chapter-2 : integral calculus

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Chapter-2 : Integral Calculus

2.1 INTRODUCTIONIn this chapter we study the integral of a function of one variable ( )f x over an interval [ , ]a b , of a function

of two variables ( , )f x y over a planer region R, and of a function of three variables ( , , )f x y z over a regionD in space. For each of the integral mentioned above, we study various methods of integration such as themethod of substitutions and the changing of order of integration. We also study geometrical meaning of theintegrals and their applications in finding length of a curve, area of a planer region and volume of a regionin space.

Subtopics in the chapter are:

Single Integral ( )b

a

f x dx :

Definition Geometrical considerations : area of a planer region, length of a curve Fundamental Theorem of Calculus Method of Substitutions

Double Integral ( , )R

f x y dxdy :

Definition Geometrical considerations : area of a planer region and volume enclosed by surfaces Methods of integration :

(a) Iterative integration (Fubini’s theorem)(b) Integration in polar coordinates(c) Method of substitutions

Triple Integral ( , , )D

f x y z dxdydz :

Definition Geometrical considerations : volume enclosed by surfaces Methods of integration:

(a) Iterative integration(b) Integration in cylindrical coordinates(c) Integration in spherical coordinates(d) Method of substitutions



2

2.2 SINGLE INTEGRAL ( )b

a

f x dx

2.2.1 Definition: For a function f defined and bounded on the interval [ , ]a b , the Riemann integral of fon [ , ]a b is defined, if the limit exists, as

max| | 0 1( ) lim ( ) ,

k

b n

k kx ka

f x dx f x

where 0 1 2 .... na x x x x b , 1( 1, 2,3,..., )k k kx x x k n , and k is any real number such

that 1[ , ].k k kx x

If the limit exists, we say that f is Riemann integrable on [ , ]a b and denote it as [ , ]f a b . Riemannintegrable and integrable are considered to be the same.

0 1 2{ , , ..., }nP x x x x is called the partition of [ , ]a b . P partition [ , ]a b into n sub-intervals

1 1 1 2 1[ , ] [ ] [ , ] ... [ ]U U U na b a x x x x b .

If a function f is continuous or piecewise continuous on [ , ]a b , then [ , ]f a b .

Piecewise continuous function: A function f is called piecewise continuous on [ , ]a b if it has finitely

many discontinuities in [ , ]a b and for any ( , )c a b ,

– –lim ( ), lim ( ), lim ( )x a x b x c

f x f x f x

, and lim ( )x c

f x

have finite values. In this chapter we study only continuous and piecewise continuous functions.2.2.2 Some Notations and Terms

( )b

a

f x dxupper limit of integration

lower limit of integrationintegrand

x: variable of integrationintegralsign

Integral of from to :f a b

2.2.3 Geometrical Interpretations: For a function f continuous or piecewise continuous on [ , ]a b

( )b

a

f x dx (signed) area between ( )y f x and x-axis ( )a x b .

Signed area: Area above the x-axis taken to be positive and area below the x-axis taken to benegative.Explanation:

By definition, max| | 0 1

( ) lim ( )k

b n

k kx ka

f x dx f x

.

Also, max | | 0kx

| | 0kx for any k = 1, 2, 3,..., n

y

x0

x1 x2a 1 2

1( f( ))1,

xk–1 xk

kb

( f( )) , k k A fk k = ( )xk

n (number of subintervals goes to )



3

( )k k kA f x = (signed) area of the shaded region as shown in the figure

Clearly, 0kA for ( ) 0kf , and 0kA for ( ) 0kf .Therefore,

max| | 0 1

lim ( )k

n

k kx kf x

1

limn

kn kA

= (signed) area between ( )y f x and x-axis ( )a x b .

(As n , we write ( ) ( )b

ka

A dA f x dx dA n dx (signed) area)

For the following regions :

xO

y (A>0)

(i)

A

( )b

a

f x dx A

xO

y (A, B>0)

(ii)

AxO

y (B>0)

(iii)

( )b

a

f x dx A B ( )b

a

f x dx B

Ba b

B .

(A and B are the areas of the shaded region as shown in the figure)

Example 1. Evaluate 3

0

(1 )x dx using geometrical meaning of the integral.

Solution:3

0

(1 )x dx

= (signed) area of the shaded region= +(area of OAD ) – (area of ABC )

A

1

0–1–2

Dy

y x=1–

2 3 xB

C(–2, 3)

1O

1 11 1 (3 1)(2)2 2

1 22

32

.

Example 2. Evaluate 1/2

2

0

1 x dx using geometrical meaning of the integral.

