Unit - 4

Line integrals

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 F= F1i + F2 j+ F3 k

Note-

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 for͞͞͞ F (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 .

Sol. Parametric eqn of circle are:

x=a cos

y=a sin

z=0

=xi+yj+zk = a cosi + b cos + 0 k

d=(-a sin i + a cos j)d

Circulation = =+zj+xk). d

=-a sin i + a cos j)d

= =

Key takeaways:

Applications of line integral:

- The work done by a force

Let the force is moving a particle along a curve c from point P1 to P2 as

When denotes velocity of a fluid, then the circulation of around a closed curve c is defined by

2. Exact differential test

For =F1 + F2 + F3 , the necessary and sufficient condition that

F1 + F2 dy+ F3 ,

Be an exact differential is that F must be conservative.

3. Area of a rectangular region bounded by the curve c

Similarly

Example: Find the work done in a moving particle in the force field = + (2xz – y) along curve c defined by from x = 0 to x = 2.

Sol:

Work done

Key takeaways:

Statement- Suppose that C is a smooth curve given by . Also suppose that f is a function whose gradient vector, is continuous on C, then

Proof:

The line integral,

We use the chain rule to simplify the integral,

To finish this off, we use the fundamental theorem of calculus for single integral,

We get

Example: Evaluate where f(x, y, z) = and C is any path that starts at (1, ½, 2) and ends at (2, 1, -1).

Sol:

Let be any path that starts at (1, ½, 2) and ends at (2, 1, -1). Then,

The integral is then,

Conservative vector fields-

If then the field is said to be conservative which means no work is done in displacement from a point a to another point in the field and back to a and the mechanical energy is conserved.

Note- every irrotational field is conservative.

Irrotational fields-

An irrotational field F is characterized by any one of the conditions given below-

- along every closed surface is zero.
- , if the domain is simply connected.

Solenoidal fields-

A solenoidal field F is characterized by any one of the conditions given below-

- Flux across every closed surface is zero.

Example: Determine whether F = ( is a conservative vector field.

Sol:

As we know that F is a conservative vector field when curl F = 0, so that

Hence we can say that F is a conservative field.

Independence of path-

If in a conservative field

Along any closed curve C.

Which is the condition of the independence of path.

Note- if is a conservative vector field then is independent of path.

Example: Check whether is independent of path C.

Sol:

The line integral of F is independent of path of integration if

And

Hence the given integral is independent of path C,

Key takeaways:

- Every irrotational field is conservative.
- If in a conservative field

Along any closed curve C.

3. if is a conservative vector field then is independent of path.

Statement- 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.

Proof:

Let the equation of the curves AEB and AFB are y = f1(x) and y = f2 (x) respectively.

Consider

….(1)

Similarly, let the equations of the curve EAF and EBF be x = f1 (y) and x = f2 (y) respectively, then

…(2)

Adding equation (1) and (2), we obtain

Green’s theorem in vector form-

Note- Area of the plane region S bounded by closed curve C cam be calculated as-

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.

Key takeaways:

1. Green’s theorem in vector form-

2.

3.

Definition- An integral which is to be evaluated over a surface is called a surface integral.

Let be a continuous vector point function. Let = (u, v) be a smooth surface such

That (u, v) possesses continuous first order partial derivatives. Then the normal surface integral

Of over S is denoted by

Where d is the vector area of an element dS and is a unit vector normal to the surface dS.

Suppose F1, F2, F3 which are the functions of x, y, z be the components of F along the coordinate axes, then surface integral

Example: Evaluate

Where S is the surface of the sphere

in the first octant.

Sol:

Here

Vector normal to the surface =

Here

Now

Key takeaways:

- An integral which is to be evaluated over a surface is called a surface integral

A surface S may be represented by F(x, y, z) = 0.

The parametric representation of surface S is of the form

R(u, v) = x(u, v)I + y(u, v)J + z(u, v)K

And the continuous function of a real parameter t represent a curve C on this surface S.

Example: What will be the parametric representation of the circular cylinder

The parametric form will be

Where the parameters u and v vary in the rectangle and .

Also u = t and v = bt represent a circular helix on the this circular cylinder.

The equation of the circular helix is R = a cost I + a sint J +btK

Definition of smooth surface and piecewise smooth surface:

If S has a unique normal at each of its points whose direction depends continuously on the points of S, then the surface is called a smooth surface.

If S is not smooth but it can be divided into infinitely many smooth portions, then it is called a piecewise smooth function.

Key takeaways:

- A surface S may be represented by F(x, y, z) = 0.
- The parametric representation of surface S is of the form

R(u, v) = x(u, v)I + y(u, v)J + z(u, v)K

Statement-

If is any continuously differentiable vector point function and S is a surface bounded by a curve C, then-

Where is the unit normal vector at any point of S.

Proof:

Let S is surface such that its projection on the xy, yz and xz planes are regions

Bounded by simple closed curves. Let equation of surface f(x, y, z) = 0, can be written as

z = f1 (x, y)

y = f2 (x, z)

x = f3 (y, z)

Let

Then we need to show that

Suppose an integral , we have

….(1)

… (2)

Also

So that

But is tangent to the surface S. Hence, it is perpendicular to

So

Hence

Hence, (ii) becomes

….(3)

But on surface S

… (4)

… (5)

Hence, (iii) with the help of (v) gives

….(6)

Where R is projection of S on xy-plane.

Now, by Green’s theorem in plane, we have

Where C1 is the boundary of R.

As at each point (x, y) of the curve C1 the value of F is same as the value of F1 at each point (x, y, z) on C and dx is same for both curves. Hence, we have

Hence

… (7)

From eqns. (vi) and (vii), we have

…(8)

Similarly, taking projection on other planes, we have

…(9)

And

…. (10)

Adding equations (viii), (ix), (x), we get

Which is the proof of stoke’s theorem

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.

Key takeaways:

Gauss divergence theorem

Statement:

The surface integral of the normal component of a vector function F taken around a closed surface S is equal to the integral of the divergence of F taken over the volume V enclosed by the surface S.

Mathematically it can be written as-

Where unit vector to the surface S.

Note- Divergence theorem is the relationship between surface integral and volume integral.

Proof:

Putting the values of in the statement of the divergence theorem, we have

..(1)

We need to prove equation (1),

Let us evaluate,

…(2)

For the upper part of the surface i.e. S2, we have

Putting these values in (2), we get

…(3)

Similarly, it can be shown that

Adding (3), (4) & (5), we have

Hence proved

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,

=

= =

.[

=

Key takeaways:

References:

1. S C Mallik and S Arora: Mathematical Analysis, New Age International.

2. Publications S Ghorpade, B V Limaye, Multivariable calculus, Springer international edition.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers.

4. Advanced engineering mathematics, by HK Dass.

5. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.