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ACCA Unit 1 Double and Triple Integrals Class Notes

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Course
18MAB102T

Institution
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY

Class notes for Course 18MAB102T - Advanced Calculus and Complex Analysis, Unit 1, focusing on Double and Triple Integrals. Designed for students seeking a comprehensive understanding of these fundamental concepts, our notes offer a clear and concise explanation of key topics including the theory...

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Uploaded on August 22, 2024
Number of pages 22
Written in 2019/2020
Type Class notes
Professor(s) Dr. anirban majumdar assistant professor
Contains All classes

class notes
class notes of unit 1 and 2 of course acca
notes
double and triple integrals
double and triple integrals class notes
vector calculus class note

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SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
Course
18MAB102T

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18MAB102T - ADVANCED CALCULUS AND
COMPLEX ANALYSIS
(Unit I - Double and Triple Integrals)

Dr. Anirban Majumdar
Assistant Professor, Department of Mathematics
SRM Institute of Science and Technology, India
Email: anirbanm@srmist.edu.in

April 8, 2021

,Double Integral in Cartesian Coordinate Double Integral in Polar Coordinate Double Integral by Changing Order of Integration

Outline

1 Double Integral in Cartesian Coordinate

2 Double Integral in Polar Coordinate

3 Double Integral by Changing Order of Integration

,Double Integral in Cartesian Coordinate Double Integral in Polar Coordinate Double Integral by Changing Order of Integration

Outline

1 Double Integral in Cartesian Coordinate

2 Double Integral in Polar Coordinate

3 Double Integral by Changing Order of Integration

,Double Integral in Cartesian Coordinate Double Integral in Polar Coordinate Double Integral by Changing Order of Integration

Double integral
Consider a function f (x, y) defined at each point in the finite region R of the xy-plane.
Divide R into n elementary areas ∆A1 , ∆A2 , · · · , ∆An . Let (xk , yk ) be any point
within the k-th elementary area ∆Ak . Consider the sum
n
X
f (x1 , y1 )∆A1 + f (x2 , y2 )∆A2 + · · · + f (xn , yn )∆An = f (xk , yk )∆Ak .
k=1

The limit of this sum, if exists, as the number of subdivision increases indefinitely and
area of each sub-division decreases to zero, is defined as the double integral of f (x, y)
over the region R and it is written as
ZZ ZZ
f (x, y) dA or f (x, y) dx dy.
R R

Therefore,
ZZ n
X
f (x, y) dA = lim f (xk , yk )∆Ak .
n→∞
R k=1

Note: The continuity of f is a sufficient condition for the existence of the double
integral, but not a necessary one. The above limit exists for many discontinuous
function as well.

,Double Integral in Cartesian Coordinate Double Integral in Polar Coordinate Double Integral by Changing Order of Integration

Properties of double integrals
Like single integrals, double integrals of continuous functions have algebraic properties
that are useful in computations.
1 ZZ ZZ
kf (x, y) dA = k f (x, y) dA, for any number k.
R R
2 ZZ ZZ ZZ
(f (x, y) ± g(x, y)) dA = f (x, y) dA ± g(x, y) dA.
R R R
3 ZZ
f (x, y) dA ≥ 0, if f (x, y) ≥ 0 on R.
R
4 ZZ ZZ
f (x, y) dA ≥ g(x, y) dA if f (x, y) ≥ g(x, y) on R.
R R
5 If R is the union of two non-overlapping regions R1 and R2 with boundaries that
are again made of a finite number of line segments or smooth curves, then
ZZ ZZ ZZ
f (x, y) dA = f (x, y) dA f (x, y) dA.
R R1 R2

,Double Integral in Cartesian Coordinate Double Integral in Polar Coordinate Double Integral by Changing Order of Integration

Evaluating double integral on rectangular domain

Theorem (First Form of Fubini’s Theorem)
If f (x, y) is continuous on the rectangular region R : a ≤ x ≤ b, c ≤ y ≤ d, then
ZZ Z bZ d Z dZ b
f (x, y) dA = f (x, y) dy dx = f (x, y) dx dy.
a c c a
R

Example
(1 − 6x2 y) dA where R = {(x, y) : 0 ≤ x ≤ 2, −1 ≤ y ≤ 1}.
RR
Calculate
R

Solution: By Fubini’s theorem,
ZZ Z 1 Z 2
(1 − 6x2 y) dA = (1 − 6x2 y) dx dy
−1 0
R Z 1 h ix=2
= x − 2x3 y dy (by keeping y fixed)
x=0
Z−1
1
= (2 − 16y) dy
h −1 i1
= 2y − 8y 2 = 4.
−1

, Double Integral in Cartesian Coordinate Double Integral in Polar Coordinate Double Integral by Changing Order of Integration

Evaluating double integral on rectangular domain
Note: Reversing the order of integration gives the same answer:
ZZ Z 2 Z 1
(1 − 6x2 y) dA = (1 − 6x2 y) dy dx
0 −1
R Z 2 h iy=1
= y − 3x2 y 2 dx (by keeping x fixed)
0 y=−1
Z 2
= 2 dx
0
h i2
= 2x
0

= 4.

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