Final report calculus 2 semester 213

Theorem - In mathematics, double integral is defined as the integrals of a function in two variables over a region in R2 ∬Ąý, þĂýā the double integral is equal to the volume under the s

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UNIVERSITY OF TECHNOLOGY FACULTY OF APPLIED SCIENCE

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Class: CC05 - Group: 04

Work Assignment

1 Tăng Gia B¿o 2152429

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TABLE OF CONTENTS

Work Assignment 2

Table of contents 3

Double integral 4

Theorems

Application of double integral 9

Triple integral 13

Theorems

Application of triple integral 15

Line integral 21

Theorems

Application of line integral 23

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DOUBLE INTEGRAL

1 Theorem

- In mathematics, double integral is defined as the integrals of a

function in two variables over a region in R2

∬Ą(ý, þ)Ăýā

the double integral is equal to the volume under the surface

ÿ = Ą (ý, þ) and above the xy-plane in the region of integration R

- In the region R is a rectangle [ ÿ, Ā ∗ [ ā, Ă ]] we can subdivide [ÿ, Ā]

into small intervals with a set of numbers { ý ,ý0 1, , ýă}

- ÿ = ý < ý0 1< ý2< < ýÿ< < ýă21< ýă= Ā

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- Similarly, a set of numbers {þ0, þ1, , þĄ}is said to be a partition of [c,d] along the y - axis, if

ā = þ0< þ1< þ2< < þĀ< < þĄ21< þĄ= Ă The Riemann sum of a function f(x,y) over this partition of [a,b] x [x,d] is

∑ ∑ Ą(Ă ,ăÿ Ā) △ ýÿ△ þĀ Ą

Ā=1

ă ÿ=1

where (Ă , ăÿ Ā) is some point in the rectangle (ý , ýÿ21 ÿ) (þĀ21, þĀ) and

ă ÿ=1

- To define the double integral over a bounded region R other than a rectangle, we choose a rectangle [a,b] x [c,d] the contains R and define the function g(x,y) so that:

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ą(ý, þ) = Ą(ý, þ), Ą(ý, þ) Ā Ă ÿĄą(ý, þ) = 0 Ą(ý, þ) + Ă ÿĄ

The the double integral of the function f(x,y) over a general region R

is defined to be :

∬Ą(ý, þ)Ăý ā

= ∬ ą(ý, þ)Ăý

[ÿ,Ā][ā,Ă]

What is double integral used for ?

- Double integrals are used to calculate the area of a region, the volume under the surface, and the average value of a function of two

variables over a rectangular region

- Beside, double integral are also used for calculating the mass of whole 2 - dimensionale things based on density function Ă(ý, þ) Does double integral order matter ?

- The order of the integration is irrelevant, yet we should take the limits of the integrals based on the chosen order

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Double integral rule

- In calculus, we usually follow the rules and formulas to perform any integration method To solve integration problems, you must have studied various ways such as integration by part, integration by substitution, or formulas In the case of double integration also, we will discuss here the rule for double integration by parts, which is given by :

++u dv/dx dx.dy = +[uv +v du/dx dx]dy

-Properties of double integral

- + + Ą(ý, þ) ÿĀ āĂ Ăþ Ăý= + + Ą(ý, þ)ĂýĂþ āĂ ÿĀ

- + + (Ą(ý, þ) ± ą(ý, þ)) = + + Ą(ý, þ) ± + + ą(ý, þ) Ăý Ăý Ăý

- ĀĄ Ą(ý, þ) < ą(ý, þ), ā/ + + Ą(ý, þ) < + + ą(ý, þ)ăĀ Ăý Ăý

- ý + + Ą(ý, þ).Ăý = + + ý Ą(ý, þ).Ăý

- + + Ă * (ý, þ).㥠Ăý = + + ĂĄ(ý, þ).Ăý+ + + (ý, þ).ĀĄ Ăý

- If f(x,y) 0 on R an R and S are non-overlapping regions the g

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How to evaluate double integral

There are two method to evaluate double integral :

Given a flat plate in the shape of the disk with center the origin and radius

2 Given that for any point M in the plate, the density of the plate at M equal

to the distance from M to the origin Evaluate the mass of the plate

o The position of M in the Cartesian coordinate system: (ý�㕀, þ�㕀)

o The function of the plate: ý2+ þ f 22 2

o The function of the density at M: Ā =ÿ�㕀 = √2ý+ þ�㕀 �㕀 2

o Mass of the plate:

