M1 PART-A

UNIT-I (2marks questions)
1. Find the characteristic equation of the matrix
1 2
0 2
 
 
 
.
Sol.
The characteristic equatin of A is 0A I 
2
2
1 2 1 0
0
0 2 0 1
1 2
0
0 2
(1 )(2 ) 0 0
2 2 0
3 2 0



 
  
 
   
    
   



   
   
  
The required characteristic equation is 2
3 2 0    .
2. Obtain the characteristic equation of
1 2
5 4
 
 
 
.
Sol.
Let A=
1 2
5 4
 
 
 
The characteristic equation of A is 2
1 2 0c c   
1
2
1 4 5
1 2
5 4
4 10
6
c sumof themaindiagonal elements
c A

  




 
 
2
2
(5) ( 6) 0
5 6 0
Hencethecharacteristicequationis
 
 
   
  
3. Find the sum and product of the eigenvalues of the matrix
1 1 1
1 1 1
1 1 1



.
Sol.
( 1) ( 1) ( 1)
3
1 1 1
1 1 1
1 1 1
1(1 1) 1( 1 1) 1(1 1)
1(0) 1( 2) 1(2)
4
sumof theeigenvalues sumofthediagonal elements
product of theeigenvalues

     
 

 

       
    

4. Two eigen values of the matrix
11 4 7
7 2 5
10 4 6
  
   
   
are 0 and 1,
find the third eigen value.
Sol.
1 2 3
1 2 3
3
3
0, 1, ?
11 ( 2) ( 6)
0 1 3
2
Given
sumof theeigenvalues sumof themaindiagonal elements
  
  


  

      
  


5. Verify the statement that the sum of the elements in the diagonal
of a matrix is the sum of the eigenvalues of the matrix
2 2 3
2 1 6
1 2 0
.
( 2) (1) (0)
1
2 2 3
2 1 6
1 2 0
2(0 12) 2(0 6) 3( 4 1)
24 12 9
45
sol sumof theeigenvalues sumof themaindiagonal elements
product of theeigenvalues
  
 
 
   

   
 
 
 
 
       
  

6. The product of the eigenvalues of the matrix
6 2 2
2 3 1
2 1 3
A
 
    
  
is
16, Find the third eigenvalue.
Sol.
1, 2 3
1 2
1 2 3
, .
16
6 2 2
2 3 1
2 1 3
let theeigenvalues of thematrix Abe
Given
weknowthat A
  
 
  



  

6(9 1) 2( 6 2) 2(2 6)
6(8) 2( 4) 2( 4)
      
    
3
3
32
16 32
2





7. Two eigenvalues of the matrix
1 2 3
1 2 3
3
3
1 2
8 6 2
6 7 4 3 0. ?
2 4 3
. 3, 0, ?
. .
8 7 3
3 0 18
15
are and what isthe product of theeigenvaluesof A
sol given
w k tThe sumof theeigenvalues sumof themaindiagonal elements
productofeigenvalues
  
  


 
 
   
  
  

    
  

 3 (3)(0)(15) 0  
8. Find the sum and product of the eigen values of the matrix
2 0 1
0 2 0
1 0 2
.
2 2 2
6
2 0 1
0 2 0
1 0 2
2(4 0) 0(0) 1(0 2)
8 2
6
sol sumof theeigenvalues sumof themaindiagonal elements
product of theeigenvalues A
 
 
 
  

  


 
   
  
    
 


9.Find the characteristic equation of the matrix
1 2
0 2
 
 
 
and get its
eigenvalues.
Sol.
Given is a upper triangular matrix.
Hence the eigenvalues are 1,2
W.k.t the chacteristic equation of the given matrix is
2
2
2
( ) ( ) 0
(1 2) (1)(2) 0
3 2 0
sumof theeigenvalues product of theeigenvalues 
 
 
  
   
  
10.Prove that if  is an eigenvalues of a matrix A, then
1

is the
eigenvalue of 1
A
1
1 1
1
1
1
1
;
,
1
1
. ,
proof
If X betheeigenvector corresponding to
then AX X
premultiplying bothsides by A weget
A AX A X
IX A X
X A X
X A X
i e A X X








 










11.Find the eigenvalues of A given
3 3 3
1 2 3
0 2 7
0 0 3
.
1 2 3
0 2 7
0 0 3
1,2,3
1,2,3
1 ,2 ,
A
sol
A
clearly given Ais aupper triangular matrix
Hencetheeigenvalues are
theeigenvalues of the givenmatrix Aare
Bythe propertytheeigenvaluesof thematrix A are
 
  
 
  
 
   
  
3
3 .
12.If  and  are cthe eigen values of
3 1
1 5
 
  
form the
matrix whose eigenvalues are 3
 and 3

Sol.
2 2 2
1 7 5
0 2 9 0
0 0 5
(1 )[(2 )(5 ) 0] 7[0 0] 5[0 0] 0
(1 )(2 )(5 ) 0
1, 2, 5
1 2 5
30
sumof theeigenvalues



  
  
  

 

        
   
  
  


13.Sum of square of the eigenvalues of
1 7 5
0 2 9
0 0 5
 
 
 
  
is……..
Sol.
The characteristic equatin of A is 0A I 
2 2 2
1 7 5
0 2 9 0
0 0 5
(1 )[(2 )(5 ) 0] 7[0 0] 5[0 0] 0
(1 )(2 )(5 ) 0
1, 2, 5
1 2 5
30
sumof theeigenvalues



  
  
  

 

        
   
  
  

14 .two eigenvalues of A=
4 6 6
1 3 2
1 5 2
 
 
 
    
are equal and they are
double the
third.Find the eigenvalues of A.
Sol.
2 2 2 2
2 ,2
2 2 (4) (3) ( 2)
5 5
1
2,2,1
2 ,2 ,1
Letthethirdeigenvaluebe
Theremainingtwoeigenvaluesare
sumftheeigenvalues sumofthemaindiagonalelements
theeigenvaluesofAare
HencetheeigenvaluesofA are

 
  



     


15.show that the matrix
1 2
2 1
 
 
 
satisfies its own characteristic
equation.
Sol.
1 2
2 1
LetA
 
  
 
The cha.equation of the given matrix is
2
1 2
1
2
2
2
2
2
0
0
1 1 2
1 2
1 4 5
2 1
2 5 0
2 5 0
.
1 2 1 2
2 1 2 1
3 4
4 3
3 4 1 2 1 0
2 5 2 5
4 3 2 1 0 1
0 0
0 0
A I
S S
S sumof main iagonal elements
S A
Thecharacteristicis
Toprove A A I
A A A
A A I

