1. Double integrals allow us to calculate the volume of a solid bounded above by a surface z=f(x,y) and below by a plane region R. This is done by setting up a Riemann sum of the volumes of thin rectangular prisms.
2. The limit of the Riemann sum as the number of rectangles goes to infinity gives the double integral, written as ∫∫R f(x,y) dA.
3. Example problems demonstrate calculating the double integral to find the volume in different plane regions R, including those bounded by curves such as y=x2 and y=x.
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
The process of finding area of some plane region is called Quadrature. In this chapter we shall find the area bounded by some simple plane curves with the help of definite integral. For solving
(i) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by,
the problems on quadrature easily, if possible first draw the rough sketch of the required area.
b
Area = y dx
a
b
= f(x) dx
a
In chapter function, we have seen graphs of some simple elementary curves. Here we introduce some essential steps for curve tracing which will enable us to determine the required area.
(i) Symmetry
The curve f(x, y) = 0 is symmetrical
about x-axis if all terms of y contain even powers.
about y-axis if all terms of x contain even
powers.
about the origin if f (– x, – y) = f (x, y).
Examples
based on
Area Bounded by A Curve
For example, y2 = 4ax is symmetrical about x-axis, and x2 = 4ay is symmetrical about y- axis and the curve y = x3 is symmetrical about the origin.
(ii) Origin
Ex.1 Find the area bounded by the curve y = x3, x-axis and ordinates x = 1 and x = 2.
2
Sol. Required Area = ydx
1
If the equation of the curve contains no constant
2
= x3 dx =
Lx4 O2 15
MN PQ=
Ans.
term then it passes through the origin.
For example x2 + y2 + 2ax = 0 passes through origin.
(iii) Points of intersection with the axes
If we get real values of x on putting y = 0 in the equation of the curve, then real values of x and y = 0 give those points where the curve cuts the x-axis. Similarly by putting x = 0, we can get the points of intersection of the curve and y-axis.
1 4 1 4
Ex.2 Find the area bounded by the curve y = sec2x,
x-axis and the line x = 4
/4
Sol. Required Area = ydx
x0
For example, the curve x2/a2 + y2 /b2 = 1 intersects the axes at points (± a, 0) and (0, ± b) .
(iv) Region
/4
= sec2
0
xdx = tan x /4 = 1 Ans.
Write the given equation as y = f(x) , and find minimum and maximum values of x which determine the region of the curve.
For example for the curve xy2 = a2 (a – x)
Ex.3 Find the area bounded by the curve y = mx, x-axis and ordinates x = 1 and x = 2
2
Sol. Required area = y dx .
1
y = a
a x
x
2
= mxdx =
LMmx2 2
Now y is real, if 0 < x a , so its region lies between the lines x = 0 and x = a.
1 MN2
= m ( 4 – 1) =
2
PQ1
FGH3 JKm Ans.
Ex.4 Find the area bounded by the curve y = x (1– x)2 and x-axis.
Sol. Clearly the given curve meets the x-axis at (0,0) and (1,0) and for x = 0 to 1, y is positive
Ex.7 Find the area bounded between the curve y2 = 2y – x and y-axis.
Sol. The area between the given curve x = 2y – y2 and y-axis will be as shown in diagram.
so required area- Y
= xb1 xg2 dx
0
1
= ex 2x2 x3 jdx
0
O (0,0) (1,0) X
Lx2
2x3
x4 O1
1 2 1 1
= M P=
– + =
MN2 3 4 PQ0
2 3 4 12
Ans.
Required Area =
e2y y2 jdy
(i
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
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
The process of finding area of some plane region is called Quadrature. In this chapter we shall find the area bounded by some simple plane curves with the help of definite integral. For solving
(i) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by,
the problems on quadrature easily, if possible first draw the rough sketch of the required area.
b
Area = y dx
a
b
= f(x) dx
a
In chapter function, we have seen graphs of some simple elementary curves. Here we introduce some essential steps for curve tracing which will enable us to determine the required area.
