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.
This document discusses several topics in calculus of several variables:
- Functions of several variables and their partial derivatives
- Maxima and minima of functions of several variables
- Double integrals and constrained maxima/minima using Lagrange multipliers
It provides examples of computing partial derivatives of functions, interpreting them geometrically, and using partial derivatives to determine rates of change. Level curves are also discussed as a way to sketch graphs of functions of two variables.
Here are the steps to solve this problem:
1. The region R is bounded by the circle r = 1 and the line x + y = 1 in polar coordinates.
2. To set up the double integral in polar coordinates, we first identify the limits of integration with respect to r. Since we are integrating r dr first, we hold θ fixed and let r vary from the minimum to maximum r value along each radial line. The minimum is r = 1/(cosθ + sinθ) and the maximum is r = 1.
3. Next, we identify the limits with respect to θ. The radial lines intersecting the region range from θ = 0 to θ = π/2.
4.
1) The document discusses directional derivatives and the gradient of functions of several variables. It defines the directional derivative Duf(c) as the slope of the function f in the direction of the unit vector u at the point c.
2) It shows that the partial derivatives of f can be computed by treating all but one variable as a constant. The gradient of f is defined as the vector of its partial derivatives.
3) It derives an expression for the directional derivative Duf(c) in terms of the partial derivatives of f and the components of the unit vector u, showing the relationship between directional derivatives and the gradient.
The document discusses calculating the area of a region R. It introduces using a ruler x to measure the span of R from x=a to x=b. It defines the cross-sectional length L(x) and partitions the interval [a,b] into subintervals. The Riemann sum of the areas of approximating rectangles is shown to approach the actual area of R, defined as the definite integral of L(x) from a to b. As an example, it calculates the area between the curves y=-x^2+2x and y=x^2 by finding the interval spans from 0 to 1 and taking the integral of the difference of the functions.
The document discusses Cartesian products, domains, ranges, and co-domains of relations and functions through examples and definitions. It explains that the Cartesian product of sets A and B, written as A×B, is the set of all ordered pairs (a,b) where a is an element of A and b is an element of B. It also defines what constitutes a relation between two sets and provides examples of relations and functions, discussing their domains and ranges. Arrow diagrams are presented to illustrate various functions along with questions and their solutions related to relations and functions.
The document discusses double integrals in polar coordinates. It defines a domain D bounded by functions r1(θ) and r2(θ) between angles A and B. It partitions the domain into tiles of widths Δr and Δθ and approximates the area and volume integrals using sums over the tiles. The volume integral over the domain D of a function z=f(r,θ) is written as a double integral in polar coordinates from r1 to r2 and θ from A to B.
1. This document discusses methods for calculating the length of an arc of a curve and the surface area of revolution. It provides formulas for finding arc length and surface area when curves are defined by rectangular coordinates, parametric equations, or polar coordinates.
2. Several examples are given of applying the formulas to find the arc length of curves and the surface area when graphs are revolved about axes. This includes revolving curves like y=x^3, y=x^2, and xy=2 about the x-axis and y-axis.
3. The key formulas presented are that arc length can be found using an integral of the form ∫√(dx/dy)^2 + 1 dy or
This document introduces vector fields and defines scalar and vector fields. It defines a vector field F over a planar region D as a function that assigns a vector to each point, given by the components P and Q. Similarly, a vector field over a spatial region E is given by components P, Q, and R. Examples of vector fields include velocity fields and gravitational and electric force fields. It also defines the gradient of a scalar field, divergence and curl of a vector field, and discusses integral calculations for scalar and vector fields along curves.
This document provides an overview of key concepts in multivariable calculus including:
- Three-dimensional coordinate systems and vectors in space. Operations on vectors such as addition, scalar multiplication, dot products, and cross products.
- Lines, planes, and quadric surfaces in space. Multiple integrals, integration in vector fields including line integrals, work, and flux.