 
 
 
  

  

    
  
  

 

  
  
 
       
         
     
 
  
 

16.If A=
1 0
4 5
 
 
 
express 3
A in terms of A and I using Cayley
Hamilton theorem.
Sol.The cha.equation of the given matrix is 0A I 
2
1 0 1 0
0
4 5 0 1
1 0
0
4 5
(1 )(5 ) 0 0
(1 )(5 ) 0
6 5 0



 
 
 
   
    
   



   
  
  
By Cayley Hamilton theorem,
2 2
3 2
3 2
6 5 0, 6 5
6 5 0
6 5
6(6 5 ) 5
36 30 5
31 30
A A I A A I
multiply Aon both sides
A A A
A A A
A I A
A I A
A I
    
  
 
  
  
 
17.Write the matrix of the quadratic form
2 2
2 8 4 10 2x z xy xz yz    .
Sol.
Q=
2
2
2
1 1
2 2
1 1
2 2
1 1
2 2
coeff of x coeff of xy coeff of xz
coeff of xy coeff of y coeff of yz
coeff of xz coeff of yz coeff of z
 
 
 
 
 
 
 
  
Q=
2 2 5
2 0 1
5 1 8
 
  
  
18.Determine the nature of the following quadratic form
  2 2
1 2 3 1 2, , 2
. .
f x x x x x
sol Thematrixof Q F is
 
Q=
2
2
2
1 1
2 2
1 1
2 2
1 1
2 2
 
 
 
 
 
 
 
  
=
1 0 0
0 2 0
0 0 0
 
 
 
  

There for the eigenvalues are 0,1,2. so find the eigenvalues one
eigenvalue is Zero another two eigenvalues are positive .so given
Q.F is positive semi definite.
19. State Cayley Hamilton theorem.
Every square matrix satisfies its own characteristic equation.
20. Prove that the Q.F 2 2 2
2 3 2 2 2x y z xy yz zx     .
Sol.The matrix of the Q.F form,
Q=
2
2
2
1 1
2 2
1 1
2 2
1 1
2 2
 
 
 
 
 
 
 
  
=
1 1 1
1 2 1
1 1 3
 
 
 
  
1 1
1 1
2
2 2
1 1 1
3 2 2 2
3 3 3
1 1( )
1 1
(2 1) 1( )
1 2
1(6 1) 1(3 1) 1(1 2) 2( )
D a ve
a b
D ve
a b
a b c
D a b c ve
a b c
   
     
         
The Q.F is indefinite.

UNIT II - SEQUENCES AND SERIES
Part A
1. Given an example for (i) convergent series (ii) divergent series
(iii) oscillatory series
Solution:
(i) The series
+ is convergent
(ii) 1+2+3+….+n+… is divergent
(iii) 1-1+1-1+…… is oscillatory
2. State Leibnitz’s test for the convergence of an alternating
series
Solution:
The series a1-a2+a3-a4+…. In which the terms are
alternately +ve and –ve and all ai’s are positive, is convergent if
(i) and
(ii)
3. State the comparison test for convergence of series
Solution:
Let ∑an and ∑bn be any two series and let a finite
quantity ≠ 0, then the two series converges or diverges together
4. State any two properties of an infinite series
Solution:
(i) The converges or diverges of an infinite series is
not affected when each of its terms is
multiplied by a finite quantity
(ii) If a series in which all the terms are positive is
convergent, the series will remain convergent
even when some or all of its terms are made
negative
5. Define alternating series
Solution:
A series whose terms are alternatively positive and
negative is called alternating series
Eg: + is an alternating series
6. Prove that the series is convergent
Solution:
The nth term of the series is an=
Then an+1 =
now = =
= =0(
Hence by D’Alembert’s test, ∑an is convergent
7. When is an infinite series is said to be (i) convergent (ii)
divergent (iii) oscillatory?
Solution:
Let ∑an be an infinite series and let Sn be the sum of the
first n terms of an infinite series then
(i) If is finite the series is said to be convergent
(ii) If If →±∞ the series is said to be divergent
(iii) If not tend to a definite limit or ±∞, then
the series is oscillatory.

8. State true or false
(i) If ∑an is convergent, ∑an
2
is also convergent.
(ii) If the nth
term of a series does not tend to zero as n→∞,
the series is divergent.
(iii) The convergence or divergence of an infinite seies is
not affected by the removal of a finite number of terms
from the beginning
(iv) An absolutely convergent series is convergent
Solution:
All are true.
9. Prove that the series is conditionally convergent
Solution:
The nth term of the series be an =
Then =1/n and =1/n+1
Since , n+1 n ,
an is decreasing and = =0
By Leibnit’z test, the given series is convergent. Also the series
formed by the absolute value of its terms is divergent. Hence
the series is conditionally convergent.
10. For what values of p, the series + +…+ +… will be (i)
convergent (ii) divergent
Solution:
The p-series is convergent if p 1 and divergent if
UNIT-III
DIFFERENTIAL CALCULUS
1) Find the curvature of 2 2
4 6 1 0x y x y    
Solution:
 
3
2 2 2
2 2
2
x y
xx y xy x y yy x
f f
f f f f f f f



 
f = 2 2
4 6 1x y x y   
   
   
   
   
3 3
2 2 2 22 2
2 2 2 2
2 4 2 6
2 2
0
2 4 2 6 2 4 2 6
2 2 6 0 2 2 4 2 2 6 2 4
x y
xx yy
xy
f x f y
f f
f
x y x y
y x y x

   
 

        
    
       
 
 
 
 
 
 
2 2 3
12 2 2
3
2 2 2
1/22 2
2 2
2 2 6 (2 4)1
2 2 6 (2 4)
2 4 (2 6)
2 2 6 (2 4)
1 2
2 6 (2 4)
y x
curvature y x
x y
y x
y x




   
        
 
   
 
    
 

  
2) What is the formula of radius of curvature in
Cartesian
form and parametric form?
Sol:
Cartesian form:
2 3/ 2
1
2
(1 )y
y



Parametric form:
   
3/ 22 2
' '
' '' ' ''
x y
x y y x

 
 


3 Find the radius of curvature at x=0 on x
y e
Solution:
Given x
y e
Radius of curvature
 
3
2 2
1
2
1 y
y



   
0
1 1 0
0
2 2 0
3/ 2 3/ 22
1
2
] 1
] 1
1 1 1
2 2
1
x
x
x
x
x
y e
y e y e
y e y e
y
y




   
   
 
   
