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Mathematics Presentation

Group MembersName ID Name ID

Md. Mouuin Uddin 12206003 Md. Raziur Rahman 12206026

Md. Shakkik Zunaed 12206004 Faysal Ahmad 12206030

Kazi Nourjahan Nipun 12206005 Md. Alamgir Hossain 12206035

Abdullah Al Naser 12206006 Abu Hossain Basri 12206036

Md.Rasheduzzaman 12206007 Manik Chandra Roy 12206039

Md. Hassan Shahriar 12206013 Md. Shariful Haque Robin 12206049

Md. Ashraful Islam 12206015 Md. Saifullah 12206051

Hossain Mohammad Jakaria 12206018 Md. Mahmuddujjaman 12206062

Imrana Khaton Eva 12206019 Arun Chandra Acharjee 12206066

Md. Saddam Hussein 12206023 Md. Oliullah Sheik 12206067

Joyantika Saha 12206025 Md. Rabiul Hasan 12206069

Rajesh Chandra Barman 12206071

Topic Topic

1. Distance Between Points, Plane and Line 9. Tangent Plane and Normal Line

2. Distance Between Two Parallel Planes 10. Curl of a vector field

3. Distance Between a Point a Line in Space 11. Curl and Divergence

4. Vector Value Function 12. Line Integral

5. Velocity and Acceleration 13. Green’s Theorem

6. Tangent vector 14. Surface Integral

7. Arc Length and Curvature 15. Divergence Theorem

8. The Gradient of a Function of Two Variables 16. Stokes Theorem

Distance Between

Points

Plane and Line

Theorem: The distance between a plane and point Q

(Not in plane) is, D= 𝑃𝑟𝑜𝑗 𝑃𝑄 = 𝑃𝑄.𝑛

𝑛, where P is a

point in the plane and n is the normal in the plane.

To find a point in the plane given by, ax+by+cz+d=0

(a≠0), let y=0 and z=0, then from the equation,

ax+d=0, we can conclude that the point

( -𝑑

𝑎, 0,0 ) lies in the plane and n is the normal to the

plane.

Ex: Find the distance the point Q=( 1, 5, -5 ) and the plane

given by 3x–y+2z=6.

soln: We know that, 𝑛=< 3, -1, 2> is normal to the give

plane. To find a point in the plane, let y=0 , z=0 and obtain

the point P=( 2, 0, 0) the vector,

𝑃𝑄=< 1-2, 5-0, -4-0 >

=< -1, 5, -4 >

Using the distance formula produces,

D=𝑃𝑄 .𝑛

𝑛

=<−1, 5,−4> .< 3,−1, 2>

32+(−1)2+22

=−1 −5 −8

9+1+4

=−16

14

=16

14

=4.276 (Answer)

Ex: Find the distance the point Q=(1, 2, 3) and plane given by 2x-y+z=4.

Soln: We know that 𝑛=<2, -1, 1> is normal to the given

plane. To find a point in the plane, let y=0, z=0 and obtain

the point P=(2, 0, 0). The vector,

𝑃𝑄=<1-2, 2-0, 3-0>

=<-1, 2, 3>

Using the distance formula produce

D= 𝑃𝑄 . 𝑛

𝑛

= <−1, 2, 3> .<2,−1, 1>

22+(−1)2+12

= −2−2+3

4+1+1

= −1

6

=1

6

=0.408 (Answer)

Distance Between Two Parallel Planes

Theorem: We can determine that the distance between

the point Q = (x0, y0, z0) and the plane given by

ax+by+cz+d=0 is

D = 𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑

𝑎2+𝑏2+𝑐2

Distance between point and plane where P= (x,y,z) is a

point on the plane and d= - (ax1 + by1 + cz1) or ax1 + by1

+ cz1 +d = 0

Ex: Find the distance between the two parallel planes

given by 3x – y + 2z – 6 = 0 and 6x – 2y + 4z + 4= 0

Soln:

Let, y=0, z=0

Q=(2,0,0)

And here we given,

a=6, b= - 2, c= 4, d = 4

Distance between two parallel plane is

D=𝑎𝑥0+𝑏𝑦0+𝑐𝑧0

𝑎2+𝑏2+𝑐2

D=6∗2 + −2∗0 + 4∗0 +4

62+(−2)2+42

=12+4

36+4+16

=16

56

=16

56

=2.138 (Answer)

Ex: Find the distance between the two parallel planes

given by x – 3y + 4z =10 , x – 3y + 4z -6 =0.

Soln: Let, y=0 and z=0

Q=(10,0,0)

And here we given

X= 1, y= -3, z= 4, d= -6

Distance between two parallel

plane

D=𝑎𝑥0+𝑏𝑦0+𝑐𝑧0

𝑎2+𝑏2+𝑐2

D=0∗1 + −3∗0 + 4∗0 +(−6)

12+(−3)2+42

=10−6

1+9+16

=4

26

=4

26

=0.7845 (Answer)

Distance Between a Point a Line in Space

Theorem: The distance between a point Q and a

line in space is given by D=𝑃𝑄×𝑈

𝑈,where 𝑈 is

the directional vector for line and P is a point on the line.

