Unit - 2

Vector Calculus

Q1) A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

A1)

Suppose

Now,

At t = 0 | 4 |

At t = π/2 | - 4 |

At t = 0 | |v|= |

At t = π/2 | |v|= |

Again acceleration-

Now-

At t = 0 | |

At t = π/2 | |

At t = 0 | |a|= (- 4 )2 = 4 |

At t = π/2 | |a|=(- 4 )2 = 4 |

Q2) Show that where

A2)

Here it is given-

Therefore-

Note-

Hence proved.

Q3) Write the properties of gradient?

A3)

Properties of gradient-

Property-1:

(

Proof:

First we will take left hand side

L.H.S = (

= ({

= ({

=

Now taking R.H.S,

R.H.S. = (

=

=

Here- L.H.S. = R.H.S.

Hence proved.

Property-2: Gradient of a constant (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

We know that the gradient-

= 0

Property-3: Gradient of the sum and difference of two functions-

If f and g are two scalar point functions, then

∇(f ± g) = ∇f ± ∇g

Proof:

∇(f ± g) =

L.H.S

=

=

=

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

Property-5: Gradient of the quotient of two functions-

If f and g are two scalar point functions, then-

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Q4) If , then show that

1. ∇(

2. Grad r =

A4)

Suppose

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Q5) If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

A5)

Grad f = (3x2y – y3z2)

Now grad f at (1 , -2, -1) will be-

= - 12

Q6) If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

A6)

Here- grad u =

Grad v =

Grad w =

Now-

Grad u (grad v grad w) =

Apply R2 R2 + R3

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

Q7) What is Divergence and curl?

A7)

Divergence (Definition)-

Suppose (x, y, z) is a given continuous differentiable vector function then the divergence of this function can be defined as-

∇. = . =

Curl (Definition)-

Curl of a vector function can be defined as-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Q8) Write Vector identities.

A8)

Vector identities:

Identity-1: grad uv = u grad v + v grad u

Proof:

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3 div (u = u div

Proof: div = ∇.(u

So that-

Div(u

Identity-4 div (

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5

Proof:

=

=

=

= (grad u) + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Q9) Show that-

1. Div

2. Curl

A9)

We know that-

Div

= 1 + 1 + 1 = 3

2. We know that

Curl

Q10) If then find the divergence and curl of .

A10)

We know that-

Div =

Now-

= I (2 z4 + 2x2y) – j (0 – 3z2x) + k(-4xyz – 0)

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Q11) Prove that

Note- here is a constant vector and

A11)

Here

So that

∇

Now-

So that-

Q12) Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xy plane from (1,0) to (0,1)

A12)

The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

Q13) 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).

A13)

F x dr =

Put x=t, y=t2, z= t3

Dx=dt ,dy=2tdt, dz=3t2dt.

F x dr =

=(3t4-6t8) dt i – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]

= - 1/15 i + 23/24 j + 43/60 k

Q14) Evaluate where =yzi+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.

A14)

= (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

Q15) Evaluate , where S is the surface of the sphere x2 + y2 + z2 = a2 in the first octant.

A15)

Here-

Which becomes-

Q16) Evaluate , where ∅ = 45 x2y and V is the closed reason bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.

A16)

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-

= 128

So that-

Q17) Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

A17)

x varies from 0 to 2

The volume will be-

= 180

Q18) 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 x2 + y2 = a2.

A18)

We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P = 2x2 – y2 and Q = x2 + y2

Now-

Q/x = 2x and P/x = -2y

So that by Green’s theorem, we have the following integral-

Q19) Evaluate by using Green’s theorem, where C is a triangle formed by y = 0, x = π/2, y = 2x/π

A19)

First we will draw the figure-

Here the vertices of triangle OED are (0,0), (π/2 , 0) and (π/2, π/2)

Now by using Green’s theorem-

Here P = y – sinx, and Q =cosx

So that-

Q/x = - sin x and P/x = 1

Now-

Which is the required answer.