4 Find the radius of curvature of the curve 2
( , )xy c at c c
Sol:
   
2
2
2 2
1 12 2
2 2
2 23 3
3/2 3/ 22 3/ 2
1
2
( , )
1
2 2 2
1 1 1 .2
2/ 2
2.
Given xy c at c c
c
y
x
c c
y y
x c
c c
y y
x c c
y c
y c
c




 
    
   
 
  

5 What is the curvature of the curve 2 2
25x y  at the
point (4,3) on it.
Sol:
Since the given curve is a circle &
We know that radius of given circle is 5 units
radius of curvature of a circle is equal to the radius of
the given circle
5
1 1
.
5
curvature


 
 
6 Find radius of curvature of the curve cos ,x a 
siny b  at any point ' '
Sol:
   
 
3/2 3/22 2 2 2 2 2
2 2
3/22 2 2 2
2 2
cos sin
' sin ' cos
'' cos '' sin
' ' sin cos
' '' '' ' sin cos
sin cos
( sin cos 1)
x a y b
x a y b
x a y b
x y a b
x y x y ab ab
a b
ab
 
 
 
 

 
 
  
 
  
   
 
 
 

  
7 Find the radius of curvature at any point on the
curve r e
 .
Sol:
3/ 22 2
1
2 2
1 22
r r
r rr r

  
 
1 2&
Given r e
r e r e

 

 

   
   
 
     
 
 
 
3/2 3/ 22 2 2
2 2 2 2 2
33/ 2 3
1
2
2
2
2 2
2
2 2
2
2.
e e e
e e e e e e e
e
e e
e
r
  
      

 




   
       
   
  

8 Find the radius of curvature at y=2a on the curve
2
4y ax
Sol:
 
 
 
 
 
 
2
3/22
1
2
1
1
1
1
2
2 1 1 2 1
4 1
1
1 . . ,
2 4
2 2
2
2
2 1
2
2 . . ,
0
Given y ax
y
Formula
y
diff w r to x
yy a
yy a
a
y
y
a
y at y a
a
diff w r to x
yy y y yy y

 



 
 
  
    
 
2
1
2
2 2 1/ 2
y
y
y
y at y a a


  
 
   
  
 
3/ 2 3/2
3/ 2
5/ 2
1 1 2
2 2
1/ 2 1/ 2
2
a
a a
a


    
 
 
i.e. 5/2
2 . 4 2a a  
9 Find the radius of curvature at (a,0) on
3 3
2 a x
y
x


Sol: Given
3 3
2 a x
y
x


3
2 2
3
1 2
3
1 2
2 2
2
a
y x
x
a
yy x
x
a x
y
x y y
 

 

 
1
2 3 3
2 2
( ,0)
.2 . 0 3
at a y
dxHence we find
dy
xy a x
dx dx
x y y x
dy dy
  
  
  
2 2
2 2
2 ( 3 ) 0
2
3
( ,0) 0
dx
xy y x
dy
dx xy
dy x y
dx
at a
dy
  



 
  
 

   
 
  
 
 
2
2 2
2
22 2 2
22 3
22 42
3/ 22
3/ 2
2
3 2 2 2 6 2
3
3 0 0 2 0 6 2
( ,0)
9 33 0
1
1 0 3
2 2
3
3
2
dx dx
x y y x xy x y
dy dyd x
dy x y
a ad x a
at a
dy a aa
dx
dy
a
d x
ady
a


   
        
   

    
  

  
  
        
 
 
 

10 Find the radius of curvature at
2
x

 on the
curve 4sin sin 2y x x  .
Sol:
1
4sin sin 2
4cos 2cos2
y x x
dy
y x x
dx
 
  
2
2 2
1
2
4sin 4sin 2
/ 2, 4(0) 2cos 2
/ 2, 4(1) 4sin 4
d y
y x x
dx
at x y
at x y
 
 
   
   
     
3/ 2 3/ 22 2
1
2
1 1 (2)
4
y
y

         

 
 
3/ 2 3/ 2 1/ 2
1 4 5 5.5 5 5
4 4 4 4
5 5
4
is ve 
 
   
  
  
11 Define the curvature of a plane curve and what is the
curvature of a straight line
Sol:
The curvature of a plane curve at K
d
ds


The curvature of a straight line is zero.
12 Find the radius of curvature at any point (x,y) on the
curve log sec
x
y c
c
  
   
  
Sol:
 
1
2
2
3/ 2 3/ 2
2 23/ 22
1
2 22
3
2
logsec
1 1
. tan .sec tan
sec
1
sec
1 tan sec
1
1 1
sec sec
sec
.
sec
x
y c
c
x x x
y c c
x c c c c
c
x
y
c c
x x
y c c
x xy
c c c c
x
c
c
x
c

 
  
 
     
      
       
 
 
 
  
 
      
                 
   
   
   
 
 
 



.sec
x
c
c
 
  
  



13 Find the radius of the curve given by 3 2cos ,x  
4 2siny  
Sol:
 
2
2
2
3
3/2 3/ 22 2 3
1
3 32
3 2cos 4 2sin
2sin 2cos
2cos
cot
2sin
1
cot
2sin
cos 1
cos
2sin 2
1 1 cot cos
2
1 1
cos cos
2 2
2
x y
dx dy
d d
dy
dx
d y d dy d d
dx d dx dx d
ec
ec
y ec
y ec ec
 
 
 





  



 

 

   
  
  

 
   
 

 

           
 

14 Write the formula for centre of curvature and
equation of circle of curvature.
Sol:
Centre of curvature:
 2
1 1
2
1y y
x x
y

 
&
2
1
2
(1 )y
y y
y

 
Circle of curvature:   
2 2
2
x x y y    
15 Find the centre of curvature of 2
y x of the origin.
Sol:
The centre of curvature is given by
 
  
 
  
2
1
21
1
2 2
2
1 2
1
2
2
2
1
1 ( ) ,
; 2 ; 2.
(0,0), 0
(0,0), 2
0
1 (0)
2
1 0 1
2 2
(0,0), 0
1
(0,0),
2
1
0,
2
yy
X x y Y y
y y
Given y x y x y
at y
at y
X x x
Y y y
at X
at Y
Centreof curvatureis

    
  


    

   


 
  
 
16 Write properties of evolutes.
Sol:
(i) The normal at any point of a curve touches the
evolute at the corresponding Centre of curvature.
(ii)The length of an of the evolute is equal to the
of curvature at the points on the original curve
corresponding to the extremities of the arc
(iii)There is only one evolute, but an infinite number
of involutes.
17 Find the envelope of the family of straight lines
2
y mx am  , m being the parameter.
Sol:
Given 2
y mx am 
Diff. partially w.r.to m, we get,

