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MODULE - VI
TOPICS IN VECTOR CALCULUS
(CONTINUED
...
If f (x, y ) and g (x, y ) are continuous and have its
first order partial derivatives on some open set containing R,
then
¢
¤
f (x, y )dx + g (x, y )dy =
C
...
-1
¡
Use Green’s theorem to evaluate C x 2 ydx + xdy where C is
the triangle with vertices (0, 0), (1, 0) and (1, 2)
...
-1
¡
Use Green’s theorem to evaluate C x 2 ydx + xdy where C is
the triangle with vertices (0, 0), (1, 0) and (1, 2)
...
:
Ans
...
Since f (x, y ) = x 2 y and g (x, y ) = x, it follows that
¢
¤
2
x ydx + xdy =
C
¢
1
[
R
¢
∂
∂ 2
(x) −
(x y )]dA
∂x
∂y
2x
(1 − x 2 )dydx
=
0
0
¢
1
(2x − 2x 3 )dx
=
0
x4
= x −
2
1
2
=
0
1
2
Qn
...
Qn
...
Fig
...
Since f (x, y ) = (xy + y 2 ) and g (x, y ) = x 2 , it follows that
¢
¤
2
2
(xy + y )dx + x dy =
C
[
R
¤
[(2x) − (x + 2y )]dxdy
=
R
¤
(x − 2y )dxdy
=
R
¢
1
¢
x
(x − 2y )dydx
=
x2
0
¢
1
xy −
=
0
∂ 2
∂
(x ) −
(xy + y 2 )]dA
∂x
∂y
x
y 2 x2
¢
1
(x 4 − x 3 )dx =
=
0
x5 x4
−
5
4
1
=
0
−1
20
ctd
...
¢
¢
¢
F
...
dr +
y =x 2
C
F
...
dr = (3x 3 + x 4 )dx
¢
¢
1
(3x 3 + x 4 )dx =
F
...
on y = x, r = xi + yj = xi + xj,
dr = (i + j)dx, F = (2x 2 )i + x 2 j, F
...
dr =
y =x
¢
1
F
...
Thus, if we wanted to emphasize the counterclockwise direction of integration, we could express the above formula as
$
¤
f (x, y )dx + g (x, y )dy =
C
[
R
∂g
∂f
−
]dA
∂x
∂y
Finding work done by using Green’s Theorem
Qn
...
Finding work done by using Green’s Theorem
Qn
...
Fig
...
Qn
...
Qn
...
Fig
...
W =
F
...
5
Use Green’s theorem to evaluate the integral C 4xydx+2xydy
where C is the rectangle bounded by x = −2, x = 4, y =
1, y = 2
Qn
...
Ans
...
6
Evaluate C xcosydx − ysinxdy where C is the square with
vertices (0, 0), (π, 0), (π, π), (0, π)
...
6
Evaluate C xcosydx − ysinxdy where C is the square with
vertices (0, 0), (π, 0), (π, π), (0, π)
...
7
Evaluate C x 2 ydx − y 2 xdy where C is the boundary of the
region in the first quadrant, enclosed between the co-ordinate
area and the circle x 2 + y 2 = 9
Qn
...
Qn
...
8
Evaluate C x 2 ydx + (y + xy 2 )dy where C is the boundary of
the region enclosed by y = x 2 and x = y 2
...
9
Qn
...
Qn
...
Ans
...
Gauss Divergence Theorem
Theorem
Let F = fi + gj + hk be a vector field with the functions
f , g , h having continuous first order partial derivaives at all
points in a solid region H having volume V whose surface S
is closed and bounded
...
ˆ dS =
n
S
...
ˆ
Qn
...
ˆ dS where F = (x 2 −
n
2
2
yz)i +(y −xz)j +(z −xy )k and S is the surface of rectangular
parallelopiped 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c
Qn
...
ˆ dS where F = (x 2 −
n
2
2
yz)i +(y −xz)j +(z −xy )k and S is the surface of rectangular
parallelopiped 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c
Fig
Ans
...
On S1 , z = 0, F = x 2 i + y 2 j − xyk, n = −k, F
...
ˆ dS1 =
n
S1
xydydx =
0
0
a2 b 2
4
On S2 , z = c, F = (x 2 − cy )i + (y 2 − cx)j + (c 2 − xy )k, n =
ˆ
k, F
...
ˆ dS2 =
n
S2
0
0
a2 b 2
4
Ans
...
On S3 , y = 0, F = x 2 i − xzj + z 2 k, n = −j, F
...
ˆ dS3 =
n
S3
xzdzdx =
0
0
a2 c 2
4
On S4 , y = b, F = (x 2 − bz)i + (b 2 − xz)j + (z 2 − bx)k, n =
ˆ
j, F
...
ˆ dS4 =
n
S4
a2 c 2
(b − xz)dzdx = acb −
4
2
0
0
2
Ans
...
On S5 , x = 0, F = −yzi + y 2 j + z 2 k, n = −i, F
...
ˆ dS5 =
n
yzdydz =
S5
0
0
b2 c 2
4
On S6 , x = a, F = (a2 − yz)i + (y 2 − az)j + (z 2 − az)k, n =
ˆ
i, F
...
ˆ dS6 =
n
S6
0
0
b2 c 2
4
Ans
...
£
2 2
2 2
2 2
2 2
F
...
2
£
Use divergence theorem to evaluate S F
...
Also verify the
result by computing the surface integral over S
...
2
£
Use divergence theorem to evaluate S F
...
Also verify the
result by computing the surface integral over S
...
Ans
...
ctd
Qn
...
4
£
Use divergence theorem to evaluate S F
...
Also verify the result
by computing the surface integral over S
...
4
£
Use divergence theorem to evaluate S F
...
Also verify the result
by computing the surface integral over S
...
Ans
...
ctd
...
5
Stoke’s Theorem
Theorem
Let S be a piecewise smooth oriented surface bounded by a
simple, closed piece wise smooth curve C with positive orientation
...
dr =
(curlF )
...
dr =
C
(curlF
...
1
Use Stoke’s theorem to evaluate C F
...
1
Use Stoke’s theorem to evaluate C F
...
Ans
...
2
Use Stoke’s theorem to evaluate C (x + y )dx + (2x − y )dy +
(y + z)dz where C is the boundary of the triangle whose
vertices are (0, 0, 0), (2, 0, 0) and (0, 3, 0)
Qn
...
Ans
...
3
£
Use Stoke’s theorem to evaluate S curlF
...
3
£
Use Stoke’s theorem to evaluate S curlF
...
Ans
...
4
Qn
...
Ans
...
ctd
...
5
Use stoke’s theorem for evaluating the line integral of F =
(x 2 + y 2 )i − 2xyj, taken round the rectangle bounded by
x = ±a, y = 0 and y = b and verify the result by finding the
integral directly
...
5
Use stoke’s theorem for evaluating the line integral of F =
(x 2 + y 2 )i − 2xyj, taken round the rectangle bounded by
x = ±a, y = 0 and y = b and verify the result by finding the
integral directly
...
Ans
...
ctd
...
ctd
...
6
Verify Stokes theorem for F = (2x − y )i − yz 2 j − y 2 zk where
S is the upper half of the sphere x 2 + y 2 + z 2 = 1 and C is
its boundary
Qn
...
Ans
...
Ans
...
Problems
Use Stoke’s theorem to evaluate C F