Ex : Find the distance between the point P= (3,−1,4) and the line given by x =−2+3t , y = −2t, z = 1+4t.

𝑆𝑜𝑙𝑛: Using the direction number 3,−2, 4

We know that the direction vector for the line is

𝑈=< 3, −2, 4 >

To find a point on the line , let t= 0 and we obtain P = (−2 , 0, 1)

point on the line. Then 𝑃𝑄=< 3− (−2) , −1−0 , 4−1 >

=< 5, −1 ,3 > and we can form

𝑃𝑄 × 𝑈 = 𝑖 𝑗 𝑘5 −1 33 −2 4

= 𝑖( −4 + 6) − 𝑗(20−9) + 𝑘(−10 + 3)

= 2 𝑖 − 11 𝑗 − 7𝑘

=< 2, −11,−7 >

D =𝑃𝑄×𝑈

𝑈

=22+(−11)2+(−7)2

9+4+16

=4+121+49

29

=174

29

=2.45 (Answer)

Vector Value Function

A function of the form

r(t)= f(t) 𝑖 +g(t) 𝑗

and

r(t)= f(t) 𝑖 +g(t) 𝑗 + h(t)𝑘 is a vector valued function,

where the component of f, g, h are real valued

function of the parameter t,

r(t)= 𝑓 𝑡 , 𝑔(𝑡)

or

r(t)= 𝑓 𝑡 , 𝑔 𝑡 , ℎ(𝑡)

Differention of a vector valued function:

Defn: The derivative of a vector valued function r is

defined by

r´(t)= lim∆𝑡→0

𝑟 𝑡+∆𝑡 −𝑟(𝑡)

∆𝑡

for all t for which the limit exists. If r´(c) exists for all c in open interval I, then r is differentiable on the interval I. Differentiability of vector valued intervals by considering one side limits.

Ex: Find the derivative of each of the

following vector-valued function

(i) r(t)= t2 𝑖 4 𝑗

Soln:

𝑑𝑟

𝑑𝑡/ r´(t)=2t 𝑖

(ii) r(t)= 1

𝑡 𝑖 + lnt 𝑗+℮2t 𝑘

Soln: 𝑑𝑟

𝑑𝑡=

1

𝑡2 𝑖 +

1

𝑡 𝑗+ 2℮2t 𝑘

Velocity and Acceleration

Defn: If x and y are twice differentiable functions of t

and r is a vector-valued function given by r(t)= x(t) 𝑖 +

y(t) 𝑗, then the velocity vector , acceleration vector , and

speed at time t are follows

Velocity = 𝑣 𝑡 = r´(t)= x´(t) 𝑖 + y´(t) 𝑗

Acceleration n= a(t)= r´´(t)= x´´(t) 𝑖 + y´´(t) 𝑗

Speed = 𝑣 𝑡 = r´(t) = [x´(t)]2+[y´(t)]2

Example: Find the velocity and acceleration vector when t=0 and t=3 for the vector valued functionr(t)= <𝑒𝑡 𝑐𝑜𝑠𝑡, 𝑒𝑡𝑠𝑖𝑛𝑡, 𝑒𝑡 >.

r(t) = 𝑒𝑡 𝑐𝑜𝑠𝑡 𝑖 + 𝑒𝑡𝑠𝑖𝑛𝑡 𝑗 + 𝑒𝑡 𝑘

Velocity = 𝑒𝑡𝑑

𝑑𝑥𝑐𝑜𝑠𝑡 + 𝑐𝑜𝑠𝑡

𝑑

𝑑𝑥𝑒𝑡 𝑖 + 𝑒𝑡 𝑘

= (−𝑒𝑡𝑠𝑖𝑛𝑡 + 𝑒𝑡 𝑐𝑜𝑠𝑡 ) 𝑖 +(𝑒𝑡 𝑐𝑜𝑠𝑡 + 𝑒𝑡𝑠𝑖𝑛𝑡) 𝑗 + 𝑒𝑡 𝑘

= < 𝑒𝑡 𝑐𝑜𝑠𝑡 − 𝑠𝑖𝑛𝑡 , 𝑒𝑡 (𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡) , 𝑒𝑡 >

(Answer)

Speed = rˊ(t)

= 𝑒𝑡𝑐𝑜𝑠𝑡 − 𝑒𝑡 𝑠𝑖𝑛𝑡 2 + (𝑒𝑡𝑠𝑖𝑛𝑡 + 𝑒𝑡𝑐𝑜𝑠𝑡)2+(𝑒𝑡)2

= 𝑒𝑡 2 𝑐𝑜𝑠𝑡 − 𝑠𝑖𝑛𝑡 2 + 𝑒𝑡 2 𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡 2 + 𝑒𝑡 2

= 𝑒𝑡 1 − 2𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝑡 + 𝑎 + 2𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝑡 + 1

= 𝑒𝑡 3 (Answer)

Tangent vector

Defn : Let C be a smooth curve represented by r on

an open interval I . The unit tangent vector T(t) at t

is Defined to be

T(t)=𝑟 ́(𝑡)

