This document provides information about an upcoming mathematics presentation including:
- A list of 12 group members presenting with their names and student IDs
- 16 topics that will be covered in the presentation ranging from distance between points and planes to surface integrals
- Sample problems and solutions for several topics like finding the distance between a point and plane, parallel planes, a point and line in space, and vector valued functions. Formulas and definitions are provided for concepts like velocity, acceleration, tangent vectors, arc length, and curvature.
Basic concepts of integration, definite and indefinite integrals,properties of definite integral, problem based on properties,method of integration, substitution, partial fraction, rational , irrational function integration, integration by parts, reduction formula, improper integral, convergent and divergent of integration
Basic concepts of integration, definite and indefinite integrals,properties of definite integral, problem based on properties,method of integration, substitution, partial fraction, rational , irrational function integration, integration by parts, reduction formula, improper integral, convergent and divergent of integration
Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
A parabola is the locus of a point which moves in such a way that its distance from a fixed point is equal to its perpendicular distance from a fixed straight line.
1.1 Focus : The fixed point is called the focus of the Parabola.
1.2 Directrix : The fixed line is called the directrix of the Parabola.
(focus)
2.3 Vertex : The point of intersection of a parabola and its axis is called the vertex of the Parabola.
NOTE: The vertex is the middle point of the focus and the point of intersection of axis and directrix
2.4 Focal Length (Focal distance) : The distance of any point P (x, y) on the parabola from the focus is called the focal length. i.e.
The focal distance of P = the perpendicular distance of the point P from the directrix.
2.5 Double ordinate : The chord which is perpendicular to the axis of Parabola or parallel to Directrix is called double ordinate of the Parabola.
2.6 Focal chord : Any chord of the parabola passing through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes through the focus of parabola then it is called as latus rectum.
2.7.1 Length of latus rectum :
The length of the latus rectum = 2 x perpendicular distance of focus from the directrix.
2.1 Eccentricity : If P be a point on the parabola and PM and PS are the distances from the directrix and focus S respectively then the ratio PS/PM is called the eccentricity of the Parabola which is denoted by e.
Note: By the definition for the parabola e = 1.
If e > 1 Hyperbola, e = 0 circle, e < 1
ellipse
2.2 Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola.
If we take vertex as the origin, axis as x- axis and distance between vertex and focus as 'a' then equation of the parabola in the simplest form will be-
y2 = 4ax
3.1 Parameters of the Parabola y2 = 4ax
(i) Vertex A (0, 0)
(ii) Focus S (a, 0)
(iii) Directrix x + a = 0
(iv) Axis y = 0 or x– axis
(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 𝑙 then coordinates of end points of this Double ordinate are
𝑙2 𝑙2
, 𝑙
and
, 𝑙 .
4a
4a
3.2 Other standard Parabola :
Equation of Parabola Vertex Axis Focus Directrix Equation of Latus rectum Length of Latus rectum
y2 = 4ax (0, 0) y = 0 (a, 0) x = –a x = a 4a
y2 = – 4ax (0, 0) y = 0 (–a, 0) x = a x = –a 4a
x2 = 4ay (0, 0) x = 0 (0, a) y = a y = a 4a
x2 = – 4ay (0, 0) x = 0 (0, –a) y = a y = –a 4a
Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and directrix is x + y – 2 = 0, then its vertex is (A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
Sol. First we find the equation of axis of parabola
Super-Structural Construction Work of a Six Storied Residential Building.Shariful Haque Robin
In this presentation I will try to uphold about my 3 month’s practicum period experience of the construction work of a six storied residential building on super structural parts.
Superstructure is building parts located above the ground level such as column, beam, floor, wall and roof.
Here I have focused on mainly Beam, Slab, Stair and Column construction process and also the problems I have faced & my contribution to solve those.
Analysis and Design of Structural Components of a Ten Storied RCC Residential...Shariful Haque Robin
This report has been prepared as an integral part of the internship program for the Bachelor of Science in Civil Engineering (BSCE) under the Department of Civil Engineering in IUBAT−International University of Business Agriculture and Technology. The Dynamic Design and Development (DDD) Ltd. nominated as the organization for the practicum while honorable Prof. Dr. Md. Monirul Islam, Chair of the Department of Civil Engineering rendered his kind consent to academically supervise the internship program.