0 2
2
x am
x
m
a
 


2
2
2 2
2
2 2 2
2
2 2
2 4
2 4 4
4
y mx am
x x
y x a
a a
x x
a
a a
x x x
y
a a a
x ay is the required envelope
  
    
    
   

 
 
  
 
18 efine envelope of a family of curves.
Definition:
A curve which touches each member of
a family of curve is called the envelope of that family
curves.
19 efine Evolute and Involute.
The locus of the centre of the given curve is called the
evolute of the curve.
The given curve is called the Involute of its evolute.
20 Find the envelope of the family of lines 2
x
yt c
t
  ,
t being the parameter.
Sol:
Given family of lines can be written as,
2
2 0yt ct x   --------- (1)
The envelope of 2
0At Bt C   is 2
4 0B AC 
From (1) we get A = y, B= -2c, C = x
Putting these values in (2) we get,
2
2
2
2
( 2 ) 4 0
4 4 0
0
c yx
c yx
c xy
xy c
  
 
 

This is required envelope.
21 Find the Envelope of the family of Straight
lines
a
y mx
m
  , where m is a parameter.
Sol:
2
2
(1)
0
a
Given y mx
m
ym m x a
m x ym a
       
  
   
This is a quadratic in ‘m
2
2
2
4 0
, ,
4 0
4
So the envelope is B AC
Here A x B y c a
y ax
y ax
 
   
 

22 Find the Envelope of the family of lines cos sin 1,
x y
a b
   
being the parameter
Sol:
 Given, cos sin 1 1
(1) . . ' '
sin cos 0 (2)
x y
a b
diff partially w r to we get
x y
a b
 

 
  

  

2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
(1) (2) ,
cos sin sin cos 1 0
cos sin 2 cos sin
0
sin cos 2 cos sin
we get
x y x y
a b a b
x y
xy
a b
x y
xy
a b
   
   
  

   
       
   

  

  

2 2
2 2 2 2
2 2
2 2
2 2
cos sin sin cos 1
1
x y
y b
x y
a b
             
  
23 Find the envelope of the straight lines
cos sin sec ,x y a    where is the parameter.
Sol:
Given  cos sin sec 1x y a    
Dividing equation (1) by cos we get,
2 2
2 2
sec
tan sec (1 tan )
cos
tan tan ( ) 0
x y a a a
a y a x

  

 
    
   
Which is a quadratic equation in tan
Here A=a, B=-y, C = (a-x).
2
2
4 0,
4 ( ) 0
B AC
y a a x
 
  
24 Find the envelope of 2 2 2
,y mx a m b   where m is a
parameter.
Sol:
2 2 2
2 2 2 2
2 2 2 2 2 2
2 2 2 2 2
( )
2
( ) 2 0
y mx a m b
y mx a m b
y m x mxy a m b
m x a mxy y b
  
  
   
    
Which is a quadratic equation in m.
Hence the envelope is 2
4 0B AC 
Here A= ( 2 2
x a ), B=-2xy, C = 2 2
y b
2 2 2 2 2 2
2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2
2 2
2 2
4 4( )( ) 0
( )( ) 0
0
. ,
1
x y x a y b
x y x a y b
x y x y b x a y a b
i e b x a y a b
x y
a b
   
   
      
 
 
25 Find the envelope of cos sin ,x y a   where  is
a parameter.
Sol:
Given cos sin (1)x y a   
Diff w.r.to 
sin cos 0 (2)x y    
Eliminate  between (1) and (2)
   
2 2
2 2 2 2
1 2 ,
( cos sin ) ( sin cos ) 0
wehave
x y x y a   

     
2 2 2 2
2
2 2 2 2
cos sin 2 sin cos
sin cos 2 sin cos
x y xy
a
x y xy
   
   
  

   

2 2 2 2 2 2 2
2 2 2
(cos sin ) (sin cos )x y a
x y a
      
 
26 Find the envelope of the family given by
1
,x my
m
 
m is parameter.
Sol:
The given equation can be written as
2
1 0,m y mx  
Which is quadratic equation in m ,
Here, , , 1A y B x c  
2
2
2
4 0
4 0
4
Hencethe envelopeis B AC
x y
x y
 
 

27 Find the envelope of 2
1y mx m   where m is a
parameter.
Sol:
Given 2
1y mx m  
2
2 2
2 2 2 2
2 2 2
1
( ) 1
2 1
( 1) 2 1 0.
y mx m
Squaring onbothsides y mx m
y mxy m x m
m x mxy y
  
  
   
    
2 2
2
2 2 2
1, 2 , 1.
4 0
( 2 ) 4( 1)( 1) 0
Here A x B xy C y
B AC
xy x y
     
 
    
2 2 2 2
4 4( 1)( 1) 0x y x y   

UNIT –IV
FUNCTIONS AND SEVARAL VARIABLE
PART-A
1. 1 1
cos , . . cot
2
x y u u
If u PT x y u
x yx y

   
   
   
.
Proof:
( , ) cos
1
hom deg ,
2
' .
x y
Given f x y u
x y
As f is ogeneous functionof reen
it is satisfiesthe Euler sequation

 


(cos ) (cos ) 1
cos .
2
1
( sin ) ( sin ) cos .
2
1 cos
.
2 sin
1
cot .
2
f f
x y nf
x y
u u
x y u
x y
u u
x u y u u
x y
u u u
x y
x y u
u u
x y u
x y
 
  
 
 
 
 
 
   
 
 
  
 
 
 
 
2.
3 3
1
tan , . . sin 2
x y u u
If u PT x y u
x y x y
    
      
.
Solution:
Given
3 3
( , ) tan
x y
f x y u
x y

 

As f is a homogeneous function of order n=2, it satisfies
the Euler’s theorem.
(tan ) (tan )
2tan .
f f
x y nf
x y
u u
x y u
x y
 
 
 
 
 
 
2 2
2
(sec ) (sec ) 2tan .
sin 1
2 .
cos sec
u u
x u y u u
x y
u u u
x y
x y u u
 
 
 
 
  
 
2sin
2 cos .
cos
2sin cos .
sin 2 .
u
u
u
u u
u u
x y u
x y
 
 
 
 
 
3.
3 3
log , . . 2
x y u u
If u PT x y
x y x y
   
      
Solution: Given
3 3
log
x y
u
x y
 
   
3 3
hom deg 2,
' .
u x y
Let f e
x y
As f is ogeneous functionof reen
it is satisfiesthe Euler sequation
f f
x y nf
x y