(i) Symmetry
The curve f(x, y) = 0 is symmetrical
about x-axis if all terms of y contain even powers.
about y-axis if all terms of x contain even
powers.
about the origin if f (– x, – y) = f (x, y).
Examples
based on
Area Bounded by A Curve
For example, y2 = 4ax is symmetrical about x-axis, and x2 = 4ay is symmetrical about y- axis and the curve y = x3 is symmetrical about the origin.
(ii) Origin
Ex.1 Find the area bounded by the curve y = x3, x-axis and ordinates x = 1 and x = 2.
2
Sol. Required Area = ydx
1
If the equation of the curve contains no constant
2
= x3 dx =
Lx4 O2 15
MN PQ=
Ans.
term then it passes through the origin.
For example x2 + y2 + 2ax = 0 passes through origin.
(iii) Points of intersection with the axes
If we get real values of x on putting y = 0 in the equation of the curve, then real values of x and y = 0 give those points where the curve cuts the x-axis. Similarly by putting x = 0, we can get the points of intersection of the curve and y-axis.
1 4 1 4
Ex.2 Find the area bounded by the curve y = sec2x,
x-axis and the line x = 4
/4
Sol. Required Area = ydx
x0
For example, the curve x2/a2 + y2 /b2 = 1 intersects the axes at points (± a, 0) and (0, ± b) .
(iv) Region
/4
= sec2
0
xdx = tan x /4 = 1 Ans.
Write the given equation as y = f(x) , and find minimum and maximum values of x which determine the region of the curve.
For example for the curve xy2 = a2 (a – x)
Ex.3 Find the area bounded by the curve y = mx, x-axis and ordinates x = 1 and x = 2
2
Sol. Required area = y dx .
1
y = a
a x
x
2
= mxdx =
LMmx2 2
Now y is real, if 0 < x a , so its region lies between the lines x = 0 and x = a.
1 MN2
= m ( 4 – 1) =
2
PQ1
FGH3 JKm Ans.
Ex.4 Find the area bounded by the curve y = x (1– x)2 and x-axis.
Sol. Clearly the given curve meets the x-axis at (0,0) and (1,0) and for x = 0 to 1, y is positive
Ex.7 Find the area bounded between the curve y2 = 2y – x and y-axis.
Sol. The area between the given curve x = 2y – y2 and y-axis will be as shown in diagram.
so required area- Y
= xb1 xg2 dx
0
1
= ex 2x2 x3 jdx
0
O (0,0) (1,0) X
Lx2
2x3
x4 O1
1 2 1 1
= M P=
– + =
MN2 3 4 PQ0
2 3 4 12
Ans.
Required Area =
e2y y2 jdy
(i
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Automobile Management System Project Report.pdfKamal Acharya
The proposed project is developed to manage the automobile in the automobile dealer company. The main module in this project is login, automobile management, customer management, sales, complaints and reports. The first module is the login. The automobile showroom owner should login to the project for usage. The username and password are verified and if it is correct, next form opens. If the username and password are not correct, it shows the error message.
When a customer search for a automobile, if the automobile is available, they will be taken to a page that shows the details of the automobile including automobile name, automobile ID, quantity, price etc. “Automobile Management System” is useful for maintaining automobiles, customers effectively and hence helps for establishing good relation between customer and automobile organization. It contains various customized modules for effectively maintaining automobiles and stock information accurately and safely.
When the automobile is sold to the customer, stock will be reduced automatically. When a new purchase is made, stock will be increased automatically. While selecting automobiles for sale, the proposed software will automatically check for total number of available stock of that particular item, if the total stock of that particular item is less than 5, software will notify the user to purchase the particular item.
Also when the user tries to sale items which are not in stock, the system will prompt the user that the stock is not enough. Customers of this system can search for a automobile; can purchase a automobile easily by selecting fast. On the other hand the stock of automobiles can be maintained perfectly by the automobile shop manager overcoming the drawbacks of existing system.
