Q.No: 1 (a)
King Saud University
Department of Mathematics
M-203[Final Examination]
(Differential and Integral Calculus)
(First-Semester 1428/29)
Max.Marks:50
Marking Scheme: Q.No:1[3+4+5+4], Q.No2:[4+4+4+4], Q.No:3[4+4+4+6]
Discuss the convergence of the sequence { ln( 2n)  ln( 2n  1)}.
  2n  
Lim[ln( 2n)  ln( 2n  1)]  Lim ln 
  Limln( 1)   0 .
n 
n 
  2 n  1   n 
Solution:

(b)
  1
Determine whether the series
n 1
n
n
2n  1
2
is absolutely convergent,
Conditional ly convergent, or divergent.
1
1
and 
is divergent
2n
2n  1 2n

n
n
is divergent.
   1
2
2n  1
n 1
Solution:
Since
n

2
Let us now try Alternating series test:
n 
Hence
(c)
n
0,
2n  1
x
(ii) f ( x) 
, f / ( x)  0 for x  1.
2
2x  1

is conditionally convergent.
 1n 2n

2n  1
n 1
(i) Lim
2
Find the interval of convergence and radius of convergence of the power

Series
  1
n 1
n
1
( x  1) n .
n
n3
( x  1) n1
n3n
1
1

 Lim x  1  x  1
n

1
n
n (n  1)3
n  3
3
( x  1)
1
x 1  1  x 1  3
For convergence
3
 3  x  1  3
 2  x  4 .
Check convergence at x  2 ,



1
1
n 1
2n
n
n





1
(

2

1
)

1
(
3
)

=
(Divergent)



n
n
n3
n3
n 1
n 1
n 1 n
Check convergence at x  4
n

n 3
n 1
n 1
n





1
(
4

1
)


1
   1
(Convergent)


n
n
n
n3
n3
n 1
Therefore Interval of convergence is (2,4]. Radius of Convergence =3.
Solution:
(d)
Lim
Find Maclaurin’s series for the function f ( x) 
1
1  x 2
.
Time:3hrs
f ( x) 
Solution:
1
 f (0)  1
1  x 2


2
f ( x)  1  x
 f / ( x)  (2)(1  x ) 3 (1)  f / (0)  2
f // ( x)  (2)(3)(1  x ) 4  f // (0)  6
f /// ( x)  (2)(3)(4)(1  x ) 5  f /// (0)  24
f iv ( x)  (2)(3)(4)( 5)(1  x ) 6  f /// (0)  120
and so on………….
6 2 24 3 120 5
x 
x 
x 
2!
3!
4!
f ( x)  1  2 x 
Therefore

Q.No: 2 (a)
2
0
Solution:
2


2
  Cos( y
0
2
2 y
 
)dydx 
x
0

Cos( y )dxdy 
2
2
  Cos( y
Sketch the region of integration and evaluate the integral


)dydx .
x
2
 yCos( y
0
2
2
)dy
0
1
1
 
 [ Sin   Sin (0)]  .
2
2
2
(b)
Find the area of the surface of the portion of paraboloid z  x 2  y 2 that
Is cut off by the plane z  1.
Solution:
Surface Area=

1  ( f x ) 2  ( f y ) 2 dA.
R
Here z  x  y  f ( x, y)  f x ( x, y )  2 x & f y ( x, y )  2 y .
2
2
  1  (2 x) 2  (2 y ) 2 dA.   1  4( x 2  y 2 )dA .
R
R
2
 

0

(c )
1

6
0
5
1
1

1  4r rdrd  2  1  4r 2 
12
0
2

5 1.
A lamina has the shape of the region bounded by y  a 2  x 2 and the
density is directly proportional to the distance from x-axis. Find the mass of the
lamina.
Solution:
Mass of a lamina=
a
a2  x2
a
0
  ( x, y)dA   
R
a


k
kydydx   a 2  x 2 dx
2 a
a
k a 2


x3 
 2  a  x 2 dx  k a 2 x  
3 0

2 0

2
 ka3 .
3

4
(d)
Evaluate the integral
 
0
Solution:
2
2
16 x 2 32 x  y
0

x 2  y 2  z 2 dzdydx .
x2  y2
On converting this integral into spherical we get:



2

4

32

2
4

32
 
   Sinddd = 
 
   Sinddd 
0
0
0
0
0
2
2
0

1
  4 4 2  Sindd
2
0

4
4
0

2

 1
 2
 256   Cos 0 4 d  256 
 1  d  37.49
2 0

0
On converting the given integral into Cylindrical we get:
4
 
0
Q.No: 3 (a)

2
2
16 x 2 32 x  y
0

x  y  z dzdydx 
2
2
2
x2  y2
  
0
Show that the integral
 ye
xy
32 r2
2 4
0
r 2  z 2 rdzdrd 
r
dx  ( xe xy  2 y )dy is independent of path
C
by finding the potential function.