All reinforced concrete beams crack, generally starting at loads well below service level, and possibly even prior to loading due to restrained shrinkage. Flexural cracking due to loads is not only inevitable, but actually necessary for the reinforcement to be used effectively. Prior to the formation of flexural cracks, the steel stress is no more than n times the stress in the adjacent concrete, where n is the modular ratio E5/Ec. For materials common in current practice, n is approximately 8.
Railway engineering is a multi-faceted engineering discipline dealing with the design, construction and operation of all types of railway systems. It encompasses a wide range of engineering discipline as Civil engineering, electrical engineering, mechanical engineering and computer engineering. A great many other engineering sub-disciplines are also called upon to combine and create railroad engineering. The civil engineers plan the railway track, design the alignment and right of way and finally they operate.
Super-Structural Construction Work of a Six Storied Residential BuildingShariful Haque Robin
The internship report in broad-spectrum contains ten chapters in which I try to explain my
three-month experience in my hosting company. The content of all chapters is broadly
explained and it is constructed from the practical basis of the site work ended all months
A natural vibration of the ground or the earth crust produced by forces is called earthquake or seismic forces.
An earthquake is what happens when two blocks of the earth suddenly slip past one another.
If everything were the same, we would have no need of statistics. But, people's heights, ages, etc., do vary. We often need to measure the extent to which scores in a dataset differ from each other. Such a measure is called the dispersion of a distribution.
The type of handpump technology suitable for a particular area depends on the groundwater level, water quality and hydrogeological conditions. There are some areas like the costal belt in the southern part of Bangladesh, where the conventional shallow and deep tubewlls technologies are not successful due to the high salinity. Alternative water supply options are needed for those areas.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
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Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
3. Group Members
Name ID Name ID
Md. Mouuin Uddin 12206003 Md. Raziur Rahman 12206026
Md. Shakkik Zunaed 12206004 Faysal Ahmad 12206030
Kazi Nourjahan Nipun 12206005 Md. Alamgir Hossain 12206035
Abdullah Al Naser 12206006 Abu Hossain Basri 12206036
Md.Rasheduzzaman 12206007 Manik Chandra Roy 12206039
Md. Hassan Shahriar 12206013 Md. Shariful Haque Robin 12206049
Md. Ashraful Islam 12206015 Md. Saifullah 12206051
Hossain Mohammad Jakaria 12206018 Md. Mahmuddujjaman 12206062
Imrana Khaton Eva 12206019 Arun Chandra Acharjee 12206066
Md. Saddam Hussein 12206023 Md. Oliullah Sheik 12206067
Joyantika Saha 12206025 Md. Rabiul Hasan 12206069
Rajesh Chandra Barman 12206071
4. Topic Topic
1. Distance Between Points, Plane and Line 9. Tangent Plane and Normal Line
2. Distance Between Two Parallel Planes 10. Curl of a vector field
3. Distance Between a Point a Line in Space 11. Curl and Divergence
4. Vector Value Function 12. Line Integral
5. Velocity and Acceleration 13. Green’s Theorem
6. Tangent vector 14. Surface Integral
7. Arc Length and Curvature 15. Divergence Theorem
8. The Gradient of a Function of Two Variables 16. Stokes Theorem
6. Theorem: The distance between a plane and point Q
(Not in plane) is, D= 푃푟표푗 푃푄 =
푃푄.푛
푛
, where P is a
point in the plane and n is the normal in the plane.
To find a point in the plane given by, ax+by+cz+d=0
(a≠0), let y=0 and z=0, then from the equation,
ax+d=0, we can conclude that the point
( -
푑
푎
, 0,0 ) lies in the plane and n is the normal to the
plane.
7. Ex: Find the distance the point Q=( 1, 5, -5 ) and the plane
given by 3x–y+2z=6.
soln: We know that, 푛=< 3, -1, 2> is normal to the give
plane. To find a point in the plane, let y=0 , z=0 and obtain
the point P=( 2, 0, 0) the vector,
푃푄=< 1-2, 5-0, -4-0 >
=< -1, 5, -4 >
9. Ex: Find the distance the point Q=(1, 2, 3) and plane given
by 2x-y+z=4.
