0 = 8π
0
2
Applications of Definite Integral
chapter
Volume by Washer Method
7
• The Region D = {( x, y) | a ≤ x ≤ b , g ( x) ≤ y ≤ h( x)}
is revolved around x-axis.
h(x)
D
g(x)
a
b
Region D
Solid of revolution
volume of a solid of revolution?
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18
Applications of Definite Integral
h(x)
b
∆x
∆V ≈ π (h 2 ( x) − g 2 ( x))∆x
g(x)
h(x)
∆x
7
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),
and height x, thus
D
a
chapter
g(x)
b
V = π (h 2 ( x) − g 2 ( x))dx
a
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19
Applications of Definite Integral
chapter
7
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
radius 1 and outer radius 1 + x2 , thus
∆V ≈ π (( x2 +1)2 −12 )∆x
y = x2
1+ x2
!
∆x
≈ π ( x4 + 2x2 +1 −1)∆x
$
%
≈ π ( x4 + 2x2 )∆x
2
V = π x 4 + 2x 2 dx = π ( 15 x 5 + 23 x3 |02 ) = π ( 325 + 163 ) = 186
15 π
0
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20
Volume by Cylindrical
Shells
The region D = {( x, y ) | a ≤ x ≤ b , 0 ≤ y ≤ f ( x)}
is revolved around y-axis
#
!
"
Region D
Solid of revolution
volume of a solid of revolution ?
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
21
Applications of Definite Integral
7
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
shell of radius x, height f(x), and
thickness x. Thus, its volume is
#
!
∆x
chapter
"
#
∆x
∆V ≈ 2π x f ( x) ∆x
#
#
b
V = 2π xf ( x)dx
a
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
22
Applications of Definite Integral
chapter
7
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
radius x, height x2 , and thickness x.
Thus,
y = x2
x2
!
#
∆x
$
∆V ≈ 2π x x2∆x ≈ 2π x3 ∆x
volume of a solid of revolution:
2
π 4 2
3
V = 2π x dx = x |0 = 8π
2
0
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
23
Applications of Definite Integral
chapter
7
Remark :
- The method of disks and washer:
The slices are perpendicular to rotating axis
- The method of Cylindrical Shells:
The slices are parallel to rotating axis
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
24
The Length of a Curve
A parameter form of curve on R2
x = f(t) ,
y = g(t)
a
t
b
(*)
Point A(f(a), g(a)) is called original point and B(f(b), g(b))
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)
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25
Applications of Definite Integral
chapter
7
Let a curve on parameter form (*), we will find length
of that curve.
Steps:
1. Divided interval [a,b] into n subintervals
a = t o < t1 < t 2 < ... < t n = b
Qi −1
t1
ti−1 ti
t n−1 "
A partition on [a,b]
(
Qi
Qn
(
Qo
(
Q1(
(
A partition on curve
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
26
Applications of Definite Integral
chapter
2. Approximate length of curve
∆si length of arc
∆si
∆wi
Qi−1
∆xi
7
Qi−1Qi
length of line segment
Qi
∆wi
∆yi
The length of arc is approximated by
length of line segment
Qi −1Qi
2
2
∆si ≈ ∆wi = (∆xi ) + (∆yi )
= [ f (ti ) − f (ti−1 )]2 + [ g(ti ) − g(ti−1 )]2
We apply mean value theorem to function f and g.