 


 
  
 

( ) ( )
2
u u
ue e
x y e
x y
 
 
 
( ) ( ) 2 .u u uu u
x e y e e
x y
 
 
 
2.
u u
x y
x y
 
 
 
Hence the proof.
4. If 2 2
( , ) logf x y x y  , show that
2 2
2 2
0
f f
x y
 
 
 
.
Solution: Given 2 2
( , ) logf x y x y 
 2 21
log
2
f x y 
2 2 2 2
1 2
2
f x x
x x y x y

 
  
   
   
2 22 2 2
2 22 2 2 2 2
.1 2x y x xf y x
x x y x y
  
 
  
Similarly,
2
2
f
y

  
2 2
22 2
x y
x y



2 2
2 2
0
f f
x y
 
 
 
5. If 1 1
sin tan
x y
u
y x
    
    
  
show that 0
u u
x y
x y
 
 
 
Solution: Here u is a homogeneous function of degree n = 0.
Using Euler’s theorem, 0
u u
x y
x y
 
 
 
6. If
x y z
u
y z x
   show that 0
u u u
x y z
x y z
  
  
  
.
Solution:
x y z
Given u
y z x
  
2
2
1
.........(1)
1
u z
x y x
u x z
x
x y x
u x
y y z

 


 

 
 

2
..........(2)
1
..........(3)
u x y
y
y y z
u y
z z x
u y z
z
z z x
 
 

 
 

 
 

.(1),(2)&(3),
0.
Add eqn we get
u u u
x y z
x y z
  
  
  
7. If ( , ) cos , sinz f x y where x r y r    .Show
that
22 2 2
2
1z z z z
x y r r 
         
        
         
Solution: Wkt,
z z x z y
r x r y r
    
 
    

cos sin
z z
x y
 
 
 
 
( sin ) ( cos )
z z x z y
x y
z z
r r
x y
  
 
    
 
    
 
  
 
1
sin cos
z z z
r x y
 

  
 
  
2 2
2
2 2
22
2 2
22
2 2
1
. .
cos sin ( sin ) (cos )
cos sin 2 sin cos
sin cos 2 sin cos
z z
R H S
r r
z z z z
x y x y
z z z z
x y x y
z z z z
x y x y
z
x

   
   
   
    
    
    
      
       
      
     
    
      
     
    
      
 
 

22
z
y
 
  
  
Thus, R.H.S = L.H.S
8. If  , , cos , sinu u
z f x y x e v y e v   show
that 2uz z z
x y e
v u y
  
 
  
.
Solution:  , , cos , sinu u
Given z f x y x e v y e v  
z z x z y
u x u y u
    
   
    
cos sinu uz z
e v e v
x y
 
   
 
 
 
2 2 2
2 2 2
cos sin
sin cos sin ....(1)
sin cos
sin cos
sin cos cos ....(2)
(1) (2)
u u
u u
u u
u u
u u
z z z
y ye v ye v
u x y
z z
e v v e v
x y
z z x z y
v x v y v
z z
e v e v
x y
z z z
x xe v xe v
v x y
z z
e v v e v
x y
z z
x y
v u
  
   
  
 
 
 
    
   
    
 
  
 
  
  
  
 
  
 
 
 

 
 2 2 2
2
sin cos
.
u
u
z
e v v
y
z
e
y
Hence proved

 




9. If log( )u x xy where 3 3
3 1x y xy   find
du
dx
.
Solution:
3 3
, log( ) & 3 1
....(1)
Given u x xy x y xy
du u u dy
dx x y dx
   
 
  
 
1
( ) log( )
u
x y xy
x xy

  


1 log( )
u
xy
x

  

3 3
1
, 3 1
u x
x x
y xy y
consider x y xy

   

  
Diff. w.r.to ‘x’,
 
2 2
2 2
3 3 3 3 0
3 3 3 3 0
dy dy
x y y x
dx dx
dy
x y y x
dx
   
   
   
 
 
2 2
2 2
2
2
3 3
3 3
(1) 1 log( )
x y x ydy
dx y x y x
x x ydu
xy
dx y y x
   
  
 

   

10. Find 3 3
3
dy
when x y axy
dx
 
Solution:
Let 3 3
( , ) 3f x y x y axy  
2 2
2 2
2 2
3 3 ; 3 3
3 3
3 3
f f
x ay y ax
x y
f
dy x ay x ayx
fdx y ax y ax
y
 
   
 

      
  

11. Find
dy
dx
when sin cosy x x y
Solution:
sin cos
sin cos 0
Given y x x y
y x x y

   
( , ) cos sin
cos cos & sin sin
Let f x y x y y x
f f
y y x x y x
x y
 
 
    
 
 
 
 
cos cos
sin sin
cos cos
sin sin
f
y y xdy x
fdx x y x
y
dy y y x
dx x y x

  
  


 

12. If 2 2 2
u x y z   and , sin , cost t t
x e y e t z e t   find
du
dt
with actual substitution.
Solution: Given 2 2 2
u x y z   , , sin , cost t t
x e y e t z e t  
 
2 2 ( sin cos ) 2 ( cos sin )
2 (sin cos ) (cos sin )
t t t t t
t
du u dx u dy u dz
dt x dt y dt z dt
xe y e t e t z e t e t
e x y t t z t t
  
     
  
    
    
 
2 2
2 2
2 sin sin cos cos sin cos
2 sin cos
t t t t t t
t t t
e e e t e t t e t e t t
e e e t t
      
    
2 2t t
e e   
13. Find
du
dt
if sin ( / )u x y , where 2
,t
x e y t  .
Solution:

. .
du u dx u dy
dt t dt y dt
 
 
 
2
2 2 3
1
cos . cos 2
2
cos
t
t t t
x x x
e t
y y y y
du e e e
dt t t t
     
      
     
   
    
   
14. If u = f( y –z , z – x , x – y ) find
u u u
x y z
  
 
  
.
Solution:  , ,Given u f y z z x x y   
,
( 1) (1) .....(1)
Let r y z s z xand t x y
u u r u s u t
x r x s x t x
u u
s t
     
      
     
      
 
  
 
(1) ( 1) .....(2)
u u r u s u t
y r y s y t y
u u
r t
      
     
      
 
  
 
( 1) (1) .....(3)
u u r u s u t
z r z s z t z
u u
r s
      
     
      
 
  
 
(1) (2) (3) 0
u u u
x y z
  
     
  
15. Find the minimum value of F = x2
+y2
subject to the
Constraint x=1.
Solution: Given F = x2
+y2
= square of the distance from the origin
The minimum of F is 1.
16. Define Jacobian.
If u and v are functions of the two independent variables
x and y, then the determinant
u u
x y
v v
y y
 
 
 
 
is called the Jacobian
of u ,v with respect to x,y.
 