Soln: We know that 푛=<2, -1, 1> is normal to the given
plane. To find a point in the plane, let y=0, z=0 and obtain
the point P=(2, 0, 0). The vector,
푃푄=<1-2, 2-0, 3-0>
=<-1, 2, 3>
10. Using the distance formula produce
D=
푃푄 . 푛
푛
=
<−1, 2, 3> .<2,−1, 1>
22+(−1)2+12
−2−2+3
=
4+1+1
=
−1
6
=
1
6
=0.408 (Answer)
12. Theorem: We can determine that the distance between
the point Q = (x0, y0, z0) and the plane given by
ax+by+cz+d=0 is
푎푥0+푏푦0+푐푧0+푑
D =
푎2+푏2+푐2
Distance between point and plane where P= (x,y,z) is a
point on the plane and d= - (ax1 + by1 + cz1) or ax1 + by1
+ cz1 +d = 0
13. Ex: Find the distance between the two parallel planes
given by 3x – y + 2z – 6 = 0 and 6x – 2y + 4z + 4= 0
Soln:
Let, y=0, z=0
Q=(2,0,0)
And here we given,
a=6, b= - 2, c= 4, d = 4
14. Distance between two parallel plane is
푎푥0+푏푦D=
0+푐푧0
푎2+푏2+푐2
D=
6∗2 + −2∗0 + 4∗0 +4
62+(−2)2+42
12+4
=
36+4+16
=
16
56
=
16
56
=2.138 (Answer)
15. Ex: Find the distance between the two parallel planes
given by x – 3y + 4z =10 , x – 3y + 4z -6 =0.
Soln: Let, y=0 and z=0
Q=(10,0,0)
And here we given
X= 1, y= -3, z= 4, d= -6
18. Theorem: The distance between a point Q and a
푃푄×푈
line in space is given by D=
푈
,where 푈 is
the directional vector for line and P is a point on
the line.
19. Ex : Find the distance between the point P= (3,−1,4) and
the line given by x =−2+3t , y = −2t, z = 1+4t.
푆표푙푛: Using the direction number 3,−2, 4
We know that the direction vector for the line is
푈=< 3, −2, 4 >
To find a point on the line , let t= 0 and we obtain P = (−2 , 0, 1)
point on the line. Then 푃푄=< 3− (−2) , −1−0 , 4−1 >
=< 5, −1 ,3 > and we can form
22. A function of the form
r(t)= f(t) 푖 +g(t) 푗
and
r(t)= f(t) 푖 +g(t) 푗 + h(t)푘 is a vector valued function,
where the component of f, g, h are real valued
function of the parameter t,
r(t)= 푓 푡 , 푔(푡)
or
r(t)= 푓 푡 , 푔 푡 , ℎ(푡)
23. Differention of a vector valued function:
Defn: The derivative of a vector valued function r is
defined by
r´(t)= lim
Δ푡→0
푟 푡+Δ푡 −푟(푡)
Δ푡
for all t for which the limit exists. If r´(c) exists for all
c in open interval I, then r is differentiable on the
interval I. Differentiability of vector valued intervals
by considering one side limits.