Q20) Verify green’s theorem in xy-plane for where C is the boundary of the region enclosed by y = x2 and y2 = x

A20)

On comparing with green’s theorem,

We get-

P = 2xy – x2 and Q = x2 + y2

Q/x = 2x and P/x = 2x

By using Green’s theorem-

And left hand side =

Now,

Along C1 : y = x2 that means dy = 2x dx where x varies from 0 to 1

Along C2 : x2 = x that means 2ydy = dx or dy = dx/2x1/2 where x varies from 0 to

1

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Q21) Prove the following by using Gauss divergence theorem-

1.

2.

Where S is any closed surface having volume V and r2 = x2 + v2 + z2

A21)

Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

Div r = 3

= 3V

2.

Because

Q22) Show that

A22)

By divergence theorem, ..…(1)

Comparing this with the given problem let

Hence, by (1)

………….(2)

Now, ∇ .

,

Hence, from (2), We get,

Q23) Show that

A23)

We have Gauss Divergence Theorem

By data, F = rnr ∇.F = ∇.(rn. r)

= nrn – 2 ( n2 + y2 + z2) + 3rn

= nrn – 2r2 + 3rn

∇. rn. r = nrn + 3rn

=(n+3)rn

Q24) Prove that

A24)

By Gauss Divergence Theorem,

Q25) Verify stokes theorem when and surface S is the part of sphere x2 + y2 + z2 = 1 , above the xy-plane.

A25)

We know that by stoke’s theorem,

Here C is the unit circle-x2 + y2 = 1, z = 0

So that-

Now again on the unit circle C, z = 0

Dz = 0

Suppose, x = cos ∅ so that dx = - sin ∅. d∅

And y = sin ∅ so that dy = cos ∅. d∅

Now

……………… (1)

Now-

Using spherical polar coordinates-

= - π ………………… (2)

From equation (1) and (2), stokes theorem is verified.

Q26) 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.

A26)

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,

The equation of the line OB is y = x

Now by stokes theorem,

Q27) Verify Stoke’s theorem for the given function-

= (x+2y) dx + (y + 3x) dy

Where C is the unit circle in the xy-plane.

A27)

Suppose-

= F1dx + F2 dy + F3 dz

Here F1 = x+2y, F2 = y+ 3x, F3 = 0

We know that unit circle in xy-plane-

Or

x = cos ∅, dx = - sin ∅ d∅

y = sin ∅, dy = cos ∅ d∅

So that,

= ½ [ 2π + 0] = π

Now

Now,

Hence the Stokes theorem is verified.

Unit - 2

Vector Calculus

Q1) A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

A1)

Suppose

Now,

At t = 0 | 4 |

At t = π/2 | - 4 |

At t = 0 | |v|= |

At t = π/2 | |v|= |

Again acceleration-

Now-

At t = 0 | |

At t = π/2 | |

At t = 0 | |a|= (- 4 )2 = 4 |

At t = π/2 | |a|=(- 4 )2 = 4 |

Q2) Show that where

A2)

Here it is given-

Therefore-

Note-

Hence proved.

Q3) Write the properties of gradient?

A3)

Properties of gradient-

Property-1:

(

Proof:

First we will take left hand side

L.H.S = (

= ({

= ({

=

Now taking R.H.S,

R.H.S. = (

=

=

Here- L.H.S. = R.H.S.

Hence proved.

Property-2: Gradient of a constant (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

We know that the gradient-

= 0

Property-3: Gradient of the sum and difference of two functions-

If f and g are two scalar point functions, then

∇(f ± g) = ∇f ± ∇g

Proof:

∇(f ± g) =

L.H.S

=

=

=

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

Property-5: Gradient of the quotient of two functions-

If f and g are two scalar point functions, then-

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Q4) If , then show that

1. ∇(

2. Grad r =

A4)

Suppose

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Q5) If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

A5)

Grad f = (3x2y – y3z2)

Now grad f at (1 , -2, -1) will be-

= - 12

Q6) If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

A6)

Here- grad u =

Grad v =

Grad w =

Now-

Grad u (grad v grad w) =

Apply R2 R2 + R3

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

Q7) What is Divergence and curl?

