Unit 4 | unit 4 line integrals

Unit - 4

Line integrals

Q1) Define line integral.

A1)

Let- F be vector function defined throughout some region of space and let C be any curve in that region. ṝ is 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

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 for͞͞͞ F (ii) the work done in moving an object in this field from (0, 1, -1) to (/ 2,-1, 2)

A3)

(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

= 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) What is exact differentia test?

A4)

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.

Area of a rectangular region bounded by the curve c

Similarly

Q5) State and prove fundamental theorem for line integrals

A5)

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

Q6) Evaluate where f(x, y, z) = and C is any path that starts at (1, ½, 2) and ends at (2, 1, -1).

A6)

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

The integral is then,

Q7) What do you understand by conservative vector field and independence of path?

A7)

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.

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.

Q8) Check whether is independent of path C.

A8)

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

And

Hence the given integral is independent of path C,

Q9) State and prove green’s theorem

A9)

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

Q10) Evaluate by using Green’s theorem, where C is a triangle formed by

A10)

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.

Q11) Verify green’s theorem in xy-plane for where C is the boundary of the region enclosed by

A11)

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

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Q12) Define surface integral.

A12)

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

Q13) State and prove stoke’s theorem.

A13)

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

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

Q14) 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.

A14)

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

The equation of the line OB is y = x

Now by stoke’s theorem,

Q15) State and prove the divergence theorem.

A15)

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-