24. Ex: Find the derivative of each of the
following vector-valued function
(i) r(t)= t2 푖 4 푗
Soln:
푑푟
/ r´(t)=2t 푖
푑푡
(ii) r(t)=
1
푡
푖 + lnt 푗+℮2t 푘
Soln:
푑푟
=
푑푡
1
푡2 푖 +
1
푡
푗+ 2℮2t 푘
26. Defn: If x and y are twice differentiable functions of t
and r is a vector-valued function given by r(t)= x(t) 푖 +
y(t) 푗, then the velocity vector , acceleration vector , and
speed at time t are follows
Velocity = 푣 푡 = r´(t)= x´(t) 푖 + y´(t) 푗
Acceleration n= a(t)= r´´(t)= x´´(t) 푖 + y´´(t) 푗
Speed = 푣 푡 = r´(t) = [x´(t)]2+[y´(t)]2
30. Defn : Let C be a smooth curve represented by r on
an open interval I . The unit tangent vector T(t) at t
is Defined to be
T(t)=
푟́(푡)
푟́(푡)
, 푟́(푡) ≠ 0
31. Ex: Find the unit tangent vector to the curve given by
r(t)=t 푖 +푡2 푗 when t=1
Slon : Given
r(t)=t 푖 +푡2 푗
푟́(푡)= 푖 + 2t 푗
푟́(푡) = 12+(2푡)2
= 1 + 4푡2
32. The tangen vector is
T(t)=
푟́(푡)
푟́(푡)
=
1
1+4푡2 (푖 + 2t 푗 )
The unit tangent vector at t=1 is
T(1) =
1
5
(푖 + 2 푗 )
(Answer)
34. Definition of Arc Length:
The arc length of a smooth plane curve c by the
parametric equations x=x(t) and y=(t), a ≤ t ≤ b is
푏
S= 푎
[푥′(푡)]2+[푦′(푡)]2 dt . Where c is given by
r(t)= x(t)î + y(t)ĵ . We can rewrite this equation for Arc
푏
Length as, s= 푟′ 푡 푑푡
푎
35. Ex: Find the arc length of the curve given by
r(t)=cost î + sint ĵ from t=0 to t=2π
Solution:
Here, x(t)=cost y(t)= sint
X’(t)=-sint y’(t)=cost
We have
푏
S= 푎
[푥′(푡)]2+[푦′(푡)]2 dt
37. Curvature: let c be a smooth curve (in the plane or in
space) given by r(s), where s is the arc length
parameter. The curvature at s is given by
푑푇
푑푠
K=||
||
K=||T’(s)||
38. Ex: Find the curvature of the line given by r(s)= (3-
3
5
s) î +
4
5
s ĵ
Solution:
3
5
r(s) = (3-
s) î +
4
5
s ĵ
r’(s) = ( 0-
3
5
)î +
4
5
ĵ
||r’(s)|| = (−
3
5
)2+(
4
5
)2
=
9
25
+
16
25
41. Definition :
Let Z=f(x,y) be a function of x and y such that 푓푥
and 푓푦 exit. Then the gradient of f, denoted by
훻푓( 푥, 푦 ) ,is the vector of
훻푓 푥, 푦 = 푓푥( x,y )푖 + 푓푦 ( x,y) 푗
We read 훻푓 푎푠 ′′푑푒푙푓′′. Another notation for the
gradient is grad 푓 푥, 푦 .
42. Ex:
Find the gradient of 푓 푥, 푦 = 푦 푙푛 푥 + 푥푦2 at
the point (1,2).
푺풐풍풏:
Grad f or 훻푓 = 푓푥 ( x,y )푖 + 푓푦 ( x,y)
푗 …………………………(i)
Now,
푓푥=
푦
푥
+ 푦2 , 푓푦 = 푙푛 푥 + 2푥푦
∴ 푓푥( 1,2)= 2 + 22 = 6
45. Definition: Let F be differentiable at the point P= (푥0 ,푦0 ,푧0 ) u the surface s
given by F(x, y, z) = 0 such that 훻퐹 푥0 ,푦0 ,푧0 ≠ 0.
1). The plane through P that is normal to 훻퐹 푥0 ,푦0 ,푧0 is called the tangent
plane to S at P.
2). The line through P having the direction of 훻퐹 푥0 ,푦0 ,푧0 is called the normal
line to S at P.
If F is differentiable at 푥0 ,푦0 ,푧0 then an equation of the tangent plane to the
surface given by
퐹 푥, 푦, 푧 = 0 at 푥0 ,푦0 ,푧0 is
퐹푥 푥0 ,푦0 ,푧0 (x−푥0)+ 퐹푦 푥0 ,푦0 ,푧0 (y−푦0) +퐹푧 푥0 ,푦0 ,푧0 (z−푧0)= 0.
46. Ex :
Find an equation of the tangent plane to the hyperboloid given by
at 푧2−2푥2 −
2푦2 = 12 the point (1,−1, 4) .
푺풐풍풏:
An the equation of the tangent plane to the Given hyperboloid
can be rewrite
as 퐹 푥, 푦, 푧 = 12 − 푧2 +2푥2 + 2푦2
53. Problem:
Find the curl F of the followings
F(x, y, z) = 푒−푥푦푧(푖 + 푗 + 푘 ) and point (3,2,0)
Solution:
F(x, y, z) = 푒−푥푦푧(푖 + 푗 + 푘 )
F x, y, z = 푒−푥푦푧푖 + 푒−푥푦푧푗 + 푒−푥푦푧푘
55. Divergence:
div F = 훻. 퐹(푥, 푦, 푧)
div F=
휕푀
휕푥
+
휕푁
휕푦
+
휕푃
휕푧
where F(x,y,z) = M푖 + 푁푗 + 푃푘
If div F=0, then it is said to be divergence free.