Event Management System Vb Net Project Report.pdfKamal Acharya
In present era, the scopes of information technology growing with a very fast .We do not see any are untouched from this industry. The scope of information technology has become wider includes: Business and industry. Household Business, Communication, Education, Entertainment, Science, Medicine, Engineering, Distance Learning, Weather Forecasting. Carrier Searching and so on.
My project named “Event Management System” is software that store and maintained all events coordinated in college. It also helpful to print related reports. My project will help to record the events coordinated by faculties with their Name, Event subject, date & details in an efficient & effective ways.
In my system we have to make a system by which a user can record all events coordinated by a particular faculty. In our proposed system some more featured are added which differs it from the existing system such as security.
Quality defects in TMT Bars, Possible causes and Potential Solutions.PrashantGoswami42
Maintaining high-quality standards in the production of TMT bars is crucial for ensuring structural integrity in construction. Addressing common defects through careful monitoring, standardized processes, and advanced technology can significantly improve the quality of TMT bars. Continuous training and adherence to quality control measures will also play a pivotal role in minimizing these defects.
Courier management system project report.pdfKamal Acharya
It is now-a-days very important for the people to send or receive articles like imported furniture, electronic items, gifts, business goods and the like. People depend vastly on different transport systems which mostly use the manual way of receiving and delivering the articles. There is no way to track the articles till they are received and there is no way to let the customer know what happened in transit, once he booked some articles. In such a situation, we need a system which completely computerizes the cargo activities including time to time tracking of the articles sent. This need is fulfilled by Courier Management System software which is online software for the cargo management people that enables them to receive the goods from a source and send them to a required destination and track their status from time to time.
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
Democratizing Fuzzing at Scale by Abhishek Aryaabh.arya
Presented at NUS: Fuzzing and Software Security Summer School 2024
This keynote talks about the democratization of fuzzing at scale, highlighting the collaboration between open source communities, academia, and industry to advance the field of fuzzing. It delves into the history of fuzzing, the development of scalable fuzzing platforms, and the empowerment of community-driven research. The talk will further discuss recent advancements leveraging AI/ML and offer insights into the future evolution of the fuzzing landscape.
TECHNICAL TRAINING MANUAL GENERAL FAMILIARIZATION COURSEDuvanRamosGarzon1
AIRCRAFT GENERAL
The Single Aisle is the most advanced family aircraft in service today, with fly-by-wire flight controls.
The A318, A319, A320 and A321 are twin-engine subsonic medium range aircraft.
The family offers a choice of engines
2. You may recall that we can do a Riemann sum to
approximate the area under the graph of a function
of one variable by adding the areas of the
rectangles that form below the graph resulting from
small increments of x (x) within a given interval
[a, b]:
x
y
y= f(x)
a b
x
3. z
Similarly, it is possible to obtain an approximation of
the volume of the solid under the graph of a
function of two variables.
y
x
a
b
c d
R
y= f(x, y)
4. a. Suppose g1(x) and g2(x) are continuous functions on [a, b] and
the region R is defined by R = {(x, y)| g1(x) y g2(x); a x b}.
Then,
2
1
( )
( )
( , ) ( , )
b g x
R a g x
f x y dA f x y dy dx
x
y
a b
y = g1(x)
y = g2(x)
R
5. Let R be a region in the xy-plane and let f
be continuous and nonnegative on R.
