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Materi kuliah matematika teknik
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8. Integral Tertentu8.1 Integral Tertentu Lipat Satu8.2 Integral Tertentu Lipat Dua Sistem Koordinat Siku, Polar8.3 Integral Tertentu Lipat Tiga Sistem Koordinat Siku, Polar
Jumlah Riemann (Riemann Sum)An ordered collection P=(x0,x1,…,xn) of points of a closed interval I = [a,b] satisfying a = x0 < x1 < …< xn-1 < xn = b is a partition of the interval [a,b] into subintervals Ik=[xk-1,xk].
Let Δxk = xk-xk-1
For a partition P=(x0,x1,…,xn), let |P| = max{Δxk, k=1,…,n}. The quantity |P| is the length of the longest subinterval Ik of the partition P.
a = x0 x1 x2 xn-1 xn =b
|P|
Choose a sample point xi* in the subinterval [xk-1,xk].
A Riemann sum associated with a partition P and a function f is defined as:
nn
n
ii xxfxxfxxfxxf
)(...)()()( *2
*21
*1
1
*
b
af x dxIntegration
Symbol
lower limit of integration
upper limit of integration
integrand
variable of integration(dummy variable)
Note that the integral does not depend on the choice of variable.
If f is a function defined on [a, b], the definite integral of f from a to b is the number
provided that this limit exists. If it does exist, we say that f is integrable on [a, b].
n
iii
x
b
axxfdxxf
i 1
*
0max
)()( lim
Definition of a Definite Integral
n.integratiooflimitupperthe
andn,integratiooflimitlowerthe,nintegratiooflimits
thecalledareand].,[onintegrablesimplyor],[
onintegrableRiemannbetosaidis)(functiontheand
b
a
bababa
xf
7
Now, with z=f(x,y), we have two variables of integration: x and y.
Double integrals:volume under the graph of the function (surface) and above the xy-plane found by
using the volume of infinitely many rectangular prisms (obtained by partition of the area R).
Double Integrals-concept
We want to know the volume defined by z=f(x,y) ≥ 0 on the rectangle R=[a,b]×[c,d]
Double Integrals-conceptFirst we explain what we mean by a partition of the rectangle R. We begin with apartition
P1 = {x0, x1, . . . , xm} of [a, b] ,and a partition
P2 = {y0, y1, . . . , yn} of [c, d] .The set
P = P1 × P2 = {(xi , yj) : xi P1, yj P2}
is called a partition of R. The set P consists of all the grid points (xi , yj ).
Double Integrals-concept
Using the partition P, we break up R into m × n nonoverlapping rectangles
Rij : xi−1 ≤ x ≤ xi , yj−1 ≤ y ≤ yj , where 1 ≤ i ≤ m, 1 ≤ j ≤ n.
10
Double Integrals-evaluated in small volumes
• Similar to the intuition behind simple integrals, we can think of the double integral as a sum of small volumes.
11
Volume of ij’s column= area x height= Ayxf ijij ),( **
m
i
n
jijij Ayxf
1 1
** ),(Total volume of all columns:
ij’s column:
Area of Rij is Δ A = Δ x Δ y
f (xij*, yij
*)
Δ y Δ xxy
z
Rij
(xi, yj)
Sample point (xij*, yij
*)x
y
Double Integrals -evaluated in small volumes
12
m
i
n
jijij AyxfV
1 1
** ),(
• Definition of a Double Integral:
m
i
n
jijij AyxfV
1 1
**
nm,
),(lim
Double Integrals-sum of small volumes
13
The double integral of f over the rectangle R is
if the limit exists.
R
dAyxf ),(
m
i
n
jijij
R
AyxfdAyxf1 1
**
nm,
,),(),( lim
• Double Riemann sum:
m
i
n
jijij Ayxf
1 1
** ),(
Double Integrals-definition
• Note 1: If f is continuous then the limit exists and the integral is defined.• Note 2: The definition of double integral does not depend on the choice of sample points.• If the sample points are upper right-hand corners then
m
i
n
jji
R
AyxfdAyxf1 1nm,
),(),( lim
14
• Let z=16-x2-2y2, where 0≤x≤2 and 0≤y≤2.• Estimate the volume of the solid above the square and below the graph• Let’s partitioned the volume in small (mxn) volumes.• Exact volume: 48.
Double Integrals- Example
m=n=4;V≈41.5 m=n=8;V≈44.875 m=n=16;V≈ 46.46875
AAA
dAyxgdAyxfdAyxgyxf ),(),()],(),([
AA
dAyxfcdAyxcf ),(),(
AA
dAyxgdAyxf ),(),(
• Linearity
• Comparison: If f(x,y)≥g(x,y) for all (x,y) in R, then
Double Integrals-Properties
, , , ,f x y g x y dx dy f x y dx dy g x y dx dy
Double Integrals-Properties
16
• Additivity: If A1 and A2 are non-overlapping regions then
2121
),(),(),(AAAA
dAyxfdAyxfdAyxf
•Mean value condition/mean value theorem: There is a point (x0, y0) in R for which
We call f (x0, y0) the average value of f on R .