Relation between curl & divergence is div(curlF)=0.
56. Problem:
Find the divergence of 퐹 푥, 푦, 푧 = 푥푦푧 푖 − 푦푗 + 푧푘 and also find the
relation between curl & divergence.
Solution:
Given that
퐹 푥, 푦, 푧 = 푥푦푧 푖 − 푦푗 + 푧푘
푐푢푟푙 퐹 푥, 푦, 푧 = 훻 × 퐹 푥, 푦, 푧
=
푖 푗 푘
휕
휕
휕푥
휕푦
휕
휕푧
푥푦푧 −푦 푧
60. De푓푛 of line integral
If f is defined in a region containing a sooth curve c of finite
length ,then the line integral off along c is given by
푐
푓 푥, 푦 푑푠 = lim
Δ →0
푛
푖=1
푓 푥푖, 푦푖 Δ푠푖 푝푙푎푛푒 표푟
푐
푓 푥, 푦, 푧 푑푠 = lim
Δ →0
푛
푖=1
푓 푥푖, 푦푖, 푧푖 Δ푠푖 푝푙푎푛푒
푠푝푎푐푒, 푝푟표푣푖푑푒푑 푡ℎ푒 푙푖푚푖푡 푒푥푖푠푡푠
61. To evaluate the line integral over a plane curve c given by
r(t)=x(t)i + y(t)j, use the fate that
ds= 푟′(푡) 푑푡
= 푥′(푡) 2 + 푦′(푡) 2 dt
A similar formula holds for a space curve as indicated in
the following theorem.
62. Evaluation of line integral as a definite integral :
Let f be continuous in a region containing a smooth curve c .If c is
given by r(t)=x(t)i + y(t)j, where ,a st sd, then.
푐
푓 푥, 푦 푑푠 =
푏
푓(푥 푡 , 푦 푡 ) (푥′ 푡 + 푦 푡 )2
푎
If c is given by
r(t)=x(t)i + y(t)j + z(t)k where a≤ 푡 ≤ 푏 푡ℎ푒푛,
푏
푓(푥, 푦, 푧) ((푥′ 푡 )2+(푦′ 푡 )2 + (푧′(푡))2
푐 푓 푥, 푦, 푧 푑푠 = 푎
67. Let R be a simply connected region with a piecewise
smooth bounding c oriented counterclockwise (that is , c
is traversed once so that the region R always lies to the
left). If M and N have continuous partial derivatives in an
open region containing R , then
푐 푀푑푥 + 푁푑푦 = 푅 (
휕푁
휕푥
−
휕푀
휕푦
)푑퐴
68. Ex : Evaluate 푐 (푡푎푛−1푥 + 푦2)푑푥 + (푒푦 − 푥2)푑푦
Where c is the pat enclosing the annular region.
푺풐풍풏: In polar coordinate
dA=rdrd휃
1≤r≤3 0≤ 휃 ≤ 휋
Here, M = (푡푎푛−1푥 + 푦2 ) , N = 푒푦 − 푥2
휕푀
휕푦
= 2푦
휕푁
휕푥
= −2x
69. x =rcos휃 , y =rsin휃
휕푁
휕푥
−
휕푀
휕푦
= −2x−2y
= −2(x+y)
= −2(rcos휃 + rsin휃)
= −2r(cos휃 + sin휃)
According to the green theorem,
푐 (푡푎푛−1푥 + 푦2)푑푥 + (푒푦 − 푥2)푑푦 = 푅 (
휕푁
휕푥
−
휕푀
휕푦
)푑퐴
73. Theorem: Let, S be a surface equation z = g(x,y) and R its projection on the xy -
plane. If g , 푔푥and 푔푦 are continuous on R and f is continuous on S, then the surface
integral of f over S
is 푆 푓 푥, 푦, 푧 푑푠 = 푅 푓 푥, 푦, 푔 푥, 푦 . 1 + [푔푥(푥, 푦)]2 + [푔푦 (푥, 푦)]2 푑퐴
If, S is the graph of y = g(x,z) and R is its projection on the xz - plane, then
푆
푓 푥, 푦, 푧 푑푠 =
푅
푓 푥, 푧, 푔 푥, 푧 . 1 + [푔푥(푥, 푧)]2 + [푔푧(푥, 푧)]2 푑퐴
If, S is the graph of x = g(y,z) and R is its projection on the yz - plane, then
푆
푓 푥, 푦, 푧 푑푠 =
푅
푓 푦, 푧, 푔 푦, 푧 . 1 + [푔푦(푦, 푧)]2 + [푔푧(푦, 푧)]2 푑퐴
74. Example: Evaluate the surface integral,
푆 (푦2 + 2푦푧)푑푠 , where S is the first-octant portion of the
2x + y + 2z = 6 .