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Plot the graph of the plate:

Example 2:

Example 2:

Consider a thin metal plate that covers the triangular region shown below

If the density of the plate is measured by Ā(ý, þ) = 6ý + 6þ + 1 then calculate the plate’s: first moments, center of mass

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Mass of the plate:

First moments of the plate:

Center of mass:

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TRIPLE INTEGRAL

A Theorem:

Just as we defined single integrals for functions

of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables Let’s first deal with the simplest case where f is defined on a rectangular box:

þ = ý, þ, ÿ{( ) | ÿ f ý f Ā, ā f þ f Ă, ÿ f ÿ f Ā} The first step is to divide B into sub-boxes We

do this by dividing the interval [a, b] into l

subintervals [ý , ýÿ21 ÿ] of equal width &ý, dividing [c, d] into m subintervals of width &þ, and dividing [r, s] into n subintervals of width

&ÿ The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into lmn sub-boxes

þÿĀā= [ýÿ21, ýÿ] × [þĀ21, þĀ] × [ÿ , ÿā21 ā] which are shown in Figure 1 Each sub-box has volume Then we form the triple Riemantriple Riemann sum triple Riemann sum

Where the sample point (ý∗ÿĀā, þÿĀā∗ , ÿ∗ÿĀā) is in þÿĀā

• Definition: The triple integrals of Ą(x, y, z) over the box B is

∭ Ą ý, þ, ÿ( )Ă�㕉 = limĂ,ă,Ą→∞∑ ∑ ∑ Ą(ýÿ,þĀ, ÿā)&�㕉

Fubini’s Theorem for Triple Integrals:

If f is continuous on the rectangular box þ = [ÿ, Ā] × [ā, Ă] × [ÿ, Ā] then

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∭ Ą ý, þ, ÿ( )Ă�㕉 = ∫ ∫ ∫ Ą(Ā ý, þ, ÿ ĂýĂþĂÿ)

ÿ

Ă ā

Ā ÿ

HOW TO EVALUATE A TRIPLE I

HOW TO EVALUATE A TRIPLE INTEGRAL NTEGRAL

There are two methods:

• The projection method

• Using spherical coordinates

I The projection method:

Given a solid E as in the figure 2, then we have:

∭ Ą ý, þ, ÿ( )Ă�㕉= ∬Ă2[∫(ý,þ) Ą(ý, þ, ÿ)Ăÿ

Ă 1 (ý,þ) ] Ăý

Explanation:

• D is the projection of E onto Oxy

• ÿ = Ă1(ý, þ is the lower surface of ) E

• ÿ = Ă (ý, þ)2 is the upper surface of E

II Using spherical coor

II Using spherical coordinates: dinates:

Relation between Cartesian Coordinates and Spherical Coordinates

Given:

M(x, y, z) | Cartesian Coordinates M(ρ, Ø, �㔃) | Spherical Coordinates Then

{ý = ĀāāĀ∅ĀÿĀ�㔃þ = ĀĀÿĀ∅ĀÿĀ�㔃

ÿ = ĀāāĀ�㔃

So Ā =√ý2+ þ + ÿ2 2

Theorem:

= ∫ ∫:12ý2 2√1 2 ý2 2:12ý 2 ĂþĂý

1 21

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2 Plot the internal volume:

Because the cone is homogenous, then the density ÿ(ý, þ, ÿ = ÿ) ą can be

taken outside the integral sign:

Āÿ= ÿą∭(ý2+ þ2)ĂýĂþĂÿ

ĂĂĂ

We have range of z: ÿā�㔻f ÿ f ÿ Then:

Āÿ= ÿą,ýĂĂ2 +þ 2 ≤ā(ý2+ þ2)1| �㔻ÿ�㔻�㕅ĂýĂþ = ÿąýĂĂ,2 +þ ≤ā 2 (ý2+ þ2) (ÿ 2

= ÿąÿ ∫ ∫ (ÿ32ÿ4Ă ) ĂÿĂ�㔑

ā 0

2�㔋 0

2�㔋 0

=ÿąÿÿĂ10 4Also, we have:

ÿ = ÿą�㕉, �㕉 =3 ÿĂ2ÿ 1

⇒ ÿą=ÿ� 㕉 =ÿĂ2ÿ3ÿ

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Thus: Āÿ=�3ă㔋ā2�10㔻.�=㔋�㔻ā3ăā102 4

It is interesting that the moment of inertia of the cone does not depend on its height

b) Matlab:

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LINE INTEGRAL

1.