- Coordinate transformations between rectangular and cylindrical coordinates. Green's theorem and its application to calculating line integrals and surface areas.
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
This document discusses exponential functions and their graphs. It defines the exponential function f(x) = ax and gives examples. It shows how to evaluate exponential functions at different values of x. It explains the graphs of exponential functions with bases greater than and less than 1, and how they have horizontal asymptotes at y = 0. It provides examples of sketching graphs of exponential functions and stating their domains and ranges. It introduces the irrational number e and the natural exponential function f(x) = ex. It concludes with formulas for compound interest and an example problem.
This document provides an overview of double integrals and their use in calculating volumes. It begins by reviewing definite integrals of single-variable functions. It then introduces double integrals, which can be used to calculate the volume of a solid under a surface defined by a two-variable function f(x,y). This is done by dividing the domain into subrectangles and approximating the volume as a double Riemann sum. Several examples are provided to demonstrate calculating volumes and average values over a region using double integrals. The document also discusses double integrals over non-rectangular regions and provides formulas for calculating them using iterated integrals for specific region types.
1) The document discusses partial derivatives, which involve differentiating functions of two or more variables with respect to one variable while holding the others fixed. It provides definitions, examples of computing partial derivatives, and interpretations as rates of change.
2) Techniques covered include implicit differentiation, using the chain rule to find derivatives of implicitly defined functions, and computing second order partial derivatives.
3) Diagrams and tables are referenced to illustrate level curves and contour maps for functions of two variables.
1) The document discusses partial derivatives, which involve differentiating functions of two or more variables with respect to one variable while holding the others constant. It provides examples of computing first and second partial derivatives.
2) Implicit differentiation is introduced as a way to find partial derivatives of functions defined implicitly rather than explicitly. The chain rule is also discussed.
3) Methods are presented for finding partial derivatives of functions of two or three variables, including using implicit differentiation and the chain rule. Examples are provided to illustrate these concepts.
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
This document discusses calculating the area under a curve using integration. It begins by approximating the area under an irregular shape using squares and rectangles. It then introduces defining the area A as a limit of approximating rectangles as their width approaches 0. This is written as the integral from a to b of f(x) dx, where f(x) is the curve. Examples are given of setting up definite integrals to calculate the areas under curves and between two curves. Steps for determining the area of a plane figure using integration are also provided.
The document discusses double integrals and their use in calculating volumes. It explains that double integrals allow calculating the volume of a solid over a domain D by integrating the height function f(x,y) over D. Three methods are provided: integrating cross-sectional areas A(x) or A(y) with respect to x or y; directly using the fundamental theorem of calculus by partitioning D into subrectangles and summing the approximate volumes; and writing the calculation as a double integral of f(x,y) over D.
Superficies regulares planos tangentes y normales EDESMITCRUZ1
This document defines regular surfaces and their parametric representations. It provides examples of parametrizing planes, spheres, cylinders, and other surfaces. It also discusses tangent planes to surfaces, calculating surface areas, and surface integrals including of vector fields using the divergence theorem. The document contains several examples of parametrizing surfaces and calculating related properties like tangent planes and surface areas.
The document discusses key concepts in calculus including:
- Differential calculus examines how quantities change by looking at their rates of change, represented by derivatives.
- Integration is used to determine quantities like material needs or structure weights by calculating the area under a curve.
- Calculus has various applications in fields like engineering, physics, and robotics where quantities change continuously over time.
- The document provides examples of how differential and integral calculus are used in applications such as space travel planning, architecture, and robotics.
Bell Crank Lever.pptxDesign of Bell Crank Leverssuser110cda
In a bell crank lever, the two arms of the lever are at right angles.
Such type of levers are used in railway signalling, governors of Hartnell type, the drive for the air pump of condensers etc.
The bell crank lever is designed in a similar way as discussed earlier.