Solution: Curve: 2 2 21 1 ( 0)y x y x y (Circle)1/2

2

0

1 x dx = (signed) area of the shaded region

= (area of the sector OABO) + (area of OBC ) ... (1) O 12

C1 x

A B

y21y x

area of 21 1 1 1 31

2 2 2 2 8OBC BC OC



4

2112tan 31 3

2

BCBOC BOCOC

2 3 6BOA

area of sector (OABO) = 2 / 612 12

Therefore, from (1), 1/2

2

0

318 12

x dx .


2

0

2 1 x dx using geometrical meaning of the integral.

Solution: Curve: 2

2 222 1 1

2yy x x (Ellipse)

12

0

2 1 x dx = (signed) area of the shaded region x10

yz22 1y x

1 24

(area of ellipse = ab , a and b are semi-major and semi-minor axes)

2

.

2.2.4 Area enclosed by curves:

1. Area A, enclosed by the curves 1( )y g x and 2 ( )y g x between x a and x b :

2 1| ( ) ( ) |b

a

A g x g x dx 2 1[ ( ) ( )]b

a

g x g x dx (if 2 1( ) ( )g x g x for [ , ]x a b ).

y

ab x

y g x= ( )2 y g x= ( )1

O

y

x

y g x= ( )2

y g x= ( )1O ba

y

x

y g x= ( )2

y g x= ( )1

O ba

y

x

y g x= ( )2

y g x= ( )1

O ba

2. Area enclosed by the curves 1( )x h y and 2 ( )x h y between y c and y d :

2 1| ( ) ( ) |d

c

A h y h y dy 2 1[ ( ) ( )]d

c

h y h y dy (if 2 1( ) ( )h y h y for [ , ]y c d ).



5

y

xx h y= ( )2

d

c

x h y= ( )1

y

x

x h y= ( )2

Oc

d

Note: Depending upon the region and integrand one of the two formulas are preferred.2.2.5 Fundamental Theorem of Calculus: If f is continuous on [ , ]a b then there exists a function Fdefined as

( ) ( )x

a

F x f t dt such that ( ) ( ) ( )x

a

dF x f t dt f xdx

for [ , ]x a b .

1. Newton-Leibniz Formula: If f is continuous at every point of [ , ]a b and F is an antiderivative of f on

[ , ]a b , then ( ) ( ) ( )b

a

f x dx F b F a .

2. Antiderivative: F is an antiderivative of f on (a, b) if ( ) ( )F x f x on [ , ]a b . This formula is used to evaluate definite integrals. This formula comes directly from the fundamental theorem of calculus and itself regarded as the fundamental theorem of calculus.3. Leibniz’s Rule: If f is continuous on [ , ]a b and if ( )u x and ( )v x are differentiable functions of x whose

values lie in [ , ]a b , then

( )

( )

( ) ( ( )) ( ( ))v x

u x

d dv duf t dt f v x f u xdx dx dx

.

Leibniz rule can be derived from the fundamental theorem of calculus.


2

2

x dx and 3

0

( cos )xe x dx

.

Solution: 33 3

2

2 23xx dx

3 33 2 19

3 3 3 (using Newton-Leibniz formula)

3 3

00

1( cos ) sin3

x xe x dx e x 31 ( 1)

3e (using Newton-Leibniz formula)

Example 5. Find the area bounded by the parabolas 24y x , 2y x and the straight line y = 2.Solution: The required area is the shaded region in the figure. The region in the first quadrant is area between

x y and 2yx , 0 2y . Similarly, the region in the second quadrant is the area between

x y and 2yx , 0 2y .



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Hence, the area, 2 1| ( ) ( ) |d

c

A h y h y dy

2 2

0 0

( )2 2y y

y dy y dy

Ox

y

y=2

x y 2y

x

2y

x

x y

y x= 2y x=4 2

2 2

0 02 2y y

dy dy

22 2 3/21/2 3/2

0 0 0

22 .22 3 / 2 3y ydy y dy

4 23

.

Note: We found the area by integrating with respect to y. We can also find the area by integrating withrespect to x, but it would involve evaluation of four integrals.

Example 6. Find the area of the region enclosed by the parabola 22y x and the line y x .

Solution: Curves: 22y x and y x

Solving the two equations,

2 22 2 0 ( 2)( 1) 0 1, 2x x x x x x x

Hence, the required area: 2

2

1

| (2 ) ( ) |x x dx

22

1

(2 )x x dx

2(2 ( 2)( 1) 0 for 1 2)x x x x x

23 2

1

923 2 2x xx

.

Note: We get the area without drawing the curves using the formula 2 10

| ( ) ( ) |b

g x g x dx for the area

between the curves 2 ( )y g x and 1( )y g x , a x b .

Alternatively: Let 22 ( ) 2y g x x and 1( )y g x x . The region is as shown in the figure.