1 Theorem: Theorem:

- Suppose we have a solid wire having a variable mass density along

the length of this wire so how can we compute the total mass of this wire without a electronic scale? Lets find out the answer

- Lets think of the above wire with the shape like this:

- The curve (C) , (C) :ÿ(ā) = ýÿ + þĀ = ą(ā)ÿ + /(ā)Ā, has the length Āand distribution density �㔇 = Ą(ý, þ) If we divide the parameter interval [a,b] into n subintervals [āÿ21, āÿ] od equal width and we let

ý = ýÿ and þ = þÿ, then the corresponding points Āÿ21 and Āÿ form a segment Āÿ21Āÿ We choose any point Āÿ∗(ýÿ∗, þÿ∗) in subintervals

[āÿ21, āÿ] If the mass density is constant, the mass of this wire is

computed by this formula ÿ = �㔇�㗥þ In our case, the mass density vary with respect to x and y, thus we have to divide (C) into the very small

�㗥Āwhich induce �㗥ÿj Ą(ýÿ∗, þÿ∗).�㗥Ā �㗥ÿ⇒ÿ=1 j ∑, þÿ∗)�㗥ĀĄ Ą(ýÿ ÿ∗

⇒ ÿ = þÿÿĄ→∞∑ Ą(ýÿ∗

Ą ÿ=1

, þÿ∗)�㗥Āÿ= ∫Ą(ý, þ)ĂĀ

þ

- (�㗥Ā)2j (�㗥ý)2+ (�㗥þ)2 , by using these two theorems :

- Two other line intergral are obtained by replacing �㗥Ā by either �㗥ý =ÿ

ý 2 ýÿ ÿ21 or �㗥þÿ= þÿ2 þÿ21 They are called the line integrals of f along C with respect to x and y:

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It frequently happens that line integrals with respect to x and y occur together When

this happens, it’s customary to abbreviate by writing:

∫ Ā(ý, þ)Ăý + ∫ ā(ý, þ)Ăþ= ∫Ā(ý, þ)Ăý+

þ þ

Application of Line Integral

Line integral has several applications A line integral is used to calculate the surface area in the three-dimensional planes Some of the applications of line integrals in the vector calculus are as follows:

• A line integral is used to calculate the mass of wire

• It helps to calculate the moment of inertia and centre of mass of wire

• It is used in Ampere’s Law to compute the magnetic field around a conductor

• In Faraday’s Law of Magnetic Induction, a line integral helps to determine the voltage generated in a loop

• Line integral helps to calculate the work done by a force on a moving object in a vector field

2 Application:

Example 1:

How can we calculate the flux �㔙ýpast through a surface whose area cannot

be computed straightforwardly For example, magnetic field þ󰇍,þ󰇍 = þýÿ+

þþĀ + þÿý󰇍, passes through a surface inside a curve (C),(C):

ý(ā) = 216 ĀÿĀ3( ā)þ(ā) =13āāĀ( ā) 2 5 āāĀ( 2ā) 2 2 āāĀ( 3ā) 2 āāĀ( 4ā)

In this case, we can use Green theorem to compute the surface inside (C)

Ă�㔙 = þý󰇍.Ăý = þ󰇍.Ā󰇍.Ăý = þÿĂý ⇒ �㔙ý= þÿÿ, 1Ăý

Green theorem: ∮ Ā(ý, þ)Ăý+ ā(ý, þ)Ăþ = , (ÿ��㔕ý㔕Ā 2�㔕�㕌�㔕ÿĂý )

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ÿā

ÿý = 1 ⇒ ā(ý, þ) = ý + ÿ1ÿĀ

ÿ�㕌 = 0 ⇒ Ā(ý, þ) = ÿ2

We choose ÿ1and ÿ2equal to 0

⇒ ýÿăÿ = ā(ý, þ)∮þ Ăþ= + ýþ′(ā)Ăā22�㔋0

If p(x,y) is the density of wire (mass per unit length), then ÿ =+ Ă(ý, þ)ĂĀ

is the mass of the wire Find the mass of a wire having the shape of a top half of circle ý2+ þ = 42 , and the mass density is described by the function Ă(ý, þ) = ý þ.6

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