Presentation on calculus

Welcome to Our
Mathematics Presentation

Group Members
Name 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= 푃푟표푗 푃푄 =
푃푄.푛
푛
, 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
( -
푑
푎
, 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, 푛=< 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,
푃푄=< 1-2, 5-0, -4-0 >
=< -1, 5, -4 >

Using the distance formula produces,
D=
푃푄 .푛
푛
=
< −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 푛=<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,
푃푄=<1-2, 2-0, 3-0>
=<-1, 2, 3>

Using the distance formula produce
D=
푃푄 . 푛
푛
=
<−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
푎푥0+푏푦0+푐푧0+푑
D =
푎2+푏2+푐2
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
푎푥0+푏푦D=
0+푐푧0
푎2+푏2+푐2
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
푎푥0+푏푦D=
0+푐푧0
푎2+푏2+푐2
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
푃푄×푈
line in space is given by D=
푈
,where 푈 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.
푆표푙푛: Using the direction number 3,−2, 4
We know that the direction vector for the line is
푈=< 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 푃푄=< 3− (−2) , −1−0 , 4−1 >
=< 5, −1 ,3 > and we can form

푃푄 × 푈 =
푖 푗 푘
5 −1 3
3 −2 4
= 푖 ( −4 + 6) − 푗 (20−9) +푘 (−10 + 3)
= 2 푖 − 11푗 − 7푘
=< 2, −11, −7 >
D =
푃푄×푈
푈
=
22+(−11)2+(−7)2
9+4+16
=
4+121+49
29
=
174
29
=2.45 (Answer)

A function of the form
r(t)= f(t) 푖 +g(t) 푗
and
r(t)= f(t) 푖 +g(t) 푗 + h(t)푘 is a vector valued function,
where the component of f, g, h are real valued
function of the parameter t,
r(t)= 푓 푡 , 푔(푡)
or
r(t)= 푓 푡 , 푔 푡 , ℎ(푡)

Differention of a vector valued function:
Defn: The derivative of a vector valued function r is
defined by
r´(t)= lim
Δ푡→0
푟 푡+Δ푡 −푟(푡)
Δ푡
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 푖 4 푗
Soln:
푑푟
/ r´(t)=2t 푖
푑푡
(ii) r(t)=
1
푡
푖 + lnt 푗+℮2t 푘
Soln:
푑푟
=
푑푡
1
푡2 푖 +
1
푡
푗+ 2℮2t 푘

Defn: If x and y are twice differentiable functions of t
and r is a vector-valued function given by r(t)= x(t) 푖 +
y(t) 푗, then the velocity vector , acceleration vector , and
speed at time t are follows
Velocity = 푣 푡 = r´(t)= x´(t) 푖 + y´(t) 푗
Acceleration n= a(t)= r´´(t)= x´´(t) 푖 + y´´(t) 푗
Speed = 푣 푡 = 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 function
r(t)= <푒푡 푐표푠푡, 푒푡푠푖푛푡, 푒푡 >.
r(t) = 푒푡 푐표푠푡 푖 + 푒푡푠푖푛푡 푗 + 푒푡 푘
Velocity = 푒푡 푑
푑푥
푐표푠푡 + 푐표푠푡
푑
푑푥
푒푡 푖 + 푒푡 푘
= (−푒푡푠푖푛푡 + 푒푡 푐표푠푡 ) 푖 +(푒푡 푐표푠푡 + 푒푡푠푖푛푡) 푗 + 푒푡 푘
= < 푒푡 푐표푠푡 − 푠푖푛푡 , 푒푡 (푠푖푛푡 + 푐표푠푡) , 푒푡 >
(Answer)

Speed = rˊ(t)
= 푒푡 푐표푠푡 − 푒푡 푠푖푛푡 2 + (푒푡푠푖푛푡 + 푒푡푐표푠푡)2+(푒푡)2
= 푒푡 2 푐표푠푡 − 푠푖푛푡 2 + 푒푡 2 푠푖푛푡 + 푐표푠푡 2 + 푒푡 2
= 푒푡 1 − 2푠푖푛푡 푐표푠푡 + 푎 + 2푠푖푛푡 푐표푠푡 + 1
= 푒푡 3 (Answer)

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)=
푟́(푡)
푟́(푡)
, 푟́(푡) ≠ 0

Ex: Find the unit tangent vector to the curve given by
r(t)=t 푖 +푡2 푗 when t=1
Slon : Given
r(t)=t 푖 +푡2 푗
푟́(푡)= 푖 + 2t 푗
푟́(푡) = 12+(2푡)2
= 1 + 4푡2