𝑟 ́(𝑡), 𝑟 ́(𝑡) ≠ 0

Ex: Find the unit tangent vector to the curve given by

r(t)=t 𝑖 +𝑡2 𝑗 when t=1

Slon : Given

r(t)=t 𝑖 +𝑡2 𝑗𝑟 ́(𝑡)= 𝑖 + 2t 𝑗

𝑟 ́(𝑡) = 12+(2𝑡)2

= 1 + 4𝑡2

The tangen vector is

T(t)=𝑟 ́(𝑡)

𝑟 ́(𝑡)

=1

1+4𝑡2( 𝑖 + 2t 𝑗)

The unit tangent vector at t=1 is

T(1) =1

5( 𝑖 + 2 𝑗)

(Answer)

Arc Length and

Curvature

Definition of Arc Length:

The arc length of a smooth plane curve c by the

parametric equations x=x(t) and y=(t), a ≤ t ≤ b is

S= 𝑎𝑏

[𝑥′(𝑡)]2+[𝑦′(𝑡)]2 dt . Where c is given by

r(t)= x(t)î + y(t)ĵ . We can rewrite this equation for Arc

Length as, s= 𝑎𝑏𝑟′ 𝑡 𝑑𝑡

Ex: Find the arc length of the curve given by

r(t)=cost î + sint ĵ from t=0 to t=2π

Solution:

Here, x(t)=cost y(t)= sint

X’(t)=-sint y’(t)=cost

We have

S= 𝑎𝑏

[𝑥′(𝑡)]2+[𝑦′(𝑡)]2 dt

S= 02𝜋

𝑠𝑖𝑛2𝑡 + 𝑐𝑜𝑠2𝑡 dt

S= 02𝜋

1dt

S= 02𝜋1dt

S=[𝑡]02𝜋

S=2π (Answer)

Curvature: let c be a smooth curve (in the plane or in

space) given by r(s), where s is the arc length

parameter. The curvature at s is given by

K=||𝑑𝑇

𝑑𝑠||

K=||T’(s)||

Ex: Find the curvature of the line given by r(s)= (3-3

5s) î +

4

5s ĵ

Solution:

r(s) = (3-3

5s) î +

4

5s ĵ

r’(s) = ( 0-3

5)î +

4

5ĵ

||r’(s)|| = (−3

5)2+(

4

5)2

=9

25+

16

25

=25

25

=1

T(s) = 𝑟′(𝑠)

||𝑟′(𝑠)||

= −3

5î +

4

5ĵ

T’(s) = 0.î+0.ĵ

||T’(S)|| = 0

K = ||T’(S)|| = 0 (Answer)

The Gradient of a

Function of Two

Variables

Definition :

Let Z=f(x,y) be a function of x and y such that 𝑓𝑥and 𝑓𝑦 exit. Then the gradient of f, denoted by

𝛻𝑓( 𝑥, 𝑦 ) ,is the vector of

𝛻𝑓 𝑥, 𝑦 = 𝑓𝑥( x,y ) 𝑖 + 𝑓𝑦 ( x,y) 𝑗

We read 𝛻𝑓 𝑎𝑠 ′′𝑑𝑒𝑙𝑓′′. Another notation for the

gradient is grad 𝑓 𝑥, 𝑦 .

Ex:

Find the gradient of 𝑓 𝑥, 𝑦 = 𝑦 𝑙𝑛 𝑥 + 𝑥𝑦2 at

the point (1,2).

𝑺𝒐𝒍𝒏:

Grad f or 𝛻𝑓 = 𝑓𝑥 ( x,y ) 𝑖 + 𝑓𝑦 ( x,y)

𝑗…………………………(i)

Now,

𝑓𝑥=𝑦

𝑥+ 𝑦2 , 𝑓𝑦 = 𝑙𝑛 𝑥 + 2𝑥𝑦

∴ 𝑓𝑥( 1,2)= 2 + 22 = 6

∴ 𝑓𝑦 ( 1,2) = ln 1 + 2 ∙ 1 ∙ 2 = 4

From (i) we get,

𝛻𝑓 = 6 𝑖 + 4 𝑗 (Ans:)

Tangent Plane and Normal Line

Definition: Let F be differentiable at the point P= (𝑥0 ,𝑦0 ,𝑧0 ) u the surface s

given by F(x, y, z) = 0 such that 𝛻𝐹 𝑥0 ,𝑦0 ,𝑧0 ≠ 0.

1). The plane through P that is normal to 𝛻𝐹 𝑥0 ,𝑦0 ,𝑧0 is called the tangent

plane to S at P.

2). The line through P having the direction of 𝛻𝐹 𝑥0 ,𝑦0 ,𝑧0 is called the normal

line to S at P.

If F is differentiable at 𝑥0 ,𝑦0 ,𝑧0 then an equation of the tangent plane to the

surface given by

𝐹 𝑥, 𝑦, 𝑧 = 0 at 𝑥0 ,𝑦0 ,𝑧0 is

𝐹𝑥 𝑥0 ,𝑦0 ,𝑧0 (x−𝑥0)+ 𝐹𝑦 𝑥0 ,𝑦0 ,𝑧0 (y−𝑦0) +𝐹𝑧 𝑥0 ,𝑦0 ,𝑧0 (z−𝑧0)= 0.