A7)

Divergence (Definition)-

Suppose (x, y, z) is a given continuous differentiable vector function then the divergence of this function can be defined as-

∇. = . =

Curl (Definition)-

Curl of a vector function can be defined as-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Q8) Write Vector identities.

A8)

Vector identities:

Identity-1: grad uv = u grad v + v grad u

Proof:

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3 div (u = u div

Proof: div = ∇.(u

So that-

Div(u

Identity-4 div (

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5

Proof:

=

=

=

= (grad u) + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Q9) Show that-

1. Div

2. Curl

A9)

We know that-

Div

= 1 + 1 + 1 = 3

2. We know that

Curl

Q10) If then find the divergence and curl of .

A10)

We know that-

Div =

Now-

= I (2 z4 + 2x2y) – j (0 – 3z2x) + k(-4xyz – 0)

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Q11) Prove that

Note- here is a constant vector and

A11)

Here

So that

∇

Now-

So that-

Q12) Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xy plane from (1,0) to (0,1)

A12)

The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

Q13) 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).

A13)

F x dr =

Put x=t, y=t2, z= t3

Dx=dt ,dy=2tdt, dz=3t2dt.

F x dr =

=(3t4-6t8) dt i – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]

= - 1/15 i + 23/24 j + 43/60 k

Q14) Evaluate where =yzi+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.

A14)

= (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

Q15) Evaluate , where S is the surface of the sphere x2 + y2 + z2 = a2 in the first octant.

A15)

Here-

Which becomes-

Q16) Evaluate , where ∅ = 45 x2y and V is the closed reason bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.

A16)

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-

= 128

So that-

Q17) Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

A17)

x varies from 0 to 2

The volume will be-

= 180

Q18) 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 x2 + y2 = a2.

A18)

We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P = 2x2 – y2 and Q = x2 + y2

Now-

Q/x = 2x and P/x = -2y

So that by Green’s theorem, we have the following integral-

Q19) Evaluate by using Green’s theorem, where C is a triangle formed by y = 0, x = π/2, y = 2x/π

A19)

First we will draw the figure-

Here the vertices of triangle OED are (0,0), (π/2 , 0) and (π/2, π/2)

Now by using Green’s theorem-

Here P = y – sinx, and Q =cosx

So that-

Q/x = - sin x and P/x = 1

Now-

Which is the required answer.

Q20) Verify green’s theorem in xy-plane for where C is the boundary of the region enclosed by y = x2 and y2 = x

A20)

On comparing with green’s theorem,

We get-

P = 2xy – x2 and Q = x2 + y2

Q/x = 2x and P/x = 2x

By using Green’s theorem-

And left hand side =

Now,

Along C1 : y = x2 that means dy = 2x dx where x varies from 0 to 1

Along C2 : x2 = x that means 2ydy = dx or dy = dx/2x1/2 where x varies from 0 to

1

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Q21) Prove the following by using Gauss divergence theorem-

1.

2.

Where S is any closed surface having volume V and r2 = x2 + v2 + z2

A21)

Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

Div r = 3

= 3V

2.

Because

Q22) Show that

A22)

By divergence theorem, ..…(1)

Comparing this with the given problem let

Hence, by (1)

………….(2)

Now, ∇ .

,

Hence, from (2), We get,

Q23) Show that

A23)

We have Gauss Divergence Theorem

By data, F = rnr ∇.F = ∇.(rn. r)

= nrn – 2 ( n2 + y2 + z2) + 3rn

= nrn – 2r2 + 3rn

∇. rn. r = nrn + 3rn

=(n+3)rn

Q24) Prove that

A24)

By Gauss Divergence Theorem,

Q25) Verify stokes theorem when and surface S is the part of sphere x2 + y2 + z2 = 1 , above the xy-plane.