Solution:
We can re-write as,
2z = 6 − 2푥 − 푦 or, z =
1
2
6 − 2푥 − 푦 or, g(x,y) =
1
2
(6 − 2푥 − 푦)
∴ 푔푥 푥, 푦 = −1
푔푦 푥, 푦 = −
1
2
81. Theorem:
Let, Q be a solid region bounded by a
closed surface S oriented by a unit normal vector
directed outward from Q. If f is a vector field
whose component functions have partial
derivatives in Q, then
푆
퐹. 푁푑푠 =
푄
푑푖푣 퐹. 푑푣
82. Example :
Let, Q be the solid region bounded by the co-ordinate planes and
the plane 2x + 2y + z = 6 and let, x i +푦2 푗 + 푍 푘 . Find
푆
퐹 . 푁푑푠 .
Where, S is the surface of Q.
83. Solution :
Here, M = x ∴
휕푀
휕푥
= 1
N = 푦2 ∴
휕푁
휕푦
= 2푦
P = Z ∴
휕푃
휕푧
= 1
∴ 푑푖푣 퐹 =
휕푀
휕푥
+
휕푁
휕푦
+
휕푃
휕푧
= 1 + 2푦 + 1
= 2 + 2푦
89. Theorem: Let Q be a solid region bounded by a closed
surface S oriented by a unit normal vector directed
outward from Q . If F is a vector field whose
component functions have conditions partial derivative
in Q . Then,
퐹. 푁 푑푠 = 푑푖푣 퐹. 푑푣
90. Example: Let Q be a solid region bounded by the co-ordinate plane and
the plane 2x+2y+z=6 and let,
F= x푖 +푦2푗 +z푘 Find 퐹.푁 푑푠 where S is the surface of Q .
Solution:
M=x
휕푀
휕푥
= 1
N=푦2 휕푁
휕푦
= 2y
P=z
휕푃
휕푧
=1
div F = 1+2y+1
=2+2y
94. Theorem: Let s be an oriented surface with unit normal
vector n, bounded by a piecewise smoth, simple closed
curve c.
If F is a vector field whose component function have
continuous partial derivatives on an open region
containing S and C then we can write according to stocks
theorem
푐 퐹. 푑푟 = 푠 푐푢푟푙 퐹 . 푛. 푑푠
95. Example Let C be the oriented triangle ying in the plane 2x+3y+z=6 as
shown in evaluating 푐 퐹 푑푟 when
E(x,y,z) = -y2 i + z j +x k =2x+2y+
Curl F =
푖 푗 푘
푑
푑푥
푑
푑푦
푑
푑푧
−푦2 푧 푥
= i(-1) -j (1) -1k (0+2y)
Consider z= 6-2x-2y=g(x,y)
96. We have 푠 퐹. 푁푑푠 = 푅 퐹. −푔푥 푥, 푦 푖 − 푔푦 푥, 푦 푗 + 푘 푗퐴
For an upcurved normal vector tobe obtain here
푔푥= -2
푔푦 =-2
N=(2i+2j+k)
Then 푠 퐹. 푁푑푠 = 푅 ( −푖 − 푗 + 2푦푘 . (2푖 + 2푗 + 푘))푗퐴
3
= 0
3−푥
0
−2 − 2 + 2푦 푑푦푑푥