Š OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880
Innovation for Solving the Problems in Social Life with Double Integrals KAY THI WIN1, KHIN MYO HTUN2, NGWE KYAR SU KHAING3 1, 2, 3 Department of Mathematics, TU (Yamethin),Myanmar Abstract- There are many different rules for mathematics field. Among them, we express the solving problems for practical life with integration methods. There are integrals, double integrals, triple integrals, etc. Among these method, we use the double integrals method to find the volume and area of a region between two curves and between line and curves that design of engineering’s. We define and evaluate double integrals over bounded regions in the plane which are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that of determining the limits of integration. We can find the volume and area of the various shape by using the integration methods. For example, volume of, area of the leaf and area of the cardioid shape, etc. We can compute number of people in the region bounded during the interval with integration method.
II.
SOME BASIC DEFINITIONS
A. Definition If R is a region like the one shown in the xy-plane in Figure.1, bounded ‘above’ and ‘below’ by the curve đ?‘Ś = đ?‘”2 (đ?‘Ľ) and đ?‘Ś = đ?‘”1 (đ?‘Ľ) and on the sides by the lines đ?‘Ľ = đ?‘Ž and đ?‘Ľ = đ?‘?.The cross-sectional area, đ?‘Ś =đ?‘”2 (đ?‘Ľ )
đ??´ đ?‘Ľ =
B. Definition If �(�, �) is positive and continuous over R, the volume of the solid region between R and the surface � = �(�, �) to be
đ?‘…
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ??´.
Integrating đ??´ đ?‘Ľ from đ?‘Ľ = đ?‘Ž to đ?‘Ľ = đ?‘? to get the volume as an iterated integral, đ?‘?
�=
đ?‘?
I.
INTRODUCTION
We consider the integral of a function of two variables � �, � over a region in the plane and the integral of a function of three variables � �, �, � over a region in space. To define the double integral of a function over a bounded, nonrectangular region R, this paper is organized with some basic definitions and notations. This paper present finding the volume with double integrals and finding the area with double integrals. And then we stated the double integral in polar form. We express finding the volume and area of the various shapes of the graph in social life. By using the integration methods to solve the problems in our social environment. In this paper, we used double integrals to solve the problems that submit the examples.
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III.
đ?‘”2 (đ?‘Ľ)
đ??´ đ?‘Ľ đ?‘‘đ?‘Ľ = đ?‘Ž
Indexed Terms- volume by double integration, area by double integration and average value of function, double integrate a polar form.
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ?‘Ś. đ?‘Ś =đ?‘”1 (đ?‘Ľ)
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ . đ?‘Ž
đ?‘”1 (đ?‘Ľ)
FUBINI’S THEOREM (STRONGER FORM)
Let đ?‘“(đ?‘Ľ, đ?‘Ś) be continuous on a region R. 1. If R is defined by đ?‘Ž ≤ đ?‘Ľ ≤ đ?‘?, đ?‘”1 (đ?‘Ľ) ≤ đ?‘Ś ≤ đ?‘”2 (đ?‘Ľ), with đ?‘”1 đ?‘Žđ?‘›đ?‘‘ đ?‘”2 continuous on [a, b], then(By using vertical cross-sections), đ?‘?
đ?‘”2 (đ?‘Ľ)
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ??´ = đ?‘…
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ . đ?‘Ž
đ?‘”1 (đ?‘Ľ)
2. If R is defined by đ?‘? ≤ đ?‘Ś ≤ đ?‘‘, đ?‘•1 (đ?‘Ś) ≤ đ?‘Ľ ≤ đ?‘•2 (đ?‘Ś), with đ?‘•1 đ?‘Žđ?‘›đ?‘‘ đ?‘•2 continuous on [c, d], then(By using horizontal cross-sections), đ?‘?