The tangen vector is
T(t)=
푟́(푡)
푟́(푡)
=
1
1+4푡2 (푖 + 2t 푗 )
The unit tangent vector at t=1 is
T(1) =
1
5
(푖 + 2 푗 )
(Answer)

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
푏
S= 푎
[푥′(푡)]2+[푦′(푡)]2 dt . Where c is given by
r(t)= x(t)î + y(t)ĵ . We can rewrite this equation for Arc
푏
Length as, s= 푟′ 푡 푑푡
푎

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
푏
S= 푎
[푥′(푡)]2+[푦′(푡)]2 dt

2휋
S= 0
푠푖푛2푡 + 푐표푠2푡 dt
2휋
S= 0
1dt
2휋
S= 0
1dt
S=[푡]0 2
휋
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
푑푇
푑푠
K=||
||
K=||T’(s)||

Ex: Find the curvature of the line given by r(s)= (3-
3
5
s) î +
4
5
s ĵ
Solution:
3
5
r(s) = (3-
s) î +
4
5
s ĵ
r’(s) = ( 0-
3
5
)î +
4
5
ĵ
||r’(s)|| = (−
3
5
)2+(
4
5
)2
=
9
25
+
16
25

=
25
25
=1
T(s) =
푟′(푠)
||푟′(푠)||
= −
3
5
î +
4
5
ĵ
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 푓푥
and 푓푦 exit. Then the gradient of f, denoted by
훻푓( 푥, 푦 ) ,is the vector of
훻푓 푥, 푦 = 푓푥( x,y )푖 + 푓푦 ( x,y) 푗
We read 훻푓 푎푠 ′′푑푒푙푓′′. Another notation for the
gradient is grad 푓 푥, 푦 .

Ex:
Find the gradient of 푓 푥, 푦 = 푦 푙푛 푥 + 푥푦2 at
the point (1,2).
푺풐풍풏:
Grad f or 훻푓 = 푓푥 ( x,y )푖 + 푓푦 ( x,y)
푗 …………………………(i)
Now,
푓푥=
푦
푥
+ 푦2 , 푓푦 = 푙푛 푥 + 2푥푦
∴ 푓푥( 1,2)= 2 + 22 = 6

∴ 푓푦 ( 1,2) = ln 1 + 2 ∙ 1 ∙ 2 = 4
From (i) we get,
훻푓 = 6푖 + 4푗 (Ans:)

Tangent Plane and
Normal Line

Definition: Let F be differentiable at the point P= (푥0 ,푦0 ,푧0 ) u the surface s
given by F(x, y, z) = 0 such that 훻퐹 푥0 ,푦0 ,푧0 ≠ 0.
1). The plane through P that is normal to 훻퐹 푥0 ,푦0 ,푧0 is called the tangent
plane to S at P.
2). The line through P having the direction of 훻퐹 푥0 ,푦0 ,푧0 is called the normal
line to S at P.
If F is differentiable at 푥0 ,푦0 ,푧0 then an equation of the tangent plane to the
surface given by
퐹 푥, 푦, 푧 = 0 at 푥0 ,푦0 ,푧0 is
퐹푥 푥0 ,푦0 ,푧0 (x−푥0)+ 퐹푦 푥0 ,푦0 ,푧0 (y−푦0) +퐹푧 푥0 ,푦0 ,푧0 (z−푧0)= 0.

Ex :
Find an equation of the tangent plane to the hyperboloid given by
at 푧2−2푥2 −
2푦2 = 12 the point (1,−1, 4) .
푺풐풍풏:
An the equation of the tangent plane to the Given hyperboloid
can be rewrite
as 퐹 푥, 푦, 푧 = 12 − 푧2 +2푥2 + 2푦2

퐹푥 푥, 푦, 푧 =4x ∴ 퐹푥 1, −1, 4 =4
퐹푦 푥, 푦, 푧 =4y ∴ 퐹푦 1, −1, 4
= −4
퐹푧 푥, 푦, 푧 = −2푧 ∴ 퐹푧 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:
The curl of the F(x,y,z)= M푖 +N푗 +p푘 is
Curl F(x,y,z)= 훻 x F(x,y,z)
= (
휕푝
휕푦
-
휕푁
휕푧
) 푖 -(
휕푝
휕푥
-
휕푀
휕푧
) 푗 +(
휕푁
휕푥
-
휕푃
휕푦
) 푘
If curl F=0, then F is irrational.
훻 x F(x,y,z)=
i 푗 푘
휕
휕
휕푥
휕푦
휕
휕푧
푀 푁 푃

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
훻 x F(x,y,z)=
i 푗 푘
휕
휕
휕푥
휕푦
휕
휕푧
2xy (x2 + z2) 2zy
= i{
휕
휕푦
휕
휕푧
. 2zy-
(x2+z2)} – j(
휕
휕푥
.2zy -
휕
휕푧
.2xy) +k{
휕
휕푥
휕
휕푦
(x2+z2) -
.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;
Curl 퐹 = 훻 × 퐹 푥, 푦, 푧
=
휕
휕푥
+
휕
휕푦
+
휕
휕푧
× 푖 + 푗 + 푘
If 푐푢푟푙 퐹 = 0 푇ℎ푒푛 퐹 is irrational.