Structural Dynamics and Earthquake Engineeringtushardatta
Slides are prepared with a lot of text material to help young teachers to teach the course for the first time. This also includes solved problems. This can be used to teach a first course on structural dynamics and earthquake engineering. The lecture notes based on which slides are prepared are available in SCRIBD.
Vijay Engineering and Machinery Company (VEMC) is a leading company in the field of electromechanical engineering products and services, with over 70 years of experience.
The Transformation Risk-Benefit Model of Artificial Intelligence: Balancing R...gerogepatton
This paper summarizes the most cogent advantages and risks associated with Artificial Intelligence from an
in-depth review of the literature. Then the authors synthesize the salient risk-related models currently being
used in AI, technology and business-related scenarios. Next, in view of an updated context of AI along with
theories and models reviewed and expanded constructs, the writers propose a new framework called “The
Transformation Risk-Benefit Model of Artificial Intelligence” to address the increasing fears and levels of
AIrisk. Using the model characteristics, the article emphasizes practical and innovative solutions where
benefitsoutweigh risks and three use cases in healthcare, climate change/environment and cyber security to
illustrate unique interplay of principles, dimensions and processes of this powerful AI transformational
model.
Modified O-RAN 5G Edge Reference Architecture using RNNijwmn
Paper Title
Modified O-RAN 5G Edge Reference Architecture using RNN
Authors
M.V.S Phani Narasimham1 and Y.V.S Sai Pragathi2, 1Wipro Technologies, India, 2Stanley College of Engineering & Technology for Women (Autonomous), India
Abstract
This paper explores the implementation of 6G/5G standards by network providers using cloud-native technologies such as Kubernetes. The primary focus is on proposing algorithms to improve the quality of user parameters for advanced networks like car as cloud and automated guided vehicle. The study involves a survey of AI algorithm modifications suggested by researchers to enhance the 5G and 6G core. Additionally, the paper introduces a modified edge architecture that seamlessly integrates the RNN technologies into O-RAN, aiming to provide end users with optimal performance experiences. The authors propose a selection of cutting-edge technologies to facilitate easy implementation of these modifications by developers.
Keywords
5G O-RAN, 5G-Core, AI Modelling, RNN, Tensor Flow, MEC Host, Edge Applications.
Volume URL: https://airccse.org/journal/jwmn_current24.html
Abstract URL: https://aircconline.com/abstract/ijwmn/v16n3/16324ijwmn01.html
Youtube URL: https://youtu.be/rIYGvf478Oc
Pdf URL: https://aircconline.com/ijwmn/V16N3/16324ijwmn01.pdf
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Here's where you can reach us : ijwmn@airccse.org or ijwmn@aircconline.com
Agricultural Profitability through Resilience: Smallholder Farmers' Strategie...IJAEMSJORNAL
This study investigated the knowledge strategies and coping utilized by smallholder farmers in Guimba, Nueva Ecija to reduce and adjust to the effects of climate change. Smallholder farmers, who are frequently susceptible to climate change, utilize various traditional and innovative methods to strengthen their ability to withstand and recover from these consequences. Based on the results of this study, farmers in Guimba, Nueva Ecija demonstrate a profound comprehension of the adverse weather conditions, such as typhoons, droughts, and excessive rainfall, which they ascribe to climate change. While they have a fundamental understanding of climate change and its effects, their knowledge of scientific intricacies is restricted, indicating a need for information that is particular to the context. Although farmers possess knowledge about climate change, they are not actively engaging in proactive actions to adapt to it. Instead, they rely on reactive coping mechanisms. This highlights the necessity for targeted educational and communicative endeavors to promote the acceptance and implementation of approaches. Furthermore, the absence of available resources poses a significant barrier to achieving successful adaptation, highlighting the importance of pushing for inexpensive and feasible measures for adaptation. Farmers recognize the benefits of agroforestry and have started integrating the growth of fruit trees, particularly mangoes, into their coping techniques.
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