đ?‘• 2 (đ?‘Ś )
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ??´ = đ?‘…
đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś. đ?‘Ž
đ?‘• 1 (đ?‘Ś )
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SOME EXAMPLES OF FINDING VOLUME
A. Example The volume of a region such as tunnel shapes that lies the function đ?‘“ đ?‘Ľ, đ?‘Ś =
đ?‘Ľ đ?‘’ 2đ?‘Ś 4−đ?‘Ś 2
and x-axis between
0 ≤ đ?‘Ľ ≤ 2 đ?‘Žđ?‘›đ?‘‘ 0 ≤ đ?‘Ś ≤ 4 − đ?‘Ľ can be computed. đ?‘Ľ đ?‘’ 2đ?‘Ś đ?‘“ đ?‘Ľ, đ?‘Ś = 4−đ?‘Ś 2 4−đ?‘Ľ 2 đ?‘Ľđ?‘’ 2đ?‘Ś dy 0 0 4−đ?‘Ś
dx =
4 4−đ?‘Ś đ?‘Ľđ?‘’ 2đ?‘Ś dx 0 0 4−đ?‘Ś
dy.
đ?‘“ đ?‘Ľ, đ?‘Ś = đ?‘Ľ − đ?‘Ś + 2 By using the vertical cross-sections, Volume,V =
1 đ?‘Ľ 0 0
đ?‘Ľ − đ?‘Ś + 2 đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
=
1 0
đ?‘Ľđ?‘Ś −
=
1 0
đ?‘Ľ2 −
1 0
đ?‘Ľ2
=
6
2 đ?‘Ľ2 2
đ?‘Ś=đ?‘Ľ
+ 2đ?‘Ś
đ?‘‘đ?‘Ľ
đ?‘Ś=0
+ 2đ?‘Ľ đ?‘‘đ?‘Ľ
+ 2đ?‘Ľ đ?‘‘đ?‘Ľ
2
đ?‘Ľ3
=
đ?‘Ś2
1
+ đ?‘Ľ2
1
0 7
6
6
= +1=
.
By using the horizontal cross-sections, By using horizontal cross-sections, 4 0 0
4−đ?‘Ś đ?‘Ľđ?‘’ 2đ?‘Ś 4−đ?‘Ś
dx dy =
đ?‘Ľ 2 đ?‘’ 2đ?‘Ś
=
dy đ?‘‘đ?‘Ś
2
0 đ?‘’ 8 −1 4
đ?‘Ľ − đ?‘Ś + 2 đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś
=
1 đ?‘Ľ2 0 2
− đ?‘Ľđ?‘Ś + 2đ?‘Ľ
=
1 5 0 2
+
5đ?‘Ś
đ?‘Ś3
đ?‘Ľ= 4−đ?‘Ś
4 0 2(4−đ?‘Ś ) đ?‘Ľ=0 4 đ?‘’ 2đ?‘Ś =0 2 4 đ?‘’ 2đ?‘Ś
=
1 1 0 đ?‘Ś
Volume,V =
.
= = =
B. Example The volume of the region that triangle shape under the ladder in the đ?‘Ľđ?‘Ś-plane bounded by the x-axis and the line đ?‘Ś = đ?‘Ľ and đ?‘Ľ = 1 and whose top lies in the plane đ?‘“ đ?‘Ľ, đ?‘Ś = đ?‘Ľ − đ?‘Ś + 2.
2 5
+
1
2 6 15+1−9 6
6
đ?‘Ś2
− =
đ?‘‘đ?‘Ś
− 3đ?‘Ś đ?‘‘đ?‘Ś
2
−
đ?‘Ľ=đ?‘Ś
3đ?‘Ś 2
1
2
0
3 2 7 6
.
