Unit-4
Vector integral calculus
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)
A1:
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).
A2:
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
=
=
+
Q3: 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)
A3:
(a) The fleld 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
Q4: Find the circulation of 
around the curve C where 
=yi+zj+xk and C is circle 
.
A4:
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
=
= 
Q5: 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 
A5:
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-
Q6: Verify green’s theorem in xy-plane for 
where C is the boundary of the region enclosed by 
A6:
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.
Q7: Evaluate 
, where S is the surface of the sphere 
in the first octant.
A7:
Here-
Which becomes-
Q8: Evaluate 
if V is the region in the first octant bounded by 
and the plane x = 2 and 
.
A8:
x varies from 0 to 2
The volume will be-
Q9: 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.
A9:
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,
Q10: Show that 
A10:
By divergence theorem, 
..…(1)
Comparing this with the given problem let 
Hence, by (1)
………….(2)
Now ,
Hence,from (2), Weget,