Ex :

Find an equation of the tangent plane to the hyperboloid given by

at 𝑧2−2𝑥2 −

2𝑦2 = 12 the point (1,−1, 4) .

𝑺𝒐𝒍𝒏:

An the equation of the tangent plane to the Given hyperboloid

can be rewrite

as 𝐹 𝑥, 𝑦, 𝑧 = 12 − 𝑧2 +2𝑥2 + 2𝑦2

𝐹𝑥 𝑥, 𝑦, 𝑧 =4x ∴ 𝐹𝑥 1,−1, 4 =4

𝐹𝑦 𝑥, 𝑦, 𝑧 =4y ∴ 𝐹𝑦 1,−1, 4

= −4

𝐹𝑧 𝑥, 𝑦, 𝑧 = −2𝑧 ∴ 𝐹𝑧 1,−1, 4 =

−8

The tangent plane is ,

4(x−1) + −4 (y+1) + ( −8)(z−4) = 0

⇒4x−4−4y−4−8z+32 = 0

⇒4x−4y−8z+24 = 0 (Answer)

Curl of a vector field

Curl of a vector field:

The curl of the F(x,y,z)= M 𝑖+N 𝑗+p 𝑘 is

Curl F(x,y,z)= 𝛻 x F(x,y,z)

= (𝜕𝑝

𝜕𝑦-𝜕𝑁

𝜕𝑧) 𝑖 -(

𝜕𝑝

𝜕𝑥-𝜕𝑀

𝜕𝑧) 𝑗+(

𝜕𝑁

𝜕𝑥-𝜕𝑃

𝜕𝑦) 𝑘

If curl F=0, then F is irrational.

𝛻 x F(x,y,z)=

i 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑀 𝑁 𝑃

Example:

Show that the vector field F is irrotational where F(x,y,z)= 2xyi+ (x2+z2)j + 2zyk

Given,

F(x,y,z)= 2xyi+ (x2+z2)j + 2zyk

𝛻 x F(x,y,z)=

i 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

2xy (x2 + z2) 2zy

= i{𝜕

𝜕𝑦. 2zy-

𝜕

𝜕𝑧(x2+z2)} – j(

𝜕

𝜕𝑥.2zy -

𝜕

𝜕𝑧.2xy) +k{

𝜕

𝜕𝑥(x2+z2) -

𝜕

𝜕𝑦.2xy}

=i(2z-2z) –j(0-0) +k(2x-2x)

=0i + 0j + 0k

=0

If curl F=0, then the vector field F is irrational.

Curl and Divergence

Curl;

Curl 𝐹 = 𝛻 × 𝐹 𝑥, 𝑦, 𝑧

=𝜕

𝜕𝑥+

𝜕

𝜕𝑦+

𝜕

𝜕𝑧× 𝑖 + 𝑗 + 𝑘

If 𝑐𝑢𝑟𝑙 𝐹 = 0 𝑇ℎ𝑒𝑛 𝐹 is irrational.

Problem:

Find the curl F of the followings

F(x, y, z) = 𝑒−𝑥𝑦𝑧( 𝑖 + 𝑗 + 𝑘) and point (3,2,0)

Solution:

F(x, y, z) = 𝑒−𝑥𝑦𝑧( 𝑖 + 𝑗 + 𝑘)

F x, y, z = 𝑒−𝑥𝑦𝑧 𝑖 + 𝑒−𝑥𝑦𝑧 𝑗 + 𝑒−𝑥𝑦𝑧 𝑘

𝑐𝑢𝑟𝑙 𝐹(𝑥, 𝑦, 𝑧) = 𝛻 × 𝐹(𝑥, 𝑦, 𝑧)

=

𝑖 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑒−𝑥𝑦𝑧 𝑒−𝑥𝑦𝑧 𝑒−𝑥𝑦𝑧

= (−𝑥𝑧𝑒−𝑥𝑦𝑧 + 𝑥𝑦𝑒−𝑥𝑦𝑧) 𝑖 – −𝑦𝑧𝑒−𝑥𝑦𝑧 + 𝑥𝑦𝑒−𝑥𝑦𝑧 𝑗 +

(−𝑦𝑧𝑒−𝑥𝑦𝑧 + 𝑥𝑧𝑒−𝑥𝑦𝑧) 𝑘

At point (3, 2, 0)

Curl F= (0+6𝑒0) 𝑖 − 0 + 6𝑒0 𝑗 + (0 + 0) 𝑘

=6 𝑖 + 6 𝑗 (Ans)

Divergence:

div F = 𝛻. 𝐹(𝑥, 𝑦, 𝑧)

div F=𝜕𝑀

𝜕𝑥+

𝜕𝑁

𝜕𝑦+

𝜕𝑃

𝜕𝑧

where F(x,y,z) = M 𝑖 + 𝑁 𝑗 + 𝑃 𝑘

If div F=0, then it is said to be divergence free.