Here, 2 1( ) ( )g x g x for [ 1, 2]x .

Hence, integrating with respect to x,2

2 11

[ ( ) ( )]A g x g x dx

x2O–1

(–1, 1)

2 y x g x=2– = ( )22

y x=–= ( )g x12

2

1

(2 )x x dx

92

.

Note: Here, we get the area by drawing the region.



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Example 7. Find the area of the region bounded by the curves 2x y and 3x y .

Solution: Solving the two equations3 2 2 (1 ) 0 0,1y y y y y

intersecting points are (0, 0) and (1, 1). The required area:1

1/3 1/2

0

| |A x x dx (integrating with respect to x)

11/2 1/3

0

( )x x dx 1/2 1/3( for [0,1])x x x

13/2 4/3

03 / 2 4 / 3x x

2 33 4

= 112

.

Alternatively : integrating with respect to y1

2 3

0

| |A y y dy (integrating w.r.t. y)

12 3 2 3

0

( ) ( for [0,1])y y dy y y y 1

12 .

Note: We got the area without drawing the region.Alternatively : By drawing the regionRegion is as shown in the figure.

12 3

0

( )A y y dy (integrating w.r.t. y)

1

12 .

or 1

1/3

0

( )A x x dy (integrating w.r.t. x)O

y x= 1/3

x

y x1

y

1

12 .

Note: Here integrating w.r.t. x or y involves almost same amount of task.

Example 8. Find the area of the region in the first quadrant that is bounded above by y x and below

by the x-axis and the line 2y x .

Solution: Curves: , 2y x y x

2 2 0x x x x

( 2)( 1) 0x x O x

2y

4

y x= –2

y x

2 4x x



8

So, intersecting point (4, 2)The required region is as shown in the figure. As seen from the figure, it is convenient to integratew.r.t. y as it requires only one integral.So, required area,

22

0

[( 2) ( )]A y y dy 22 3

0

8 102 2 42 3 3 3y yy

.

Alternatively: Integrating with respect to x.The required area,

4

2 10

[ ( ) ( )]A g x g x dx O x

2

y

4y x= –2

y x

A

B C

2R1

R2

Here, lower curve 1( )g x changes

For 10 2, ( ) 0x g x and for 12 4, ( ) 2x g x x

So, the required area 1 2A R R (R1, R2 as shown in the figure)2 4

0 2

( 0) ( ( 2))A x dx x x dx 2 4 43/2 3/2 2

0 2 2

23 / 2 3 / 2 2x x x x

4 2 2(8 2 2) 12 4

3 3 2

16 1023 3

.

Note: Here, it was easier to find the area by integrating w.r.t. y.Example 9. Compute the area of the figure bounded by the line 1y x and cosy x and the x-axis.Solution: The required region is as shown in the figure.

As seen from the figure, it seems to be easier to find area by integrating w.r.t. y as it would requireonly one integral-evaluation.

So, the area, 1

1

0

[(cos ) ( 1)]A y y dy

Here, the integral 1

1

0

cos ydy is somewhat difficult to evaluate. x

2O–1

yx=+1

1

y x=cos

Instead, if we integrate w.r.t. x, then0 /2

1 0

(( 1) 0) cosA x dx xdx

1 312 2

.

Note: To find the area the preference should be given to integrating w.r.t. that variable (x or y) which giveseasier integration although it may require subdividing the region into several subregions thus increasingthe number of integrals to evaluate.



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Example 10. Find the area of the region bounded by the line 1,y y x and the parabola 2

4xy in the

first quadrant such that 1y .Solution: The required region is as shown in the figure. As seen from the figure, it is better to find the area

by integrating w.r.t. y, the area,4

1

( 4 )A y y dy 43/2 2

1

23 / 2 2y y

4 15(8 1)3 2

x01 2 4

y x=

4

y= x2

4

18 153 2

116

.

Alternatively:2 4 2

1 2

( 1)4xA x dx x dx

(integrating with respect to x)

2 42 2 3

1 22 2 12x x xx

3 12 64 812 2 12

x1 2 4

y x= /42

0

4

y

y=1

116

Note: Here, integration w.r.t. x and w.r.t. y are easier. So, we can choose any method. But first methodis somewhat better as it requires evaluation of only one integral.

Example 11. Find 2

7

1xd t dt

dx .

Solution: 2

7

1xd t dt

dx 2 2 7( 1 ). ( 1 7 ).dx dx

dx dx (using Liebniz’s rule)

21 x .