There are tˆi , ti ∈ (ti−1, ti ) , such that
f (ti ) − f (ti−1 ) = f ′(ti )∆t ,
g (ti ) − g (ti−1 ) = g′(tˆi )∆t
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
27
Applications of Definite Integral
chapter
where ∆ti = ti − ti−1
so
7
∆wi = [ f ' (ti )∆ti ]2 + [ g' (tˆi )∆ti ]2
= [ f ' (ti )]2 + [ g ' (tˆi )]2 ∆ti
The length of curve is approximated by length
of polygonal arc,
n
[ f ' (ti )]2 + [ g ' (tˆi )]2 ∆ti
L≈
i =1
If ||P||
0, the length of curve is
b
[ f ' (t )]2 + [ g ' (t )]2 dt
L=
a
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
28
Applications of Definite Integral
chapter
Remark:
7
For curve y = f(x), a ≤ x ≤ b
b
b
2
L=
2
[ f ' (t )] + [ g' (t )] dt
=
[
a
a
2
b
dx 2 dy 2
] + [ ] dt
dt
dt
b
dx 2
dy
dy
( ) (1+
)dt = 1+
dt
dx
dx
a
=
a
2
dx
For curve x = g(y), c ≤ y ≤ d
d
d
[ f ' (t )]2 + [ g ' (t )]2 dt =
L=
c
d
=
c
[
c
dy
dx
( )2 1 +
dt
dy
2
d
dt =
c
dx 2 dy 2
] + [ ] dt
dt
dt
dx
1+
dy
2
dy
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
29
Applications of Definite Integral
chapter
7
Example: Find length of curve
1. x = t 3 , y = t 2 ; 0 ≤ t ≤ 4
x′(t ) = 3t 2 , y' (t ) = 2t
The length of curve
4
4
(3t 2 ) 2 + (2t ) 2 dt =
L=
4
t 2 (9t 2 + 4)dt
9t 4 + 4t 2 dt =
0
0
4
0
4
2
d
t
(
9
+ 4)
2
2
1/ 2
= t 9t + 4 dt = t (9t + 4)
18t
0
0
= 181 23 (9t 2 + 4) 3 / 2 |04
= 271 (40 40 − 8) = 271 (80 10 − 8)
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
30
Applications of Definite Integral
chapter
7
1
2. y = 2x , ≤ x ≤ 7
3
3/ 2
Answer :
dy
= 3x1 / 2
dx
7
L=
(
1 + 3x
1/ 3
) dx =
1/ 2 2
7
7
1 + 9xdx = 19 (1 + 9x)1/ 2 d (1 + 9x)
1/ 3
1/ 3
= 272 (1 + 9x) 3 / 2 |17/ 3 = 272 (512 − 8) = 37 13
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
31
Problem Set 1
For problem 1 – 5, sketch and find area of region that
is bounded by
1. y = x2 and y = x + 2
2. y = x3 , y = −x, and y = 8
3. y = x , y = 4x , and y = -x +2
4. y = sin x, y = cos x, x = 0 , and x = 2π.
5. x − 2 y + 7 = 0 and y 2 − 6 y − x = 0
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
32
Problem Set 2
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. y = x3 , y = 0, and x = 2
2
y
=
9
−
x
7.
and y = 0
2
8. y = x and y = 4x
9. y = sin x, y = cos x, x = 0 , and x = π/4
3
10. y = x and y = x, in the first kuadrant.
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
33
Problem Set 3
11. Region D is bounded by y = x and x = 2y. Find the
volume of solid that is generated by rotating region
D around
(a) x-axis
(d) y-axis
(b) line x = -1
(e) line y = -2
(c) line y = 4
(f) line x = 4
12. Region D is bounded by parabola y = 4x − x 2 and line
x+ y = 4. Find the volume of solid that is generated
by rotating region D around
(a) x-axis
(b) line x = 6
(c) y-axis
(d) line y = -1
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology
34
Problem Set 4
For problem 13 – 18, find length of curves.
13. x = 4 sin t, y = 4 cost − 5; 0 ≤ t ≤ π
14. x = 3t 2 + 2, y = 2t 3 − 1/ 2; 1 ≤ t ≤ 4
1 2
y
=
( x + 2)3 / 2 , 0 ≤ x ≤ 1
15.
3
x 2 ln x
16. y = − , 2 ≤ x ≤ 4
2
4
17. y = ln(1 − x 2 ), 0 ≤ x ≤ 1/ 2
18.
x=
1
y ( y − 3), 0 ≤ y ≤ 9
3
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35