Solution:

We need to show F ( x, y )  f x ( x, y) i  f y ( x, y) j





 F ( x, y)  ye xy i  ( xe xy  2 y) j  f x ( x, y) i  f y ( x, y) j
 f x ( x, y )  ye xy
………………………..(1)
 f y ( x, y)  xe  2 y …………………….(2)
Integrating (1) with respect to x and (2) with respect to y , we get
1
f ( x, y )  ye xy  g ( y )......................(3)
y
1
f ( x, y )  xe xy  y 2  h( y )...............(4)
x
xy
Comparing (3) and (4), we get
g ( y )   y 2  h( y )
Therefore f ( x, y)  e xy  y 2  c which is a potential function.
(b)
Evaluate the integral
y
2
dx  3xydy , where C is the boundary of the region
C
that lies inside the circle x 2  y 2  4 and outside the circle x 2  y 2  1 in
upper half plane.
Solution:
We apply Green’s theorem and we have
 (3xy) ( y 2 ) 
2
y
dx

3
xydy


R  dx  dy dA  R (3 y  2 y)dA  R ydA
C
 2


r3 
7
14
   rSin ( )rdrd      Sin ( )d   Sin ( )d 
3 1
30
3
0 1
0 

(c )
Use Stokes’s theorem to evaluate
2

 F  dr , where C is the boundary of the
C
Portion of z  4  x  y above the xy-plane oriented upward and
2

2

F ( x, y , z )  ( x 2 e x  y ) i 


y2 1 j z3 k .
Solution:
 
i



 
curl  F   

x
 
 2 x
 x e  y
We find




k

z 

z 3 

j

y
y2 1




 i (0)  j (0)  k (1)  ( M 1 i  N1 j  P1 k )
2
2
z  4  x  y  g ( x, y)  g x  2 x, g y  2 y
 
curl  F   n dS   ( M 1 g x  N1 g x  P1 )dA   (0  0  1)dA  dA
 
R
R


S
2
2
2
2
0
0
0
0

(d)
  dA    rdrd 4
Verify the divergence theorem by evaluating both the surface integral and the




Triple integral for the function F ( x, y, z )  x i  y j  z k , and the surface
S which is the solid bounded by the graphs of z  x 2  y 2 and z  4.
Solution:
 
 M N z 
F  n dS   

 dV .......(1)
y z 
 x
Q

Here we have to prove
S
L.H.S=R.H.S…………………………………………(1)
First we calculate L.H.S. of (1) i.e.

Q
 M N z 


 dA.   (1  1  1)dV
y z 
 x
Q
  dV
Q

Now we evaluate the L.H.S. of (1) i.e.


2
2
4
0
0
r2
  
rdzdrd  8

F  n dS
S
The surface S is divided into two parts namely:
S1: Plane z  4  g1 ( x, y) this portion has upper normal vector
S2: Paraboloid z  x 2  y 2  g 2 ( x, y) this portion has lower normal.
Hence,







F  n dS =  F  n dS1   F  n dS 2
S1
S
S2
  ( Mg 1x  Ng1 y  P)dA   Mg 2 x  Ng 2 y  P) dA
Rxy
Rxy
  ( Mg 1x  Ng1 y  P)dA   Mg 2 x  Ng 2 y  P) dA
Rxy
Rxy
  ( Mg 1x  Ng1 y  P)dA   Mg 2 x  Ng 2 y  P) dA
Rxy
Rxy
  ( x(0)  ( y)(0)  z )dA   2 x 2  2 y 2  z ) dA
Rxy
Rxy
  zdA   (2 x 2  2 y 2  x 2  y 2 )dA
Rxy
Rxy
2

2
 
0
 16 
0
rdrd 

( x 2  3 y 2 ) dA
Rxy
2
2
0
0
 
(r 2 Cos 2  3r 2 Sin 2 )rdrd  16  8  8