R
R
AyxfdAyxf ),(),( 00
R
R dxdyAR of area
Average (Mean) Value
If f is integrable on [a,b], its average (mean) value on [a,b] is:
av(f) =
Find the average value of f(x) = 4 – x2 on [0,3] . Does f actually take on this value at some point in the given interval?
1( )
b
af x dx
b a
Applying the Mean Value
Av(f) =
= 1/3(3) = 1
4 – x2 = 1 when x = ± √3 but only √3 falls in the interval from [0,3], so x = √3 is the place where the function assumes the average.
3 2
0
1(4 )
3 0x dx
19
• If f(x,y) is continuous on rectangle R=[a,b]×[c,d] then the double integral is equal to the iterated integral
a bx
y
c
d
x
y
b
a
d
c
d
c
b
aR
dydxyxfdxdyyxfdAyxf ),(),(),(
fixed fixed
Double Integrals: Computation
20
• Example: given f(x,y)=x+y, for 0x1 and 0y1. Find the double integral (volume under the graph, V) of the f(x,y) over the area R on xy plane.
• Answer: V=1
Double Integrals: Computation
21
Double Integrals: Computation (General Case)
• If f(x,y) is continuous onA={(x,y) | x in [a,b] and h (x) ≤ y ≤ g (x)} then the double integral is equal to the iterated integral
a bx
y
h(x)
g(x)
x
b
a
xg
xhA
dydxyxfdAyxf)(
)(
),(),(
A
22
Double Integrals: Computation (General Case)
• Similarly, if f (x,y) is continuous onA={(x,y) | y in [c,d] and h (y) ≤ x ≤ g (y)} then the double integral is equal to the iterated integral
d
x
y d
c
yg
yhR
dxdyyxfdAyxf)(
)(
),(),(
c
h(y) g(y)y
A
23
• If f (x, y) = φ (x) ψ(y) then
d
c
b
a
d
c
b
aR
dyydxxdxdyyxdAyxf )()()()(),(
Double Integrals: Note
],[],[ ,2
1
],2/[]1,2/1[ ,)sin(
2
)(
2
)(22
Rdxdyee
AdAxy
R
yx
R
yx
• Examples:
Problem 1
1. Describe the region of integration and evaluate this:
2. One of applications of double integral is finding a volume. Knowing z=4x2 + 9y2, find the volume under the surface and above the rectangle with vertices (0,0),(3,0),(3,2), and (0,2).
Note: draw/sketch the problem would be helpful.
4
0
cos
sin
x
xxydydx
Changing Variables in double Integration using Jacobian
It is shown a basic region Γ in a plane that we are calling the uv-plane. (In this plane we denote the abscissa of a point by u and the ordinate by v.) Suppose that
x = x(u, v), y = y(u, v)
are continuously differentiable functions on the region Γ.
“Mapping”
Changing Variables in double Integration using Jacobian
where |J(u,v)| is the absolute value of the Jacobian*:
So, calculating the area we have:
The formula for a change of variables in double integrals from (x, y) to (u, v) is
u
y
v
x
v
y
u
x
v
y
u
yv
x
u
x
vu
yxvuJ
,
,),(
Taken from the name of German mathematician, Carl Gustav Jacob Jacobi (1804-1851) who contributed in elliptic functions, partial differential equations, and mechanics.
The Double Integral in Polar Coordinates
The changing variables from xy-plane to r-plane in polar coordinate can use the Jacobian, in which x= r cos and y=r sin . Then, using the Jacobian formula, J = r.
To calculate the area:
and
Triple Integration
• The triple integral is a generalization of the double integral.• Considering a function f of three variables,
that defined in a bounded closed region T in space in three dimensions (3D).
• Subdividing the region T by planes parallel to the three coordinates planes creates small boxes. You can imagine this as a rubic toy.
• Subdividing the region T into n partitions, we can choose an arbitrary point in box kth (for k=1,2,...,n), (xk, yk, zk), and form the sum:
),,( zyxf
k
n
kkkkn VzyxfV
1
),,(
Triple Integration• Assuming that f(x,y,z) is continuous in a domain containing T
and T is bounded by many smooth surfaces, and partitioning T with larger n more untill n approaches infinity, the sum will converge to a limit that is independent of the choice of subdivisions and corresponding point.
• The limit is called the triple integral of f(x,y,z) over region T:
• And to calculate the volume of T:
TT
dzyxfdxdydzzyxf V),,(or ),,(
Problem 2
1. A mass distribution of density in a region T in space is defined by =x2+y2+z2 with T: |x|1, |y|3, |z|2.Find the total mass.
Application of Definite Integral
• Volumes using cross sections• Volumes using cylindrical shell• Arc length• Areas of surface• Work and fluid forces• Moment and centre of mass