Problem:
Find the curl F of the followings
F(x, y, z) = 푒−푥푦푧(푖 + 푗 + 푘 ) and point (3,2,0)
Solution:
F(x, y, z) = 푒−푥푦푧(푖 + 푗 + 푘 )
F x, y, z = 푒−푥푦푧푖 + 푒−푥푦푧푗 + 푒−푥푦푧푘

푐푢푟푙 퐹(푥, 푦, 푧) = 훻 × 퐹(푥, 푦, 푧)
=
푖 푗 푘
휕
휕
휕푥
휕푦
휕
휕푧
푒−푥푦푧 푒−푥푦푧 푒−푥푦푧
= (−푥푧푒−푥푦푧 + 푥푦푒−푥푦푧)푖 – −푦푧푒−푥푦푧 + 푥푦푒−푥푦푧 푗 +
(−푦푧푒−푥푦푧 + 푥푧푒−푥푦푧) 푘
At point (3, 2, 0)
Curl F= (0+6푒0)푖 − 0 + 6푒0 푗 + (0 + 0)푘
=6푖 + 6푗 (Ans)

Divergence:
div F = 훻. 퐹(푥, 푦, 푧)
div F=
휕푀
휕푥
+
휕푁
휕푦
+
휕푃
휕푧
where F(x,y,z) = M푖 + 푁푗 + 푃푘
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 퐹 푥, 푦, 푧 = 푥푦푧 푖 − 푦푗 + 푧푘 and also find the
relation between curl & divergence.
Solution:
Given that
퐹 푥, 푦, 푧 = 푥푦푧 푖 − 푦푗 + 푧푘
푐푢푟푙 퐹 푥, 푦, 푧 = 훻 × 퐹 푥, 푦, 푧
=
푖 푗 푘
휕
휕
휕푥
휕푦
휕
휕푧
푥푦푧 −푦 푧

= (0-0)푖 − 0 − 푥푦 푗 + (0 − 푥푧)푘
Curl F(푥, 푦, 푧) = 푥푦푗 − 푥푧푘
We know that
div F=
휕푚
휕푥
+
휕푛
휕푦
+
휕푝
휕푧
Here, m=xyz n= -y p= z
휕푚
휕푥
= 푦푧
휕푛
휕푥
= −1
휕푝
휕푥
= 1
푑푖푣퐹 푥, 푦, 푧 = 푦푧 − 1 + 1 = 푦푧
At the point (1,2,1)
푑푖푣 퐹 푥, 푦, 푧 = 2 × 1 = 2 (Ans)

Now
div (curl F) =
휕푚
휕푥
+
휕푛
휕푦
+
휕푝
휕푧
Here, 푚 = 0 푛 = 푥푦 푝 = −푥푧
휕푚
휕푥
= 0
휕푛
휕푦
= 푥
휕푝
휕푧
= −푥
Relation between curl & divergence
div (curl F)=0 + 푥 − 푥 = 0

De푓푛 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
푐
푓 푥, 푦 푑푠 = lim
Δ →0
푛
푖=1
푓 푥푖, 푦푖 Δ푠푖 푝푙푎푛푒 표푟
푐
푓 푥, 푦, 푧 푑푠 = lim
Δ →0
푛
푖=1
푓 푥푖, 푦푖, 푧푖 Δ푠푖 푝푙푎푛푒
푠푝푎푐푒, 푝푟표푣푖푑푒푑 푡ℎ푒 푙푖푚푖푡 푒푥푖푠푡푠

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= 푟′(푡) 푑푡
= 푥′(푡) 2 + 푦′(푡) 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.
푐
푓 푥, 푦 푑푠 =
푏
푓(푥 푡 , 푦 푡 ) (푥′ 푡 + 푦 푡 )2
푎
If c is given by
r(t)=x(t)i + y(t)j + z(t)k where a≤ 푡 ≤ 푏 푡ℎ푒푛,
푏
푓(푥, 푦, 푧) ((푥′ 푡 )2+(푦′ 푡 )2 + (푧′(푡))2
푐 푓 푥, 푦, 푧 푑푠 = 푎