Relation between curl & divergence is div(curlF)=0.

Problem:

Find the divergence of 𝐹 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑧 𝑖 − 𝑦 𝑗 + 𝑧 𝑘 and also find the

relation between curl & divergence.

Solution:

Given that

𝐹 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑧 𝑖 − 𝑦 𝑗 + 𝑧 𝑘

𝑐𝑢𝑟𝑙 𝐹 𝑥, 𝑦, 𝑧 = 𝛻 × 𝐹 𝑥, 𝑦, 𝑧

=

𝑖 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑥𝑦𝑧 −𝑦 𝑧

= (0-0) 𝑖 − 0 − 𝑥𝑦 𝑗 + (0 − 𝑥𝑧) 𝑘

Curl F(𝑥, 𝑦, 𝑧) = 𝑥𝑦 𝑗 − 𝑥𝑧 𝑘

We know that

div F=𝜕𝑚

𝜕𝑥+

𝜕𝑛

𝜕𝑦+

𝜕𝑝

𝜕𝑧

Here, m=xyz n= -y p= z

𝜕𝑚

𝜕𝑥= 𝑦𝑧

𝜕𝑛

𝜕𝑥= −1

𝜕𝑝

𝜕𝑥= 1

𝑑𝑖𝑣𝐹 𝑥, 𝑦, 𝑧 = 𝑦𝑧 − 1 + 1 = 𝑦𝑧

At the point (1,2,1)

𝑑𝑖𝑣 𝐹 𝑥, 𝑦, 𝑧 = 2 × 1 = 2 (Ans)

Now

div (curl F) = 𝜕𝑚

𝜕𝑥+

𝜕𝑛

𝜕𝑦+

𝜕𝑝

𝜕𝑧

Here, 𝑚 = 0 𝑛 = 𝑥𝑦 𝑝 = −𝑥𝑧

𝜕𝑚

𝜕𝑥= 0

𝜕𝑛

𝜕𝑦= 𝑥

𝜕𝑝

𝜕𝑧= −𝑥

Relation between curl & divergence

div (curl F)=0 + 𝑥 − 𝑥 = 0

Line Integral

De𝑓𝑛 of line integral

If f is defined in a region containing a sooth curve c of finite

length ,then the line integral off along c is given by

𝑐

𝑓 𝑥, 𝑦 𝑑𝑠 = lim∆ →0

𝑖=1

𝑛

𝑓 𝑥𝑖, 𝑦𝑖 ∆𝑠𝑖 𝑝𝑙𝑎𝑛𝑒 𝑜𝑟

𝑐

𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 = lim∆ →0

𝑖=1

𝑛

𝑓 𝑥𝑖, 𝑦𝑖, 𝑧𝑖 ∆𝑠𝑖 𝑝𝑙𝑎𝑛𝑒

𝑠𝑝𝑎𝑐𝑒, 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡 𝑒𝑥𝑖𝑠𝑡𝑠

To evaluate the line integral over a plane curve c given by

r(t)=x(t)i + y(t)j, use the fate that

ds= 𝑟′(𝑡) 𝑑𝑡

= 𝑥′(𝑡) 2 + 𝑦′(𝑡) 2 dt

A similar formula holds for a space curve as indicated in

the following theorem.

Evaluation of line integral as a definite integral :

Let f be continuous in a region containing a smooth curve c .If c is

given by r(t)=x(t)i + y(t)j, where ,a st sd, then.

𝑐

𝑓 𝑥, 𝑦 𝑑𝑠 = 𝑎

𝑏

𝑓(𝑥 𝑡 , 𝑦 𝑡 ) (𝑥′ 𝑡 + 𝑦 𝑡 )2

If c is given by

r(t)=x(t)i + y(t)j + z(t)k where a≤ 𝑡 ≤ 𝑏 𝑡ℎ𝑒𝑛,

𝑐 𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 = 𝑎𝑏𝑓(𝑥, 𝑦, 𝑧) ((𝑥′ 𝑡 )2+(𝑦′ 𝑡 )2 + (𝑧′(𝑡))2

Ex-Evaluate 𝑐 𝑥2 − 𝑦 + 3𝑧 𝑑𝑠,where c is the line segment show

Sol-To find the parametric from of equation of a line

X=t,

Y=2t,

Z=t,

Again

X’(t)=1,

Y’(t)=2,

Z’(t)=1,

Hear

12 + 22 + 12

= 6

Thus line integral take the fowling form 𝑐 𝑥2 − 𝑦 + 3𝑧 𝑑𝑠,where c is a line

segment

= 01𝑡2 − 𝑡 + 3𝑡 6 𝑑𝑡

= 6𝑡3

3−

2𝑡2

𝑡+

3𝑡2

2 0

1

= 61

3−

2

2+

3

2– (0 − 0 + 0)

= 6(2−6+9

6)

=6∗5

6

=5

6(Ans)

Green’s Theorem

Let R be a simply connected region with a piecewise

smooth bounding c oriented counterclockwise (that is , c

is traversed once so that the region R always lies to the

left). If M and N have continuous partial derivatives in an

open region containing R , then

𝑐 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 𝑅 (𝜕𝑁

𝜕𝑥−

𝜕𝑀

𝜕𝑦)𝑑𝐴

Ex : Evaluate 𝑐 (𝑡𝑎𝑛−1𝑥 + 𝑦2)𝑑𝑥 + (𝑒𝑦 − 𝑥2)𝑑𝑦

Where c is the pat enclosing the annular region.