C. Example We will design a can to build at a high place. Therefore, we should volume of the under a can. The volume of a region between the curves đ?‘“ đ?‘Ľ, đ?‘Ś = 3 cos đ?‘Ľ bounded by the x-axis on (0 ≤ đ?‘Ś ≤ đ?œ‹ đ?œ‹ sec đ?‘Ľ đ?‘Žđ?‘›đ?‘‘ − ≤ đ?‘Ľ ≤ ) can find. 3
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đ?‘Ľ=1
3
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Š OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880
đ?œ‹ 3 đ?œ‹ − 3
sec đ?‘Ľ 0
3 cos đ?‘Ľ đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ = =
=
đ?œ‹ 3 đ?œ‹ − 3
=3 đ?‘Ľ
đ?œ‹ 3 đ?œ‹ − 3
đ?œ‹ 3 đ?œ‹ − 3
3 cos đ?‘Ľ . đ?‘Ś
đ?‘Ś=sec đ?‘Ľ đ?‘Ś =0
đ?‘‘đ?‘Ľ
3 cos đ?‘Ľ. sec đ?‘Ľ đ?‘‘đ?‘Ľ
3 đ?‘‘đ?‘Ľ đ?œ‹ 3
đ?œ‹ 3
−
đ?‘Ľ = đ?‘Ś 2 And đ?‘Ś = đ?‘Ľ + 2 đ?‘Ś = đ?‘Ś2 + 2 đ?‘Ś2 − đ?‘Ś + 2 = 0 đ?‘Śâˆ’2 đ?‘Ś+1 =0 đ?‘Ś = 2 đ?‘Žđ?‘›đ?‘‘ đ?‘Ś = −1. Region:đ?‘Ś 2 ≤ đ?‘Ľ ≤ đ?‘Ś − 2 , −1 ≤ đ?‘Ś ≤ 2. Area, A=
= 2Ď€. V.
= =
AREA BY DOUBLE INTEGRATION
=
A. Definition The area of a closed, bounded plane region R is đ??´đ?‘&#x;đ?‘’đ?‘Ž, đ??´ =
=
=
đ?‘…
B. Definition For an integrable function of two variables defined on a bounded region in the plane, the average value is the integral over the region divided by the area of the region.
VI.
1 đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘…
− 2đ?‘Ś −
đ?‘‘đ??´. đ?‘…
A. Example An engineering design for a building that parabola shape region. He think enclosed area by using the integration method. The area of the region R enclosed by the parabola shape đ?‘Ľ = đ?‘Ś 2 and the line đ?‘Ś = đ?‘Ľ + 2 ,can be computed.
−14 3
−
17 6
8 3
=
đ?‘‘đ?‘Ś
3 −1 1
−
+2+
2 −28−17 6
=
−45 6
1 3
.
B. Example The area of the regions R can be computed. The following function 1
0
4
đ?‘Ľ
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ + 0
SOME EXAMPLE OF FINDING AREA
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= 2−4−
đ?‘‘đ??´.
đ??´đ?‘Łđ?‘’đ?‘&#x;đ?‘Žđ?‘”đ?‘’ đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’ đ?‘œđ?‘“ đ?‘“ đ?‘œđ?‘Łđ?‘’đ?‘&#x; đ?‘… =
đ?‘‘đ??´.
đ?‘… 2 đ?‘Śâˆ’2 đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś −1 đ?‘Ś 2 2 đ?‘Śâˆ’2 đ?‘Ľ đ?‘Ś 2 đ?‘‘đ?‘Ś −1 2 đ?‘Ś − 2 − đ?‘Ś2 −1 2 đ?‘Ś2 đ?‘Ś3
= =
= −
3
1
+đ?‘Ľ
= − + 1+ =
đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ
đ?‘Ľ 2 −1 0 0 1 4 0 đ?‘Ś đ?‘‘đ?‘Ľ + đ?‘Ś 0 đ?‘Ľ đ?‘‘đ?‘Ľ đ?‘Ľ 2 −1 0 0 1 4 −đ?‘Ľ 2 + 1 đ?‘‘đ?‘Ľ + 0 đ?‘Ľ đ?‘‘đ?‘Ľ 0 3 4 1 đ?‘Ľ3 đ?‘Ľ2 0
+
3 2
0
16
3 3 −1+3+16 3
= 6.
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Š OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880 C. Example Finding the average values of đ?‘“ đ?‘Ľ, đ?‘Ś = cos(đ?‘Ľ + đ?‘Ś) over the rectangle 0 ≤ đ?‘Ľ ≤ đ?œ‹ , 0 ≤ đ?‘Ś ≤ đ?œ‹ . đ?œ‹
đ?œ‹
đ?‘‘đ??´ =
cos(đ?‘Ľ + đ?‘Ś) đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ 0
đ?‘…
đ?œ‹ 0 1 0
= =
0
sin(đ?‘Ľ + đ?‘Ś)
Ď€
=2
6
+
sin π
Ď€
−0 = .