Then, the volume of the solid under a
surface bounded above by z = f(x, y) and
below by R is given by
( , )
R
V f x y dA
6. z
To find the volume of the solid under the surface, we
can perform a Riemann sum of the volume Si of
parallelepipeds with base Ri = x ☓ y and height
f(xi, yi):
y
x
a
b
c d
x
y
z = f(x,
y)
R
7. z
To find the volume of the solid under the surface, we
can perform a Riemann sum of the volume Si of
parallelepipeds with base Ri = x ☓ y and height
f(xi, yi):
y
x
a
b
c d
x
y
Si
z = f(x,
y)
R
8. y
a
b
c d
z
x
To find the volume of the solid under the surface, we
can perform a Riemann sum of the volume Si of
parallelepipeds with base Ri = x ☓ y and height
f(xi, yi):
z = f(x,
y)
9. z
The limit of the Riemann sum obtained when the number of
rectangles m along the x-axis, and the number of
subdivisions n along the y-axis tends to infinity is the value
of the double integral of f(x, y) over the region R and is
denoted by
z = f(x,
y)
y
x
a
b
c d
x
y
R
( , )
R
f x y dA
y ·n
x
·m
10. b. Suppose h1(y) and h2(y) are continuous functions on [c, d]
and the region R is defined by R = {(x, y)| h1(y) x
h2(y); c y d}.
Then, 2
1
( )
( )
( , ) ( , )
d h y
R c h y
f x y dA f x y dx dy
x
y
c
d
x = h1(y) x = h2(y)
R
11. 4
3
2
1
1 2 3 4
Evaluate ∫R∫f(x, y)dA given that f(x, y) = x2 + y2 and
R is the region bounded by the graphs of g1(x) = x
and g2(x) = 2x for 0 x 2.
Solution
The region under consideration is:
x
g2(x) = 2x
R
y
g1(x) = x
Example 2, page 593
12. Evaluate ∫R∫f(x, y)dA given that f(x, y) = x2 + y2 and
R is the region bounded by the graphs of g1(x) = x
and g2(x) = 2x for 0 x 2.
Solution
Using Theorem 1, we find:
2 2
2 2
0
( , ) ( )
x
R x
f x y dA x y dy dx
2
2
2 3
0
1
3
x
x
x y y dx
2
3 3 3 3
0
8 1
2
3 3
x x x x dx
2
3
0
10
3
x dx
2
4
0
5
6
x
1
3
13
Example 2, page 593
13. 1
1
Evaluate ∫R∫f(x, y)dA, where f(x, y) = xey and R is the
plane region bounded by the graphs of y = x2 and y = x.
Solution
The region under consideration is:
x
g1(x) = x2
R
y
g2(x) = x
The points of intersection of
the two curves are found by
solving the equation x2 = x,
giving x = 0 and x = 1.
Example 3, page 593
14. Evaluate ∫R∫f(x, y)dA, where f(x, y) = xey and R is
the plane region bounded by the graphs of y = x2
and y = x.
Solution
Using Theorem 1, we find:
2
1
0
( , )
x
y
R x
f x y dA xe dy dx
2
1
0
x
y
x
xe dx
2
1
0
( )
x x
xe xe dx
2
1 1
0 0
x x
xe dx xe dx
2
1
0
1
( 1)
2
x x
x e e
1 1 1
1 (3 )
2 2 2
e e
Integrating by parts on
the right-hand side
Example 3, page 593
15. Find the volume of the solid bounded above by the
plane z = f(x, y) = y and below by the plane
region R defined by
Solution
The graph of the region R is:
1
1
x
y
2
1
y x
R
Observe that f(x, y) = y > 0 for (x, y) ∈ R.
2
1 (0 1)
y x x
Example 4, page 594
16. Find the volume of the solid bounded above by the
plane z = f(x, y) = y and below by the plane
region R defined by
Solution
Therefore, the required volume is given by
2
1 1
0 0
x
R
V ydA ydy dx
2
1
1
2
0
0
1
2
x
y dx
1
2
0
1
(1 )
2
x dx
1
3
0
1 1
2 3
x x
1
3
2
1 (0 1)
y x x
Example 4, page 594
17. Find the volume of the solid bounded above by the
plane z = f(x, y) = y and below by the plane region
R defined by
Solution
2
1 (0 1)
y x x
z
x
y
The graph of the solid
in question is:
2
1
y x
R
z = f(x, y) = y
Example 4, page 594