Example 12. Find 2

ln2 cos

sin

( )x

t

x

d t e dtdx (ln log )ex x

Solution:2

ln2 cos

sin

( )x

t

x

d t e dtdx

22

2 cos(ln ) 2 2 cos(sin ) sin(ln ) . (ln ) (sin ) .x xd d xx e x x edx dx

22 cosln 2 2 2 cossin1 ln 2 cos .sinx xx e x x x ex

.



10

Example 13: Find the limit

2

2

020

4lim

x

x

t dt

x

.

Solution:

2

2

020

4lim

x

x

t dt

x

( 0

0 indeterminate form)

2

2

0

0 2

4lim

x

x

d t dtdx

d xdx

(applying L’Hospital’s rule)

2

0

( 4 ).(2 )lim(2 )x

x xx

(using Leibhiz’s rule for numerator)

2

0lim 4x

x

= 2.

2.2.6 Method of Substitution: For a continuous function f on [ , ]a b , making the substitution ( )x g t ,where g is one to one continuous differentiable function on [ , ] and ( )g a , ( )g b , gives

( ) ( ( )) ,b

a

dxf x dx f g t dtdt

i.e., ( ) ( ( )) ( )

b

a

f x f g t g t dt

Note: Substitution is made to make the integration easier.

Example14. Evaluate /2

sin

0

cos xxe dx

.

Solution: Making the substitution: sin x tClearly, sin x in continuous, differentiable and one to one on [0, / 2] . So method of substitutionis applicable,sin cosx t xdx dt

Therefore, /2

sin

0

cos xx e dx

1

10

0

[ ] 1t te dt e e . 0 0/ 2 1

x t


1

ln x dxx (ln log log )ex x x .

Solution: Making the substitution lnt x .ln x is continuously differentiable and one-to-one on [1, 2], therefore the method of substitution isapplicable.

ln x t dx dtx .

So, ln 22 ln 2 2

1 0 0

ln2

x tdx tdtx

2(ln 2)2

. 1 02 ln 2

x t



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2.2 DOUBLE INTEGRAL ( , )R

f x y dxdy

2.2.1 Definition: Riemann integral of a function ( , )f x y over a region R is defined in a similar fashion asthe Riemann integral of a function of single variable.For a function ( , )f x y defined and bounded over a region R in the xy-plane, the Riemann integral is defined,if the limit exists, as

,max| | 0( , ) lim ( , )

kk k kA

R

f x y dA f A

.

Here, k k kA x y .

O

R

( , ) k k

x

y

xk

yk

( , ) k k

kth element

The whole region R is divided into rectangular sub-regions kA (k = 1, 2, ...,n) and ( , )k k is any point

in the kth subregion, kA .

max| | 0 askA n

Therefore, ( , ) lim ( , )k k knR

f x y dA f A

Any function continuous or piecewise continuous over the region R is Riemann integrable. In this

chapter only such Riemann integrable functions would be considered.

We write dA dxdy = area element2.2.2 Some Notations and Terms:

( , )R

f x y dx dy

integrandvariables of integration

region of integration

2.2.3 Geometrical Considerations:1. ( , )z f x y represents a surface in three-dimensional space.

By definition of the double integral,

( , )R

f x y dx dy = signed volume under the surface ( , )z f x y upon the region R.

Signed volume: The volume above the xy-plane is taken as positive and the volume below the xy-plane istaken as negative.

2.R

dxdy = dA area of the region R



12

2.2.4 Method of integration: Iterative integration (Fubini’s theorem)

1. For a region, 1 2 1 2: , ( ) ( ), ,R a x b g x y g x g g are continuous

on [ , ]a b , and a function f continuous over R,

2

1

( )

( )

( , ) ( , )g xb

R a g x

f x y dxdy f x y dy dx

.

y

O x

R

a b

y g x= ( )1

y g x= ( )2

Here, integration is done iteratively, i.e. one after another, in order. So, first the integration would bedone for inner integral with respect to y assuming x to be a constant and then limits of y are put. Thenintegration is done with respect to x and appropriate limits of x are put and evaluated.

2. For a region, 1 2: , ( ) ( )R c y d h y x h y , where 1 2,h h are continuous on [ , ]c d , and a continuousfunction f over R,

2

1

( )

( )

( , ) ( , )h yd

R c h y

f x y dxdy f x y dx dy

d

c

yh y2( )h y1( )

x

R

Again, the integration is done in order. First with respect to x (keeping y constant) putting the limits

1( )x h y and 2 ( )x h y , then integrating w.r.t. y putting the limits andy c y d .