A25)

We know that by stoke’s theorem,

Here C is the unit circle-x2 + y2 = 1, z = 0

So that-

Now again on the unit circle C, z = 0

Dz = 0

Suppose, x = cos ∅ so that dx = - sin ∅. d∅

And y = sin ∅ so that dy = cos ∅. d∅

Now

……………… (1)

Now-

Using spherical polar coordinates-

= - π ………………… (2)

From equation (1) and (2), stokes theorem is verified.

Q26) 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.

A26)

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,

The equation of the line OB is y = x

Now by stokes theorem,

Q27) Verify Stoke’s theorem for the given function-

= (x+2y) dx + (y + 3x) dy

Where C is the unit circle in the xy-plane.

A27)

Suppose-

= F1dx + F2 dy + F3 dz

Here F1 = x+2y, F2 = y+ 3x, F3 = 0

We know that unit circle in xy-plane-

Or

x = cos ∅, dx = - sin ∅ d∅

y = sin ∅, dy = cos ∅ d∅

So that,

= ½ [ 2π + 0] = π

Now

Now,

Hence the Stokes theorem is verified.

Unit - 2

Unit - 2

Unit - 2

Unit - 2

Unit - 2

Vector Calculus

Q1) A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

A1)

Suppose

Now,

At t = 0 | 4 |

At t = π/2 | - 4 |

At t = 0 | |v|= |

At t = π/2 | |v|= |

Again acceleration-

Now-

At t = 0 | |

At t = π/2 | |

At t = 0 | |a|= (- 4 )2 = 4 |

At t = π/2 | |a|=(- 4 )2 = 4 |

Q2) Show that where

A2)

Here it is given-

Therefore-

Note-

Hence proved.

Q3) Write the properties of gradient?

A3)

Properties of gradient-

Property-1:

(

Proof:

First we will take left hand side

L.H.S = (

= ({

= ({

=

Now taking R.H.S,

R.H.S. = (

=

=

Here- L.H.S. = R.H.S.

Hence proved.

Property-2: Gradient of a constant (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

We know that the gradient-

= 0

Property-3: Gradient of the sum and difference of two functions-

If f and g are two scalar point functions, then

∇(f ± g) = ∇f ± ∇g

Proof:

∇(f ± g) =

L.H.S

=

=

=

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

Property-5: Gradient of the quotient of two functions-

If f and g are two scalar point functions, then-

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Q4) If , then show that

1. ∇(

2. Grad r =

A4)

Suppose

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Q5) If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

A5)

Grad f = (3x2y – y3z2)

Now grad f at (1 , -2, -1) will be-

= - 12

Q6) If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

A6)

Here- grad u =

Grad v =

Grad w =

Now-

Grad u (grad v grad w) =

Apply R2 R2 + R3

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

Q7) What is Divergence and curl?

A7)

Divergence (Definition)-

Suppose (x, y, z) is a given continuous differentiable vector function then the divergence of this function can be defined as-

∇. = . =

Curl (Definition)-

Curl of a vector function can be defined as-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Q8) Write Vector identities.

A8)

Vector identities:

Identity-1: grad uv = u grad v + v grad u

Proof:

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3 div (u = u div

Proof: div = ∇.(u

So that-

Div(u

Identity-4 div (

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5

Proof:

=

=

=

= (grad u) + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Q9) Show that-

1. Div

2. Curl

A9)

We know that-

Div

= 1 + 1 + 1 = 3

2. We know that

Curl

Q10) If then find the divergence and curl of .

A10)

We know that-

Div =

Now-

= I (2 z4 + 2x2y) – j (0 – 3z2x) + k(-4xyz – 0)

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Q11) Prove that

Note- here is a constant vector and

A11)

Here

So that

∇

Now-

So that-

Q12) Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xy plane from (1,0) to (0,1)

A12)

The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

Q13) 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).