 
,
It is denoted by
,
x y
u v


.
17. Find the Jacobian
( , )
cos , sin
( , )
x y
if x r y r
r
 


 

.
Solution: Given cosx r  siny r 
cos
sin
x
r
y
r
r








sin
cos
y
r
y
r









 
 
cos sin,
sin cos,
x x
rx y r
y y rr
r
 
 

 
  
  
 
 
2 2
cos sinr r  

 
 
 
2 2
cos sin
,
,
r
x y
r
r
 

 



18. 2 2
2 , cos , sin ,If u xy v x y and x r y r     
( , )
( , )
u v
evaluate
r 


Solution:
 
 
 
 
 
 
2 2
, , ,
, , ,
2
2 2
2 2
u v u v x y
r x y r
u u x x
x y r
v v y y
x y r
Given u xy v x y
u v
y x
x x
u v
x y
y y
 


  
 
  
   
   
 
   
   
  
 
 
 
 
  
 
Given cosx r  siny r 
cos
sin
x
r
y
r
r








sin
cos
y
r
y
r









 
 
2 2 cos cos,
2 2 sin cos,
y x ru v
x y rr
 
 

 

   
   
2 2 2 2
2 2 2 2
4 4 cos sin
4 cos sin
y x r r
x y r
 
 
    
    
 
 
2
3
4
,
4
,
r r
u v
r
r 
  

 

19. If
2 2
( , )
,
( , )
y x u v
u v then find
x y x y

 

.
Solution:
2 2
2
2
2
2
2
2
y x
Given u v
x y
u y v x
x x x y
u y v x
y x y y
 
 
 
 
 
  
 
 
 
2
2
2
2
2
,
, 2
u u y y
x yu v x x
v vx y x x
x y y y
 

 
 
 

 
2 2
2 2
2 2y x y x
x y x y
      
           
      

 
 
1 4 3
,
3
,
u v
x y
   

 

20. If (1 ),x u v y uv   compute &J J, and prove . 1J J  .
Solution:  1Given x u v and y uv  
 1
x y
v v
u u
 
  
 
 
 
,
,
x y
u u
v v
x x
x y u v
J
y yu v
u v
 
 
 
 
  
 
 
 
1 v u
v u
 

 
 
 
 
'
(1 ) ( )
, , 1
&
, ,
u v uv
u uv uv
x y u v
J u J
u v x y u
   
  
 
   
 
To prove: J .J’
= 1
 
 
 
 
'
'
, , 1
, ,
1
x y u v
J J u
u v x y u
J J
 
    
 
  
21. If sin cos , sin sin , cosx r y r z r       .Find J.
Solution:
sin cos , sin sin , cosGiven x r y r z r      
 
 
, ,
, ,
x x x
r
x y z y y y
J
r r
y y y
r
 
   
 
  
  
   
 
   
  
  
 
2 2 2 2
2 2 2 2
2 2 2 2 2 3 2 2
2 2 2
sin cos cos cos sin sin
sin sin cos sin sin cos
cos sin 0
cos ( cos sin cos cos sin sin )
sin ( sin cos sin sin )
sin cos sin cos sin (sin cos )
sin sin cos
r r
r r
r
r r
r r r
r r
r
     
     
 
      
    
      
 



 
 
 
   
  
2
sinJ r


22. Expand ( , ) xy
f x y e in Taylor’s series at (1, 1) up to
second degree.
Solution:
 
   
   
   
   2
, int 1, 1
, 1,1
, 1,1
, 1,1
, 1,1
xy
xy
xy
x x
xy
y y
xy
xx xx
Given f x y e and the po a b
f x y e f e
f x y e y f e
f x y e x f e
f x y e y f e
  
  
   
   
   

   
   2
, (1) ( ) 1,1 2
, 1,1
xy xy
xy xy
xy
yy yy
f x y e y e x f e e e
f x y e x f e
     
  
The Taylor’s series is
          
       
  
2
2
1
, , , ,
1
, 2 ,1
...
2 ,
x y
xx xy
yy
f x y f a b f a b x a f a b y b
f a b x a f a b y b x a
f a b y b
      
    
  
   
   
      
2 2
1 1 1
1
1 4 1 1 1
2
xy
x y
e e
x x y y
       
  
          
23. Find the Taylor’s series expansion of ex
sin y near the
point 1,
4
 
 
 
up to the first degree terms.
Solution:
 
 
 
1
1
1
1
, sin 1, sin
4 4 2
1
, sin 1, sin
4 4 2
1
, cos 1, cos
4 4 2
x
x
x x
x
y y
f x y e y f e
e
f x y e y f e
e
f x y e y f e
e
 
 
 



 
     
 
 
     
 
 
     
 
The required expansion is
          
1
, , , ,
1
x yf x y f a b f a b x a f a b y b      
   
 
, 1, 1 1, 1,
4 4 4 4
1
sin 1 1
42
x y
x
f x y f x f y f
e y x y
e
   

       
              
       
  
      
  
24. Write condition for finding maxima and minima.
Necessary Conditions:
The necessary conditions for f(x, y) to have a maximum
or minimum at (a, b) are that 0 0 ( , )
f f
and at a b
x y
 
 
 
Sufficient Conditions:
     , ; , ,xx xy yyLet r f a b s f a b and t f a b  
2
() 0 0i If rt s and r   at (a, b) , then f is maximum and
f (a, b) is maximum value
2
( ) 0 0 ( , )ii If rt s and r at a b   , then f is minimum and
f(a, b) is minimum value.
2
( ) 0iii If rt s  , then f is neither maximum nor
minimum
at (a, b).
(iv) If rt – s2
= 0 , in this case further investigation are
required.
25. Find the stationary points
of 3 3
( , ) 3 12 20f x y x y x y     .