3. Special cases: For the region, : ,R a x b c y d (rectangular region)

( , ) ( , )d b b d

c a a c

f x y dxdy f x y dydx .

d

c

y

xa b

R

2.2.4 Some notations:

( , ) ( , ) ( , )R R R

f x y dxdy f x y dydx f x y dA

2 2

1 1

( ) ( )

( ) ( )

( , ) ( , )g x g xb b

a g x a g x

f x y dy dx f x y dydx

2

1

( )

( )

( , )y g xb

a y g x

f x y dydx

2

1

( )

( )

( , )g xb

a y g x

f x y dydx

2 2

1 1

( ) ( )

( ) ( )

( , ) ( , )g x g xb b

a g x a g x

f x y dydx f x y dxdy (Notice the order of integration)

Example 16. Evaluate ( 2)R

xy dA for : 0 2,0 1R x y .

Solution: Consider R as, 1 2: 0 1, ( ) ( )R y h y x h y

Here, 1( ) 0h y and 2 ( ) 1h y

Therefore,

y

O x

R

y=1

x=h y1( )

y=0 2

1

Region-R

x=h y2( )

( 2)R

xy dA = 2

1

( ) 21

0 ( ) 0

( 2) )h yy

y h y

xy dx dy

21 2

0 0

22

x

x

x y x dy

1 2 22

0

2 02 2 02 2

y y dy

1

0

(2 4)y dy = 1 + 4 = 5.



13

Alternatively: : 0 2R x and 1 2( ) ( )g x y g x where 1( ) 0g x and 2 ( ) 1g x .

So, 2

1

( )2

0 ( )

( 2) ( 2)y g xx

R x y g x

xy dA xy dy dx

12 2

0 0

22x

yx y dx

22 2

0 0

2 2 1 4 52 4x

x xdx x

.

y

xRx=0

0

x=2

y g x= ( )=12

2Note: For rectangular region, the order of integration does not matter.

Example 17. Evaluate sin cos , : 0 ,2 2 2R

x y dxdy R x y .

Solution. R is a rectangular region

Therefore, sin cosR

I x ydxdy /2 /2

/2 0

sin cosy x

x ydx dy

(Integrating first w.r.t. x, then w.r.t. y)

y

O xR

2

2–

2

/2/2

0/2

[ cos ] cosxxx ydy

/2

/2

(1)cos ydy

/2/2[sin ]y

sin sin 22 2

.

Alternatively: Integrating first w.r.t. y and then w.r.t. x

sin cosR

I x ydxdy /2 /2

0 /2

sin (cos )x y

x y dydx

(in the inner integral, the limit of y is put)

/2 /2/2 /2

/2 00 0

sin [sin ] (2sin ) [ 2cos ] 2x y dx x dx x

.

Example 18. Evaluate 3 2

2

0 0

(4 )y dydx

.

Solution: Let 3 2

2

0 0

(4 )I y dydx

23 3

0 0

43yy dx

(integrating iteratively, first w.r.t., y, then w.r.t. x)

3 3 3

0

( 2) 04( 2) 4 03 3

dx

3

0

883

dx

3

0

163

dx 30

16 [ ]3

x = –16.

Example 19. Evaluate sin

0 0

x

ydydx

.



14

Solution: Let sin

0 0

x

I ydydx

.

Integrating iteratively, i.e. first with respect to y and then with respect to x,

I sin2

0 02

y x

y

y dx

2

0

1 sin2

x dx

2

0

1 2sin4

xdx

0

1 (1 cos 2 )4

x dx

0

1 1 sin 24 2

x x

14 4

.

Note: Here we must integrate in the order in which the integral is given without changing the limits. If wewant to change the order of integration, then we must change the corresponding limits. If we integratein the reverse order keeping the limits unchanged, such mistakes may happen:

sin

0 0

x

ydxdy

sin0

0

[ ]y

xy x dy 2

2

0 0

sin sin / 22 4x

yy x dy x

.


0 0

y

ydxdy

. Then, find the region of integration.

Solution: Let 242

0 0

y

I ydxdy

2

24

00

[ ] yxyx dy

22

0

(4 )y y dy 2

3

0

(4 )y y dy

242

0

24yy

= 8 – 4 = 4

Region of integration, 2: 0 2,0 4R y x y .

y

O x4y=0

Rx=0

2x y=4– 2

Boundaries: 20, 4 , 0, 2x x y y y .


0 0

y

I ydxdy

of the example-20 by changing the order of integration.

Solution: Region, : 0 4, 0 4R x y x .

242

0 0

y

I ydxdy

44

0 0

y xx

x y

I ydydx

y

O x4

224 4 ( 0)x y y x y

Region R



15

44 2

0 02

y x

y

y dx

4

0

42

x dx

42

0

1 42 2

xx

21 44 4

2 2

1 8 42

.

Example 22. Evaluate 2( )R

I x y dxdy where R is the region enclosed by the curves x y and 2x y .