6
3
B.Example The area of the region R of a piece of moon. The area of the region that lies inside the cardioid r = 1 + π cos θ over the region R:0 ≤ θ ≤ and outside the 2
đ?œ‹ 0
circle r = 1 can be stated.
đ?‘‘đ?‘Ľ
sin đ?‘Ľ + đ?œ‹ + sin đ?‘Ľ đ?‘‘đ?‘Ľ
y
= −cos đ?‘Ľ + đ?œ‹ − cos đ?‘Ľ đ?œ‹0 = −cos 2đ?œ‹ − cos đ?œ‹ − −cos đ?œ‹ − 1 = −1 + 1 − 1 − 1 = 0. Area of rectangle =đ?œ‹ Ă— đ?œ‹ = đ?œ‹ 2 . đ??´đ?‘Łđ?‘’đ?‘&#x;đ?‘Žđ?‘”đ?‘’ đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’ đ?‘œđ?‘“ đ?‘“ đ?‘œđ?‘Łđ?‘’đ?‘&#x; đ?‘… = Average value of f over R= VII.
1 đ?œ‹2
1 đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘…
x
Figure 6. đ?‘‘đ??´.
Area,A=
đ?‘…
Ď€ 2
1+cos θ r dr dθ 1 π 1+cos θ 2 r 2 dθ 0 2 1
=2
Ă— 0 = 0.
DOUBLE INTEGRAL IN POLAR FORM
Area differential in polar coordinates,
dA.
R
0
=2 π 2
=
0
=
0
Ď€ 2
1 + 2 cos θ + cos 2 θ − 1 dθ 2cosθ +
đ?‘‘đ??´ = đ?‘&#x; đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ.
θ
1+cos 2θ
= 2sinθ + + 2
đ?‘&#x;2
đ?œƒ2
đ?‘“(đ?‘&#x;, đ?œƒ)đ?‘‘đ??´ = đ?‘…
đ?‘&#x;1
=
A.Example The area enclosed by the one leaf of the rose Ď€ Ď€ r = 2 cos 3θ Over the region R:− ≤ θ ≤ , can be 6
6
stated. y x
0
Figure 5. Area,A= = = =
R π 6
− Ď€ 6
− 1
dA. π 6
r2 π 6 π 6
r dr dθ
2 cos 3θ
dθ
2 0
4(cos 3θ)2 dθ
π 6 π 1+cos 6θ 6
2 −
=4
2 cos 3θ 0
0
2 π
=2 θ+
sin 6θ 6 6 0
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dθ
dθ
π sin 2θ 2
4
0
Ď€
2+ +0 −0
=
đ?‘“ đ?‘&#x;, đ?œƒ đ?‘&#x;đ?‘‘đ?‘&#x; đ?‘‘đ?œƒ đ?œƒ1
2
4
8+Ď€ 4
. CONCLUSION
There are many different rules for mathematics researcher. Among them, integration method is very important method. Especially for engineerings because they are compute in imaginary for their design that is, volume of system and area of design for their practical life. Define and evaluate double integrals over bounded regions in the plane which are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that of determining the limits of integration. By using the integration method: integrations, double integrations and triple integrations, computed for solving the problems in daily life. ACKNOWLEDGMENT Firstly, I would like to thank my senior in the department that all of my duty region. I want to thank
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© OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880 all of my teacher for their teaching, kindness and her urge wrote this research paper well.I am also indebted to all the teachers at my department who permitted me and friends who gives me all kind of encouragement and necessary help to write this research paper. I wish to report my family for their helping. To get a journal provide all of teachers and all of my friends. REFERENCES [1] George B.Thomas, Jr., ‘Thamos calculus early transcendentals’, America, vol-12, 1914. [2] Erwin Kreyszig,’Advance Engineering Mathematics, Wiley International Edition. [3] D.W.Jordan and P.Smith, ‘Nonlinear ordinary differential equations’, Oxford University Press Inc New York, vol-4, 2007.
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