Solution: Given 3 3
( , ) 3 12 20f x y x y x y    
2 2
3 3 3 12x yf x f y   
2 2
2 2
For stationary points 0, 0
3 3 0 1 1
3 12 0 1 2
x yf f
x x x
y y y
 
     
     
The stationary points are (1,2), (1,-2),(-1,2) & (-1,-2).
26. Find the stationary points of 2 2
2z x xy y x y     .
Solution: Given 2 2
2z x xy y x y    
2 2 , 2 1x yz x y z x y     
For stationary points 0, 0x yf f 
2 2 2 1x y and x y     Solving x =1, y =0
 The stationary point is (1,0)
27. Find the maximum and minimum values of
2 2
2x xy y x y   
Solution: 2 2
( , ) 2Given f x y x xy y x y    
2 2 2 1
2 2
1
x y
xx yy
xy
f x y f x y
f f
f
     
 
 
At maximum and minimum point: fx = fy = 0
(1,0) may be maximum point or minimum point.
At (1,0): fxx . fyy –( fxy)2
= 4-1 = 3 > 0 & fxx =2 > 0
(1,0) is a minimum point
 Minimum value = f(1,0) = -1
28.
A flat circular plate is heated so that the temperature
at any point (x,y) is u(x,y) = x2
+2y2
-x. Find the coldest
point on the plate.
Solution: 2 2
2Given u x y x  
2 1 4
2 4
0
x y
xx yy
xy
u x u y
u u
u
  
 

For stationary points
1
0 2 1 0
2
xu x x     
0 4 0 0yu y y    
 The point is
1
,0
2
 
 
 
At
1
,0
2
 
 
 
 
2
8 0& 2 0xx yy xy xxu u u u     
The point
1
,0
2
 
 
 
is the minimum point.
Hence the point
1
,0
2
 
 
 
is the coldest point.
29. Find the shortest distance from the origin to the curve
2 2
8 7 225x xy y   .
Solution:
Let 2 2 2 2
& 8 7 225f x y x xy y     

   2 2 2 2
8 7 225f x y x xy y       
 
 
0 1 4 0 (1)
0 4 1 7 0 (2)
x x
y y
f x y
f x y
  
  
      
      
Solving (1) & (2)  = 1,  = 1
9

2
1 2 & 5 225
( )
1 2 5, 20
9
If x y y
noreal valueof y
If y x x y


    
     
1. 1 2
3
3
x x
y
x

(12. (i) Find the Jacobian
 
 
, ,
, ,
u v w
x y z


, if
, ,x y z u y z u v z u v w     
(ii) If 2 2
, 2 . ( , ) ( , )u x y v xy f x y u v    show that
2 2 2 2
2 2
2 2 2 2
4( )
f f
x y
x y u v
     
    
    

UNIT-V
MULTIPLE INTEGRALS
PART-A
1. Evaluate
1
0 0
x
y x
dx e dy 
Sol:
Let I =
1
0 0
x
y x
e dydx
=
1
0 0
1
xy x ax
axe e
dydx e dx
x a
   
  
   
 
1
0
0
1
0
0
1
0
1
0
1
0
( )
( )
( 1)
( 1)
xy x
x x
xe dx
xe xe dx
xe x dx
x e dx
e x dx
   
 
 
 
 





12
0
( 1)
2
1
( 1) 0
2
1
2
x
e
e
e
 
   
 
 
    


2. Evaluate
1 1
b a
dxdy
xy
Sol:

   
  
1 1
1 1
1 1
log
log log
log log1 log log1
(log 0)(log 0) ( log1 0)
log log
b a
a b
a b
dxdy dx
Let I x
xy x
dx dy
x y
x y
a b
a b
a b
 
  
 
  
      

  
   

 
 


3. Evaluate
2 2
0 0
a a x
dydx

 
Sol:
 
 
2 2
2 2
0 0
0
0
2 2
0
2 2
0
2 2 2
1
0
2 2 2 2
1 1
1
1
0
sin
2 2
sin (1) 0 sin (0)
2 2 2
sin 0 0 sin (0) 0,
sin 1 sin (1)
2 2
a a x
a
a x
a
a
a
Let I dydx
y dx
a x dx
a x dx
x a x a x
a
a a a a a
 



 




  
 
 
  
  
   
          
   
 
    
 
 




 
2
2
0 0 0
2 2
4
a
a


  
     
  

4. Evaluate
1
0 0
( )
x
xy x y dxdy 
Sol:

 
   
1
0 0
1
2 2
0 0
1
2 2
0 0
1 2 2 3
0 0
2 3
21
0
( )
( ) ( )
2 3
2 3
x
x
x
x
Let I xy x y dxdy
x y xy dxdy
x y xy dydx correct form
x y xy
dx
x x x x
dx
 
 
 
 
  
 
 
  
 
  
 
 
 


 
1 3 5 2
3 2 5 2
0
1 13 5 2
0 0
1 14 7 2
0 0
2 3
2 3
1 1
2 4 3 7 2
x x
dx x x x
x x
dx dx
x x
 
   
 
 
   
    
   

 

 1 1 1 2
0 1 0
2 4 3 7
1 2
8 21
21 16
168
37
168
   
      
   
 



5. Evaluate
2
2 2
0 0
y
dxdy
x y
Sol:
2
2 2
0 0
2
2 2
0 0
2
1
1 0
2
1 1
1
2
1
1
2
1
1
tan
1
tan tan (0)
1
tan (1)
1
4
y
y
y
dxdy
Let I
x y
dx
dy
x y
x
dy
y y
y
dy
y y
dy
y
dy
y


 





  
   
  
  
   
  








 
2
1
1
4
log2 log1 log2 ( log1 0)
4 4
dy
y

 

   


6. Evaluate
2 cos
2
0 0
a
r drd
 
 
Sol:

2 cos
2
0 0
cos2 3
0 0
2 3 3
0
2 23
3
0 0
3
3
3
cos
3
1 3
......1,
2
cos cos
1 33
...... ,
2 2
3 1
3 3
2
9
a
a
n
Let I r drd
r
d
a
d
n n
if nis odd
a n n
d
n n
if niseven
n n
a
a
 


 




  


 
  
 

    
       
   
    
 
   

 


 
7. Evaluate
sin2
0 0
r d dr


 
Sol:
 
 
sin2
0 0
sin2
0 0
sin
22
0 0
22
0
22
2
0 0
2
sin
0
2
1 3
......1,
1 2
sin sin
1 32
...... ,
2 2
1
.
2
n
Let I r d dr
r dr d correct form
r
d
d
n n
if nisodd
n n
d
n n
if niseven
n n













  



 
  
 
 
  
  
    
       
   
    

 
 


 
1
.
2 2 8
 

8. Evaluate
cos
0 0
r dr d

 
Sol:

 
 
cos
0 0
cos2
0 0
2
0
2
0
2
0
0
0
2
cos
0
2
cos
2
1
cos
2
1 1 cos2
2 2
1 sin 2
4 2
1 sin 2 0 1
0 0
4 2 2 4 4
r
r
Let I r dr d
r
d
d
d
d
d