Solution: If we want to integrate first w.r.t. x, then w.r.t. y, then for inner integral y is considered constantand x is varied. So, limits of x would be x = y and x = y2.

Hence, 2

12

0

( )x y

x y

I x y dxdy

2

1 3

0 3

x y

x y

x xy dy

y

1

Ox

y x=

R

1

x y= 2

Region R

y

O xy x=

x y=2

1 6 33 2

0 3 3y yy y dy

1 6 32

0

23 3y yy dy

17 3 4

0

221 3 3 4y y y

1 1 1 5 .21 3 6 42

Alternatively: Changing the order of integration:

2( )R

I x y dy dx

For the limits of inner integral, x should be kept constant, then y varies from y = x to ( 0)y x y .

So, 1

2

0

( )y x

x y x

I x y dy dx

( : 0 1, )R x x y x

1 22

0 2

y x

y x

yx y dx

1 22 3

0 2 2x xx x x dx

y

1

O xy x=

R

1

x y=2

1 25/2 3

0 2 2x xx x dx

1 1 1 1

7 / 2 4 4 6

2 17 6

= 5

42.



16


0 1

xe

dydx .

Solution: Let 1

0 1

xe

I dydx 1

10

[ ]xey dx

1

0

( 1)xe dx 10[ ]xe x 0( 1) ( 0)e e = 2e .

Alternatively:

R

I dydx , : 0 1, 1 xR x y e

Region, : 0 1, 1 xR x y e can also be written as

y

Ox

1

1 R

ey e= x (1, )e

y=1

:1 , ln 1R y e y x

Therefore, 1

1 ln 1

(1 ln )e e

y

I dxdy y dy

1ln ey y y y 12 ln ey y y (2 ) (2 0)e e 2e .


sin

x

ydydxy

.

Solution: Let 0

sin

x

yI dydxy

If we try to solve the integral, then we get stuck to integrate first sin y

y w.r.t. y, hence cannot proceed.

So, we should try by changing the order of integration.Region of integration, : 0 ,R x x y

Boundaries of : 0, , ,R x x y x y

R can be also be written as : 0 , 0R y x y

y

Ox

y=

R

yx=

Hence, 0 0

siny yI dxdyy

(changing the order of integration)

00

sin [ ]yy x dyy

0

sin. yy dyy

0

sin y dy

0[ cos ]y = 2.

Note: Without changing the order of integration, it was extremely difficult to evaluate this integral. Sousefulness of changing the order of integration is observed here.


2 ln 3 ln 3

0 /2

x

y

e dxdy .



17

Solution: Let 2 2

2 ln3 ln3

0 /2

x x

y R

I e dxdy e dx

It is very difficult to integrate 2xe w.r.t. x so we cannot proceed to integrate in the order given.So, we should try to integrate by changing (reversing) the order of integration.

: 0 2 ln 3, ln 32yR y x

Boundaries of : 0, 2 ln 3,2yR y y x , ln 3x

So, R can be written as : 0 ln 3, 0 2R x y x .

y

Oxy=0 ln3

Rx=ln3

(ln3, 2ln3)

yx=2

2ln3

Region R

So, 2

ln 3 2

0 0

xxI e dydx (changing the order of integral)

2ln 3

0

(2 ) xx e dx 2 ln 3

0

xe 2(ln3) 0e e ln 3.ln 3 1e ln 33 1 .

2.2.5 Transformation in polar coordinates

Transformation from the rectangular to polar coordinates: ( , ) ( cos , sin )R H

f x y dydx f r r rdrd H : transformed region in r -plane.

O

Rx

y

O

r

H

Area elements:

dy

dx

x

y

dA dxdy=

Rectangulary

dr

rd

x

Polar

dA rd dr rdrd=( )( ) =

dA

dA

O

Polar coordinates: Polar coordinates of a point P is ( , )r as shown in the figure.

If cartesian coordinates of P be ( , )x y then

2 2 2

cos , sin, tan /

x r y rr x y y x

y

xO

r

P r( , )

For one-to-one correspondence from rectangular to the polar coordinates we take 0r and0 2 .



18

Simple geometries in polar coordinates:

1. 0r r (a constant)geometry: a circle with centre (0, 0) and radius |r0|

O

| |r 0

r r= 0

y

x

2. 0 (a constant)

geometry: a line through (0, 0) and making

an angle 0 from positive x-axis.

y

x0

0=

O

Integration: ( , )H

g r rdrd 2 2

1 1

( )

( )

( , )g

g

g r rdr d

y

x

H

O1

r g= ( )2 =2

= r g= ( )1

2.2.6 Substitution in double integration: Method of substitution from one coordinate-system to anotheris done to simplify the integration. Simplification may be of the integrand or the region of integration or both.

