Unit - 5
Vector integral calculus
The Line Integral- Let- F be vector function defined throughout some region of space and let C be any curve in that region. ṝis the position vector of a point p (x,y,z) on C then the integral ƪ F .dṝ is called the line integral of F taken over
Now, since ṝ =xi+yi+zk
And if F͞ =F1i + F2 j+ F3 K
Q1. Evaluate 
where F= cos y.i-x siny j and C is the curve y=
in the xy plae from (1,0) to (0,1)
Solution: The curve y=
i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.
= 
= 
=
=-1
Q2. Evaluate 
where 
= (2xy +z2) I +x2j +3xz2 k along the curve x=t, y=t2, z= t3 from (0,0,0) to (1,1,1).
Solution : F x dr = 
Put x=t, y=t2, z= t3
Dx=dt ,dy=2tdt, dz=3t2dt.
F x dr = 
=(3t4-6t8) dti – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k
=
t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k
=
=
+
Example 3: Prove that ͞͞͞F = [y2cos x +z3] i+(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞F (ii) the work done in moving an object in this field from (0, 1, -1) to (
/ 2,-1, 2)
Sol: (a) The field is conservative if cur͞͞͞͞͞͞F = 0.
Now, curl͞͞͞F = 
̷̷
X 
/ 
y 
/ 
z
Y2COS X +Z3 2y sin x-4 3xz2 + 2
; Cur 
= (0-0) – (3z2 – 3z2) j + (2y cos x- 2y cos x) k = 0
; F is conservative.
(b) Since F is conservative there exists a scalar potential ȸ such that
F = ȸ
(y2cos x=z3) i + (2y sin x-4) j + (3xz2 + 2) k = 
i + 
j + 
k
= y2cos x + z3, 
= 2y sin x – 4, 
= 3xz2 + 2
Now, 
= 
dx + 
dy + 
dz
= (y2cos x + z3) dx +(2y sin x – 4)dy + (3xz2 + 2)dz
= (y2cos x dx + 2y sin x dy) +(z3dx +3xz2dz) +(- 4 dy) + (2 dz)
=d(y2 sin x + z3x – 4y -2z)
ȸ = y2 sin x +z3x – 4y -2z
(c) now, work done =
.d ͞r
= 
dx + (2y sin x – 4) dy + ( 3xz2 + 2) dz
= 
(y2 sin x + z3x – 4y + 2z) (as shown above)
= [ y2 sin x + z3x – 4y + 2z ]( 
/2, -1, 2)
= [ 1 +8 
+ 4 + 4 ] – { - 4 – 2} =4
+ 15
Sums Based on Line Integral
1. Evaluate 
where 
=yz i+zx j+xy k and C is the position of the curve.
= (a cost)i+(b sint)j+ct k , from y=0 to t=π/4.
Soln. 
= (a cost)i+(b sint)j+ct k
The parametric eqn. Of the curve are x= a cost, y=b sint, z=ct (i)
=
Putting values of x,y,z from (i),
Dx=-a sint
Dy=b cost
Dz=c dt
=
=
=
= 
2. Find the circulation of 
around the curve C where 
=yi+zj+xk and C is circle 
.
Soln. Parametric eqn of circle are:
x=a cos
y=a sin
z=0
=xi+yj+zk = a cos
i + b cos
+ 0 k
d
=(-a sin
i + a cos
j)d
Circulation =
=
+zj+xk). d
=
-a sin
i + a cos
j)d
=
= 
Surface integrals-
An integral which we evaluate over a surface is called a surface integral.
Surface integral = 
Volume integrals-
The volume integral is denoted by 
And defined as-
If 
, then
Note-
If in a conservative field 
Then this is the condition for independence of path.
Example: Evaluate 
, where S is the surface of the sphere 
in the first octant.
Sol. Here-
Which becomes-
Example: Evaluate 
, where 
and V is the closed reason bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.
Sol.