𝑺𝒐𝒍𝒏: In polar coordinate

dA=rdrd𝜃

1≤r≤3 0≤ 𝜃 ≤ 𝜋

Here, M = (𝑡𝑎𝑛−1𝑥 + 𝑦2 ) , N = 𝑒𝑦 − 𝑥2

𝜕𝑀

𝜕𝑦= 2𝑦

𝜕𝑁

𝜕𝑥=−2x

x =rcos𝜃 , y =rsin𝜃

𝜕𝑁

𝜕𝑥−

𝜕𝑀

𝜕𝑦= −2x−2y

= −2(x+y)

= −2(rcos𝜃 + rsin𝜃)

= −2r(cos𝜃 + sin𝜃)

According to the green theorem,

𝑐 (𝑡𝑎𝑛−1𝑥 + 𝑦2)𝑑𝑥 + (𝑒𝑦 − 𝑥2)𝑑𝑦 = 𝑅 (𝜕𝑁

𝜕𝑥−

𝜕𝑀

𝜕𝑦)𝑑𝐴

= − 0

𝜋

1

3

2𝑟2 (cos𝜃 + sin𝜃) 𝑑𝑟𝑑𝜃

= 0

𝜋

−2𝑟3

3

3

(cos𝜃 + sin𝜃)𝑑𝜃

= 0

𝜋 −2 × 33

3+2 × 13

3(cos𝜃 + sin𝜃)𝑑𝜃

= 0

𝜋

−18 +2

3(cos𝜃 + sin𝜃)𝑑𝜃

= 0

𝜋

−52

3(cos𝜃 + sin𝜃)𝑑𝜃

= −52

3sin𝜃 − cos𝜃 0

𝜋

= −52

3sin𝜋 − cos𝜋 − sin0 + cos0

= −52

30 − 1 − 0 + 1

= −52

3× 2

= −104

3

Surface Integral

Theorem: Let, S be a surface equation z = g(x,y) and R its projection on the xy -

plane. If g , 𝑔𝑥and 𝑔𝑦 are continuous on R and f is continuous on S, then the surface

integral of f over S

is 𝑆𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 = 𝑅

𝑓 𝑥, 𝑦, 𝑔 𝑥, 𝑦 . 1 + [𝑔𝑥(𝑥, 𝑦)]2 + [𝑔𝑦(𝑥, 𝑦)]

2 𝑑𝐴

If, S is the graph of y = g(x,z) and R is its projection on the xz - plane, then

𝑆

𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 =

𝑅

𝑓 𝑥, 𝑧, 𝑔 𝑥, 𝑧 . 1 + [𝑔𝑥(𝑥, 𝑧)]2 + [𝑔𝑧(𝑥, 𝑧)]

2 𝑑𝐴

If, S is the graph of x = g(y,z) and R is its projection on the yz - plane, then

𝑆

𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 =

𝑅

𝑓 𝑦, 𝑧, 𝑔 𝑦, 𝑧 . 1 + [𝑔𝑦(𝑦, 𝑧)]2 + [𝑔𝑧(𝑦, 𝑧)]

2 𝑑𝐴

Example: Evaluate the surface integral,

𝑆(𝑦2 + 2𝑦𝑧)𝑑𝑠 , where S is the first-octant portion of the

2x + y + 2z = 6 .

Solution:

We can re-write as,

2z = 6 − 2𝑥 − 𝑦 or, z = 1

26 − 2𝑥 − 𝑦 or, g(x,y) =

1

2(6 − 2𝑥 − 𝑦)

∴ 𝑔𝑥 𝑥, 𝑦 = −1

𝑔𝑦 𝑥, 𝑦 = −1

2

Now, 1 + [𝑔𝑥(𝑥, 𝑦)]2 + [𝑔𝑦(𝑥, 𝑦)]

2

= 1 + (−1)2+(−1

2)2

= 1 + 1 +1

4

=4+4+1

4

=9

4

=3

2

For limit,

z = 1

2(6 − 2𝑥 − 𝑦)

or, 0 = 1

2(6 − 2𝑥 − 𝑦)

or, 0 = 6 − 2𝑥 − 𝑦

or, y = 6 − 2𝑥

or, y = 2(3 − 𝑥)

Again,

y = 2 3 − 𝑥

or, o = 2(3 − 𝑥)

or, o = 3 − 𝑥

or, x = 3

So, the limit is,

0 ≤ 𝑦 ≤ 2(3 − 𝑥) and 0 ≤ 𝑥 ≤ 3

∴

𝑆

(𝑦2 + 2𝑦𝑧)𝑑𝑠 =3

2

𝑅

𝑦2 + 6𝑦 − 2𝑥𝑦 − 𝑦2 𝑑𝐴

= 3

2 𝑜3 02(3−𝑥)