Transformation: ( , ) ( , )( , ), ( , )

( , ) ( ( , ), ( , )) | ( , ) |R G

x y u vx g u v y h u v

f x y dydx f g u v h u v J u v dudv

x

y

R

v

uG

G: The transformed region in uv-coordinate-systemJ(u,v) is “Jacobian determinant” or simply “Jacobian” of the transformation, defined as

( , )( , )( , )

x xx yu vJ u v

y y u vu v

.

Note: 1. g, h must be continuously differentiable function and g, h must be one-one correspondence.

2. ( , ). ( , ) 1J u v J x y , i.e. ( , ) ( , ). 1( , ) ( , )x y u vu v x y

3. Usually, the reverse substitution is made, i.e. 1( , )u g x y and 2 ( , )v g x y and ( , )( , )( , )u vJ x yx y

is calculated first, then ( , ) 1( , )( , ) ( , )x yJ u vu v J x y

is evaluated

4. Jacobian of transformation for polar coordinates: cos , sin ( 0)x r y r r

( , )( , )( , )

x xx y rJ r

y yrr

cos sinsin cos

rr

( 0)r

Therefore, ( , ) ( cos , sin )R G

f x y dydx f r r rdrd



19

2.2.7 Some properties of double integral: If ( , )F x y and ( , )G x y are continuous, then

1.R G

kFdA k FdA (k, a constant)

2. ( )R G G

F G dA FdA GdA

3. if 0 onR

FdA F R x

y

R2R1

1 2R R R

4.R R

FdA GdA if F G on R

5.1 2R R R

FdA FdA FdA (R is the union of two non-overlapping regions R1 and R2)

2.3.4 Method of integration : Iterative method

( , , )D

f x y z dV is integrated iteratively, i.e. first with respect to inner most variable, then with respect to

middle variable of integration and finally with respect to outermost variable of integration.

For a region D enclosed between two surfaces 1( , )z g x y and 2 ( , )z g x y which has projection R inthe xy plane as shown in the figure.

Here, 1 2( , ) ( , )g x y g x y for region R.

Then, ( , , )D

f x y z dV 2

1

( , )

( , )

( , , ) )g x y

R g x y

f x y z dz dxdy

First iterative integration would make the integral adouble integral over R which can be integrated as before.

yR

x O

D

z

z g x y = ( , )2

z g x y = ( , )1

dV dzdydx dzdxdy dxdydz dxdzdy dydzdx dydxdz

One of the above order of integration is adopted according tothe convenience of the integration which depends upon theregion D and the integrand f.

2.3.5 Some Notations:

( , , )D

f x y z dz dydxintegrand

region of integration

variables of integration

2.3 TRIPLE INTEGRAL ( , , )D

f x y z dxdydz

2.3.1 Definition: For a function ( , , )f x y z defined and bounded over a region D in the space, the triple(Riemann) integral of f over D is defined if the limit exists, as



20

|| || 0 1

( , , ) lim ( , , )n

k k k kP kD

f x y z dv f V

.

Here, D is divided into n cuboidal sub-regions with dimensions k k kx y z ,

|| || max{ , , }( 1, 2,3,..., )k k kP x y z k n ,

( , , )k k k is any point in the sub-region kV .

2.3.2 Volume element

dV dzdydx dzdxdy dxdydz dxdzdy

dydxdz dydzdx

( )( )( )V x y z

yk

xk

zk y

x

z

Cuboidal sub-region

Vk

2.3.3 Geometric Interpretation:From the definition of the triple integral

D

dV volume of the region D.

2.3.4 Cylindrical coordinatesCylindrical coordinates of a point P is ( , , )r z as shown in the figure. If cartesian coordinates of P be( , , )x y z then

2 2 2

cos , sin ,and tan /

x r y r z zr x y y x

For one-to-one correspondence from rectangular to

z

P( , , )r z

z

y

xrx

xP

O

ycylindrical coordinates we take 0r and 0 2 .

Although, there can be six order of integration, similar to the rectangular coordinates. We take onlyrdzdrd as it gives easier integration.

Simple geometries in cylindrical coordinates:1. 0r r (a positive constant)

geometry: a right circular cylinder with z-axis its axis and radius as r0.y

xr r= 0

z

2. 0 0(0 2 ) (a constant)

geometry : a plane containing z-axis

and making an angle 0 with positive x-axis.

z

y0

O

= 0x

3. 0z z (a constant)

geometry: a plane perpendicular to thez-axis and cutting z-axis at (0, 0, z0)

Ox

y

z

z0



21

Volume element: Cartesian and cylindrical

y

x

z

dz

dydx

Cartesian

z

d

rd

rdr

dz

Cylindrical

dV dxdydz ( )( )dV rd dr dz rdzdrd dzrdrd

(Cuboid) (Cylindrical wedge)

Integration: For the region, G enclosed by 2 ( , )z g r

and 1( , )z g r and having projection R in the xy-plane

( , , )G

g r z rdzdrd 2

1

( , )

( , )

( , , )g r

R g r

g r z rdz drd

R

z

z g r = ( , )2

G

z g r = ( , )1

yxG : transformed region in coordinate system.