Here- 4x + 2y + z = 8
Put y = 0 and z = 0 in this, we get
4x = 8 or x = 2
Limit of x varies from 0 to 2 and y varies from 0 to 4 – 2x
And z varies from 0 to 8 – 4x – 2y
So that-
So that-
Example: Evaluate 
if V is the region in the first octant bounded by 
and the plane x = 2 and 
.
Sol.
x varies from 0 to 2
The volume will be-
Stoke’s theorem (without proofs) and their verification-
If 
is any continuously differentiable vector point function and S is a surface bounded by a curve C, then-
Example-1: Verify stoke’s theorem when 
and surface S is the part of sphere 
, above the xy-plane.
Sol.
We know that by stoke’s theorem,
Here C is the unit circle- 
So that-
Now again on the unit circle C, z = 0
Dz = 0
Suppose, 
And 
Now
……………… (1)
Now-
Curl 
Using spherical polar coordinates-
………………… (2)
From equation (1) and (2), stoke’s theorem is verified.
Example-2: If 
and C is the boundary of the triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0), then evaluate 
by using Stoke’s theorem.
Sol. Here we see that z-coordinates of each vertex of the triangle is zero, so that the triangle lies in the xy-plane and 
Now,
Curl 
Curl 
The equation of the line OB is y = x
Now by stoke’s theorem,
Example-3: Verify Stoke’s theorem for the given function-
Where C is the unit circle in the xy-plane.
Sol. Suppose-
Here 
We know that unit circle in xy-plane-
Or
So that,
Now
Curl 
Now,
Hence the Stoke’s theorem is verified.
Gauss divergence theorem
If V is the volume bounded by a closed surface S and 
is a vector point function with continuous derivative-
Then it can be written as-
Where
unit vector to the surface S.
Example-1: Prove the following by using Gauss divergence theorem-
1. 
2. 
Where S is any closed surface having volume V and 
Sol. Here we have by Gauss divergence theorem-
Where V is the volume enclose by the surface S.
We know that-
= 3V
2.
Because 
Example – 2 Show that 
Sol
By divergence theorem, 
..…(1)
Comparing this with the given problem let
Hence, by (1)
………….(2)
Now ,
Hence, from (2), We get,
Example Based on Gauss Divergence Theorem
1. Show that
Soln. We have Gauss Divergence Theorem
By data, F=
=(n+3)
2. Prove that 
=
Soln. By Gauss Divergence Theorem,
=
=
=
.[
= 
Green’s theorem in a plane
If C be a regular closed curve in the xy-plane and S is the region bounded by C then,
Where P and Q are the continuously differentiable functions inside and on C.
Green’s theorem in vector form-
Example-1: Apply Green’s theorem to evaluate 
where C is the boundary of the area enclosed by the x-axis and the upper half of circle 
Sol. We know that by Green’s theorem-
And it it given that-
Now comparing the given integral-
P = 
and Q = 
Now-
and 
So that by Green’s theorem, we have the following integral-
Example-2: Evaluate 
by using Green’s theorem, where C is a triangle formed by 
Sol. First we will draw the figure-
Here the vertices of triangle OED are (0,0), (
Now by using Green’s theorem-
Here P = y – sinx, and Q =cosx
So that-
and
Now-
= 
Which is the required answer.
Example-3: Verify green’s theorem in xy-plane for 
where C is the boundary of the region enclosed by 
Sol.
On comparing with green’s theorem,
We get-
P = 
and Q = 
and 
By using Green’s theorem-
………….. (1)
And left hand side= 
………….. (2)
Now,
Along 
Along 
Put these values in (2), we get-
L.H.S. = 1 – 1 = 0
So that the Green’s theorem is verified.
References:
1. Higher Engineering Mathematics: B. S. Grewal
2. Applied Mathematics Volume I & II: J. N. Wartikar
3. Textbook of Engineering Mathematics: Bali, Iyenger (Laxmi Prakashan)