(6𝑦 − 2𝑥𝑦) 𝑑𝑦 𝑑𝑥

= 3 03[3𝑦2

2−

𝑥𝑦2

2]0

2(3−𝑥)

𝑑𝑥

= 3 03[3

22 3 − 𝑥 2 −

𝑥

2{2 3 − 𝑥 }2] 𝑑𝑥

= 3 03[6(9 − 6𝑥 + 𝑥2) − 2𝑥(9 − 6𝑥 + 𝑥2)] 𝑑𝑥

= 3 03[54 − 36𝑥 + 6𝑥2 − 18𝑥 + 12𝑥2 − 2𝑥3] 𝑑𝑥

= 3 03[−2𝑥3 + 18𝑥2 − 54𝑥 + 54] 𝑑𝑥

= 3 [−2𝑥4

4+

18𝑥3

3−

54𝑥2

2+ 54]0

3𝑑𝑥

= 3 [−81

2+ 162 − 243 + 162]

= 121.5 (Answer)

Divergence Theorem

Theorem:

Let, Q be a solid region bounded by a

closed surface S oriented by a unit normal vector

directed outward from Q. If f is a vector field

whose component functions have partial

derivatives in Q, then

𝑆

𝐹.𝑁𝑑𝑠 =

𝑄

𝑑𝑖𝑣 𝐹. 𝑑𝑣

Example :

Let, Q be the solid region bounded by the co-ordinate planes and

the plane 2x + 2y + z = 6 and let, x i +𝑦2 𝑗 + 𝑍 𝑘 . Find

𝑆

𝐹 .𝑁𝑑𝑠 .

Where, S is the surface of Q.

Solution :

Here, M = x ∴𝜕𝑀

𝜕𝑥= 1

N = 𝑦2 ∴𝜕𝑁

𝜕𝑦= 2𝑦

P = Z ∴𝜕𝑃

𝜕𝑧= 1

∴ 𝑑𝑖𝑣 𝐹 =𝜕𝑀

𝜕𝑥+

𝜕𝑁

𝜕𝑦+

𝜕𝑃

𝜕𝑧

= 1 + 2𝑦 + 1

= 2 + 2𝑦

For limit, 𝑍 = 6 − 2𝑥 − 2𝑦

or, 0 = 6 − 2𝑥 − 2𝑦

or, 2𝑦 = 6 − 2𝑥

or, 𝑦 = 3 − 𝑥

Again, 𝑦 = 3 − 𝑥

or, 0 = 3 − 𝑥

or, 𝑥 = 3

So, the limit is,

0 ≤ 𝑧 ≤ 6 − 2𝑥 − 2𝑦 , 0 ≤ 𝑦 ≤ 3 − 𝑥 𝑎𝑛𝑑 0 ≤ 𝑥 ≤ 3

∴

𝑆

𝐹.𝑁𝑑𝑠 =

𝑄

𝑑𝑖𝑣 𝐹. 𝑑𝑣

= 03 03−𝑥

06−2𝑥−2𝑦

(2 + 2𝑦) 𝑑𝑧 𝑑𝑦 𝑑𝑥

= 03 03−𝑥

[2𝑧 + 2𝑦𝑧]06−2𝑥−2𝑦

𝑑𝑦 𝑑𝑥

= 03 03−𝑥

[2 6 − 2𝑥 − 2𝑦 + 2𝑦 6 − 2𝑥 − 2𝑦 ] 𝑑𝑦 𝑑𝑥

= 03 03−𝑥

[12 − 4𝑥 − 4𝑦 + 12𝑦 − 4𝑥𝑦 − 4𝑦2] 𝑑𝑦 𝑑𝑥

= 03 03−𝑥

[−4𝑦2 − 4𝑥 + 8𝑦 − 4𝑥𝑦 + 12] 𝑑𝑦 𝑑𝑥

= 03[−

4𝑦3

3− 4𝑥𝑦 +

8𝑦2

2−

4𝑥𝑦2

2+ 12𝑦]0

3−𝑥𝑑𝑥

= 03[−

4

33 − 𝑥 3 − 4𝑥 3 − 𝑥 + 4 3 − 𝑥 2 − 2𝑥 3 − 𝑥 2 + 12 3 − 𝑥 ] 𝑑𝑥

= 03[−

4

327 − 27𝑥 + 9𝑥2 − 𝑥3 − 12𝑥 + 4𝑥2 + 4 9 − 6𝑥 + 𝑥2 −

2𝑥 9 − 6𝑥 + 𝑥2 + 36 − 12𝑥]