 




 
 



 
  
 
 
  
  




 
   
    
          
    
 





9. Why do we change the order of integration in multiple
integrals?Justify your answer with an example?
Sol :
Some of the problems connected with double integrals,which
seen to be complicated,can be made easy to handle by a change in
the order of integration.
Example:
0
y
x
e
dxdy is difficult to solve
y
  

But by changing the order we get,
 
0 0
0
0
0
0
( 0)
y y
y
y
y
y
e
dxdy
y
e
x dy
y
e
y dy
y
e dy
 
 
 



 
  
 
 





 
0
0
0
1
( ) (0 1) 1
y
y
e
e
e e



 
  
 
 
      
10. Express
2
2 2
0
( )
a a
y
x
dxdy
x y in polar co-ordinates
Sol:

The region of integration is bounded by
0, , , .y y a x y x a   
Let us transform this integral in polar co-ordinates by taking
cos , sin ,x r y r dxdy rdrd     .
Consider the limits , , 0x y x a y   .
0 sin 0 0,sin 0
0, 0
If y r r
r
 

     
  
cos
cos sin 1
sin
tan 1
4
cos
cos
sec
If x y r r
a
If x a r a r
r a

 







    
 
 
    
 
   
sec2 24
3 22 2 2 2
0 0 0
4 sec 3 2
3 22 2 2
0 0
4 sec
2
0 0
( cos )
cos sin
cos
(cos sin )
cos
a a a
y
a
a
x r rdrd
x y r r
r drd
r
drd


 
 
 
 
 
 
 

  
 

  

  
 
 
11 .Find dxdy over the region bounded by 0, 0, 1x y x y   
Sol:
Given 0, 0 & 1x y x y   
The region of integration is the triangle.
Here x varies from 0 1x to x y  
y varies from 0 1y to y 
 
11
0 0
1
1
0
0
1
0
12
0
(1 )
2
1 1
1
2 2
R
y
y
I dxdy
dxdy
x dy
y dy
y
y





 
 
  
 
  

 


12. Find the area of a circle of radius “a” by double integration in
polar
Co-ordinates
Sol:

The equation of circle whose radius is “a” is given by
2 cosr a 
The limits for
: 0 2 cos
: 0 2
r r to r a
to

   
 
 
2 2 cos
0 0
2 cos2 2
0 0
2
2 2
0
2
2 2
0
2
2 2
2
2
2
2
4 cos
4 cos
2 1
4
2 2
1
4
2 2
a
a
Area upper area
rdrd
r
d
a d
a d
a
a a
 





 
 



 

 
  
 


 
  
 
 
  
 
 



13. Define Area in polar Co-ordinates
Sol:
Area=
R
rdrd
14. Express the Volume bounded by
0, 0, 0 1x y z and x y z     
in triple integration.
Sol:
For the given region
2 22
2 2
2
11 1
0 0 0
var 0 1
var 0 1
var 0 1
x yx
z ies from to x y
y ies from to x
x ies from to
I dzdydx
 
 

    
15. Evaluate
2 3 2
2
0 1 1
xy z dzdydx
Sol:
2 3 2
2
0 1 1
2 3 2
2
0 1 1
2 3 22 3 2
0 1 1
2 3 2
4 27 1 4 1
0
2 3 3 2 2
26 3
(2) 26
3 2
Let I xy zdxdydz
xdx y dy zdz
x y z

     
      
    
     
      
     
   
      
   
  
   
  

  
16. Find the volume of the region bounded by the surface
2 2
,y x y x  and the planes 0, 3z z 

Sol:
 
2
2
4
4
3
(1)
(2)
(2) (1)
0
1 0
0,1
y x
x y
Substituting in we get
x x
x x
x x
x
     
     

  
  
 
 
 
2
2
2
2
1 3
0 0
1
3
0
0
1
0
1
0
1
2
0
Re
3
3
3
x
x
x
x
x
x
x
x
quired volume dzdydx
z dydx
dydx
y dx
x x dx




   
  
 
 


13 2 3
0
13 2 3
0
3
3 2 3
2
3
3 3
2(1) 1
3
3
2 1 1
x x
x x
 
  
 
 
  
 
 
   
  
17. Sketch roughly the region of integration for
1
0 0
( , ) .
x
f x y dy dx
Sol:
The region of integration is bounded by 0, 1, 0,x x y y x   
var 0 1
var 0
Here x ies from x to x
y ies from y to y x
 
 
18. Sketch the region of integration
2 2
20
.
a a x
ax x
dydx


 
Sol:
Given x varies from x = 0 to x = a
y varies from 2 2 2
y a x to y ax x   
i.e., 2 2
y x a  which is a circle with centre (0,0)
and radius a.

2 2
2 2 2
2 2
2
0
2 4
. .,
2 4
a a
x y ax x y
a a
i e x y
 
       
 
 
   
 
This is a circle with centre (a/2,0)
and radius a/2.
19. Change the order of integration in
0 0
( , )
a x
f x y dydx
Sol:
Given
0 0
( , )
a x
f x y dydx
The region of integration is bounded by
0, , 0,x x a y y x   
. ., var 0
var 0
i e x ies from x to x a represents Vertical path
y ies from y to y x represents Vertical strip
 
 
Now changing the order of integration we get
var
var 0
x ies from x y to x a represents Horizontal strip
y ies from y to y a represents Horizontal path
 
 
0 0 0
( , ) ( , )
a x a a
y
f x y dydx f x y dx dy  
20. Sketch roughly the region of integration for the following double
integral
2 2
0 0
( , )
a a x
f x y dxdy

 
Sol:
2 2
2 2 2
2 2 2
var 0
var 0
. .,
Given that x ies from x to x a
y ies from y to y a x
i e y x a
x y a
 
  
 
 
Which is a circle with centre (0,0) and radius a

21.Change the order of integration in
11
0 0
( , )
y
f x y dxdy

 
Sol:
var 0 1 . ., 1
var 0 1
Given x ies from x to x y i e x y represents Horizontal strip
y ies from y to y represents Horizontal path
    
 
The region of integration is bounded by 0, 1, 0, 1y y x x y    
var 0 1
var 0 1
x ies from x to x represents Vertical path
y ies from y to y x represents Vertical strip
 
  
After changing the order of integration limits of x and y becomes
0, 1, 0 1x x y and y x     .
11 1 1
0 0 0 0
. ., ( , ) ( , )
y x
i e f x y dxdy f x y dydx
 
   

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