A13)

F x dr =

Put x=t, y=t2, z= t3

Dx=dt ,dy=2tdt, dz=3t2dt.

F x dr =

=(3t4-6t8) dt i – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]

= - 1/15 i + 23/24 j + 43/60 k

Q14) Evaluate where =yzi+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.

A14)

= (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

Q15) Evaluate , where S is the surface of the sphere x2 + y2 + z2 = a2 in the first octant.

A15)

Here-

Which becomes-

Q16) Evaluate , where ∅ = 45 x2y and V is the closed reason bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.

A16)

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-

= 128

So that-

Q17) Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

A17)

x varies from 0 to 2

The volume will be-

= 180

Q18) 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 x2 + y2 = a2.

A18)

We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P = 2x2 – y2 and Q = x2 + y2

Now-

Q/x = 2x and P/x = -2y

So that by Green’s theorem, we have the following integral-

Q19) Evaluate by using Green’s theorem, where C is a triangle formed by y = 0, x = π/2, y = 2x/π

A19)

First we will draw the figure-

Here the vertices of triangle OED are (0,0), (π/2 , 0) and (π/2, π/2)

Now by using Green’s theorem-

Here P = y – sinx, and Q =cosx

So that-

Q/x = - sin x and P/x = 1

Now-

Which is the required answer.

Q20) Verify green’s theorem in xy-plane for where C is the boundary of the region enclosed by y = x2 and y2 = x

A20)

On comparing with green’s theorem,

We get-

P = 2xy – x2 and Q = x2 + y2

Q/x = 2x and P/x = 2x

By using Green’s theorem-

And left hand side =

Now,

Along C1 : y = x2 that means dy = 2x dx where x varies from 0 to 1

Along C2 : x2 = x that means 2ydy = dx or dy = dx/2x1/2 where x varies from 0 to

1

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Q21) Prove the following by using Gauss divergence theorem-

1.

2.

Where S is any closed surface having volume V and r2 = x2 + v2 + z2

A21)

Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

Div r = 3

= 3V

2.

Because

Q22) Show that

A22)

By divergence theorem, ..…(1)

Comparing this with the given problem let

Hence, by (1)

………….(2)

Now, ∇ .

,

Hence, from (2), We get,

Q23) Show that

A23)

We have Gauss Divergence Theorem

By data, F = rnr ∇.F = ∇.(rn. r)

= nrn – 2 ( n2 + y2 + z2) + 3rn

= nrn – 2r2 + 3rn

∇. rn. r = nrn + 3rn

=(n+3)rn

Q24) Prove that

A24)

By Gauss Divergence Theorem,

Q25) Verify stokes theorem when and surface S is the part of sphere x2 + y2 + z2 = 1 , above the xy-plane.

A25)

We know that by stoke’s theorem,

Here C is the unit circle-x2 + y2 = 1, z = 0

So that-

Now again on the unit circle C, z = 0

Dz = 0

Suppose, x = cos ∅ so that dx = - sin ∅. d∅

And y = sin ∅ so that dy = cos ∅. d∅

Now

……………… (1)

Now-

Using spherical polar coordinates-

= - π ………………… (2)

From equation (1) and (2), stokes theorem is verified.

Q26) 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.

A26)

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,

The equation of the line OB is y = x

Now by stokes theorem,

Q27) Verify Stoke’s theorem for the given function-

= (x+2y) dx + (y + 3x) dy

Where C is the unit circle in the xy-plane.

A27)

Suppose-

= F1dx + F2 dy + F3 dz

Here F1 = x+2y, F2 = y+ 3x, F3 = 0

We know that unit circle in xy-plane-

Or

x = cos ∅, dx = - sin ∅ d∅

y = sin ∅, dy = cos ∅ d∅

So that,

= ½ [ 2π + 0] = π

Now

Now,

Hence the Stokes theorem is verified.