=

03[−36 + 36𝑥 − 12𝑥2 +

4

3𝑥3 − 12𝑥 + 4𝑥2 + 36 − 24𝑥 + 4𝑥2 − 18𝑥 +

= 03[+

4

3𝑥3 − 2𝑥3 + 8𝑥2 − 30𝑥 + 36] 𝑑𝑥

= [4𝑥4

3.4−

2𝑥4

4+

8𝑥3

3−

30𝑥2

2+ 36𝑥 ]0

3

= [𝑥4

3−

𝑥4

2+

8𝑥3

3− 15𝑥2 + 30𝑥 ]0

3

= [34

3−

34

2+

8∗33

3− 15 ∗ 32 + 36 ∗ 3 ]

= 27 − 40.5 + 72 − 135 + 108

= 31.5 (Answer)

Divergence Theorem

Theorem: Let Q be a solid region bounded by a closed

surface S oriented by a unit normal vector directed

outward from Q . If F is a vector field whose

component functions have conditions partial derivative

in Q . Then,

𝐹.𝑁 𝑑𝑠 = 𝑑𝑖𝑣 𝐹. 𝑑𝑣

Example: Let Q be a solid region bounded by the co-ordinate plane and

the plane 2x+2y+z=6 and let,

F= x 𝑖+𝑦2 𝑗+z 𝑘 Find 𝐹.𝑁 𝑑𝑠 where S is the surface of Q .

Solution:

M=x 𝜕𝑀

𝜕𝑥= 1

N=𝑦2𝜕𝑁

𝜕𝑦= 2y

P=z 𝜕𝑃

𝜕𝑧=1

div F = 1+2y+1

=2+2y

𝐹.𝑁 𝑑𝑠 = 𝑑𝑖𝑣 𝐹. 𝑑𝑣

= 03 03−𝑥

06−2𝑥−2𝑦

(2 − 2𝑦) 𝑑𝑧𝑑𝑦𝑑𝑥

= 03 03−𝑥

2𝑧 + 2𝑦𝑧 06−2𝑦−2𝑧

=4 03 03−𝑥

(1 + 𝑥)(3 − 𝑥 − 𝑦)𝑑𝑦𝑑𝑥

=4 03 03−𝑥

3 − 𝑥 − 𝑦 + 𝑥𝑦 − 𝑦2 𝑑𝑦𝑑𝑥

=4 033𝑦 − 𝑥𝑦 − 𝑦2 −

𝑥

2𝑦2 −

𝑦3

3 0

3−𝑥

𝑑𝑥

=4 03[3 3 − 𝑥 − 𝑥 3 − 𝑥 − 3 − 𝑥 2 −

𝑥

23 − 𝑥 2 −

1

3(3 − 𝑥)3]𝑑𝑥

=4 03[9 − 3𝑥 + 𝑥2 − 9 + 6𝑥 − 𝑥3 −

9𝑥−6𝑥2+𝑥3

2−

27−𝑥3−27𝑥+9𝑥2

3]𝑑𝑥

=4 03[9𝑥

2− 9 −

𝑥3

6]𝑑𝑥

=4 03 9𝑥2

4− 9𝑥 −

𝑥4

24 0

3

=4(81

4− 27 −

91

6)

=81−108−91

6

=−253

6

=−42.16(Ans)

STOKES THEOREM

Theorem: Let s be an oriented surface with unit normal

vector n, bounded by a piecewise smoth, simple closed

curve c.

If F is a vector field whose component function have

continuous partial derivatives on an open region

containing S and C then we can write according to stocks

theorem

𝑐 𝐹. 𝑑𝑟 = 𝑠𝑐𝑢𝑟𝑙 𝐹 . 𝑛. 𝑑𝑠

Example Let C be the oriented triangle ying in the plane 2x+3y+z=6 as

shown in evaluating 𝑐 𝐹 𝑑𝑟 when

E(x,y,z) = -y2 i + z j +x k =2x+2y+

Curl F =

𝑖 𝑗 𝑘𝑑

𝑑𝑥

𝑑

𝑑𝑦

𝑑𝑑𝑧

−𝑦2 𝑧 𝑥

= i(-1) -j (1) -1k (0+2y)

Consider z= 6-2x-2y=g(x,y)

We have 𝑠𝐹.𝑁𝑑𝑠 = 𝑅

𝐹. −𝑔𝑥 𝑥, 𝑦 𝑖 − 𝑔𝑦 𝑥, 𝑦 𝑗 + 𝑘 𝑗𝐴

For an upcurved normal vector tobe obtain here

𝑔𝑥= -2

𝑔𝑦 =-2

N=(2i+2j+k)

Then 𝑠𝐹.𝑁𝑑𝑠 = 𝑅

( −𝑖 − 𝑗 + 2𝑦𝑘 . (2𝑖 + 2𝑗 + 𝑘))𝑗𝐴

= 03 03−𝑥

−2 − 2 + 2𝑦 𝑑𝑦𝑑𝑥

= 03𝑦2 − 4𝑦

= 03(3 − 𝑥2 − 4(3 − 𝑥) 𝑑𝑥

= 03(9 − 6𝑥 + 𝑥2 − 12 + 4𝑥 dx

= 03𝑥2 − 2𝑥 − 3 𝑑𝑥

=𝑥3

3− 𝑥2 − 3𝑥

=33

3− 32 − 3.3

= 9- 9- 9

=-9 (Answer)