Matematika Teknik

Pertemuan 8:Integral Tertentu dan aplikasinya

Outline...

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

• When we deal with single variable integral, the result is an area under the curve.

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

The Double Integral in Polar Coordinates

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

kV is the volume of box kth

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 ),,(

Triple Integration

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

Volume using cross section

Slicing method

Disk method

Reference..

Salas, Hille, Etgen Calculus: One and Several Variables, Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Erwin Kreizig, Advanced Engineering Mathematics 8th edition, Willey 1999

www.bauer.uh.edu/rsusmel/phd/MR-12.ppt Thomas’ Calculus