Transformation to cylindrical coordinates: Transformation from rectangular to cylindrical coordinates.

( , , ) ( cos , sin , )D G

f x y z dzdxdy f r r z rdzdrd

G : transformed region in r z coordinates

2.3.6 Spherical coordinates: Spherical coordinates of a point P is ( , , ) as shown in the figure. Pis the perpendicular projection of P on the xy-plane.

| |OP

= angle between OP

and positive z-axis

= angle between OP

and positive x-axis

Ox

z

x P

z= cos

P( , , )

y

0 , 0 2

If cartesian coordinates of P be ( , , )x y z , cylindrical coordinates of P be ( , , )r z , and spherical coordinates

of P be ( , , ) , then

Rectangular Cylindricalsin cos sinsin sin coscos

x ry zz

Simple geometries in spherical coordinates:

1. 0 (a positive constant)

geometry: a sphere of radius 0 and centre (0, 0, 0) O

x

z

0

y



22

2. 0 (a constant such that 00 )

geometry: a right circular single cone with axis

z-axis and vertex (0, 0, 0) and half angle 0

Special case:0 positive z axis y

x

z

/ 2 xy–plane

negative z-axis

3. 0 (a constant)

geometry: half plane containing z-axis

z

0

x

y

and making an angle 0 with positive x-axis.

Volume elements:

y

x

z

dz

dydx

Rectangular

z

SdS

Spherical

O

sin d sin d

dV dxdydz (Cuboid) 2 sin cosdV ds d ( sin ) ( sin )d d d (Spherical wedge)

Integration:

( , , ) ( sin cos , sin sin , cos )D G

I f x y z dxdydz f 2 sin d d d

2( , , ) sin

G

g d d d .

Integration in spherical coordinates is useful if G is as

1 2 1 2 1 2{( , , ) : ( , ) ( , ), , }G g g

Then, 2 2 2

1 1 1

( , )2

( , )

( , , ) sing

g

I g d d d

.

Rx

= 1

y=

2

Projection = ( , )g1

= ( , )g2

z12

in plane xy

2.3.7 Substitutions in Triple IntegrationTransforming an integral in ( , , )x y z to new coordinates ( , , )u v w as

( , , ), ( , , ) and ( , , )x g u v w y h u v w z k u v w ,



23

( , , ) ( ( , , ), ( , , ), ( , , )) | ( , , ) |D G

f x y z dxdydz f g u v w h u v w k u v w J u v w dudvdw .

G: The transformed region in u, v, w coordinate system,( , , )J u v w is “Jacobian determinant or simply “Jacobian” defined as

( , , )( , , )( , , )

x x xu v wy y y x y zJ u v wu v w z u v wz z zu v w

.

Note:1. The , ,g h k must be differentiable functions and the transformation must to one-to-one.

2. ( , , ). ( , , ) 1J u v w J x y z i.e., ( , , ) ( , , ). 1( , , ) ( , , )x y z u v wu v w x y z

3. Usually, the reverse substitution is made, i.e., 1 2( , , ), ( , , )u g x y z v g x y z and 3 ( , , )w g x y z ,

and ( , , )J x y z is calculated first, then 1( , , )( , , )

J u v wJ x y z

is evaluated.

4. Jacobian of transformations to cylindrical coordinates: cos , sin ,x r y r z z

cos sin 0( , , ) sin cos 0

0 0 1

x x xr z ry y yJ r z rr zz z zr z

cos sin

( 0)sin cos

rr

r

Hence, dV dxdydz rdzdrd .

5. Jacobian of transformation to spherical coordinates: sin cos , sin sin , cosx y z ,

sin cos cos cos sin sin

( , , ) sin sin cos sin sin coscos sin 0

x x x

y y yJ

z z z

2 sin .

Hence, 2 sindV d d d .



24

2.3.8 Some properties of triple integral: If ( , , )F x y z and ( , , )G x y z are continuous, then

1.D D

kFdV k FdV (k, a constant)

2. ( )D D D

F G dV FdV GdV

3. 0 if 0 onD

FdV F D

4. if onD

FdV GdV F G D

5.1 2D D D

FdV FdV FdV (D is the union of two non-overlapping regions D1 and D2)

chapter-2 : integral calculus

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