Ex-Evaluate 푐 푥2 − 푦 + 3푧 푑푠,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 푐 푥2 − 푦 + 3푧 푑푠,where c is a line
segment
1
= 0
푡2 − 푡 + 3푡 6 푑푡
= 6
푡3
3
−
2푡2
푡
+
1
3푡2
2 0
= 6
1
3
−
2
2
+
3
2
– (0 − 0 + 0)
= 6(
2−6+9
6
)
6∗5
=
6
=
5
6
(Ans)

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
푐 푀푑푥 + 푁푑푦 = 푅 (
휕푁
휕푥
−
휕푀
휕푦
)푑퐴

Ex : Evaluate 푐 (푡푎푛−1푥 + 푦2)푑푥 + (푒푦 − 푥2)푑푦
Where c is the pat enclosing the annular region.
푺풐풍풏: In polar coordinate
dA=rdrd휃
1≤r≤3 0≤ 휃 ≤ 휋
Here, M = (푡푎푛−1푥 + 푦2 ) , N = 푒푦 − 푥2
휕푀
휕푦
= 2푦
휕푁
휕푥
= −2x

x =rcos휃 , y =rsin휃
휕푁
휕푥
−
휕푀
휕푦
= −2x−2y
= −2(x+y)
= −2(rcos휃 + rsin휃)
= −2r(cos휃 + sin휃)
According to the green theorem,
푐 (푡푎푛−1푥 + 푦2)푑푥 + (푒푦 − 푥2)푑푦 = 푅 (
휕푁
휕푥
−
휕푀
휕푦
)푑퐴

= −
휋
0
3
2푟2 (cos휃 + sin휃) 푑푟푑휃
1
=
휋
0
−
2푟3
3
3
(cos휃 + sin휃)푑휃
=
휋 −2 × 33
0
3
+
2 × 13
3
=
휋
0
−18 +
2
3
=
휋
0
−
52
3

= −
52
3
휋
sin휃 − cos휃 0
= −
52
3
sin휋 − cos휋 − sin0 + cos0
= −
52
3
0 − 1 − 0 + 1
= −
52
3
× 2
= −
104
3

Theorem: Let, S be a surface equation z = g(x,y) and R its projection on the xy -
plane. If g , 푔푥and 푔푦 are continuous on R and f is continuous on S, then the surface
integral of f over S
is 푆 푓 푥, 푦, 푧 푑푠 = 푅 푓 푥, 푦, 푔 푥, 푦 . 1 + [푔푥(푥, 푦)]2 + [푔푦 (푥, 푦)]2 푑퐴
If, S is the graph of y = g(x,z) and R is its projection on the xz - plane, then
푆
푓 푥, 푦, 푧 푑푠 =
푅
푓 푥, 푧, 푔 푥, 푧 . 1 + [푔푥(푥, 푧)]2 + [푔푧(푥, 푧)]2 푑퐴
If, S is the graph of x = g(y,z) and R is its projection on the yz - plane, then
푆
푓 푥, 푦, 푧 푑푠 =
푅
푓 푦, 푧, 푔 푦, 푧 . 1 + [푔푦(푦, 푧)]2 + [푔푧(푦, 푧)]2 푑퐴

Example: Evaluate the surface integral,
푆 (푦2 + 2푦푧)푑푠 , where S is the first-octant portion of the
2x + y + 2z = 6 .
Solution:
We can re-write as,
2z = 6 − 2푥 − 푦 or, z =
1
2
6 − 2푥 − 푦 or, g(x,y) =
1
2
(6 − 2푥 − 푦)
∴ 푔푥 푥, 푦 = −1
푔푦 푥, 푦 = −
1
2

Now, 1 + [푔푥(푥, 푦)]2 + [푔푦(푥, 푦)]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 − 2푥 − 푦)
or, 0 =
1
2
(6 − 2푥 − 푦)
or, 0 = 6 − 2푥 − 푦
or, y = 6 − 2푥
or, y = 2(3 − 푥)

Again,
y = 2 3 − 푥
or, o = 2(3 − 푥)
or, o = 3 − 푥
or, x = 3
So, the limit is,
0 ≤ 푦 ≤ 2(3 − 푥) and 0 ≤ 푥 ≤ 3

∴
푆
(푦2 + 2푦푧)푑푠 =
3
2
푅
푦2 + 6푦 − 2푥푦 − 푦2 푑퐴
=
3
2
3
표
2(3−푥)
0
(6푦 − 2푥푦) 푑푦 푑푥
3
[
= 3 0
3푦2
2
−
푥푦2
2
2(3−푥)
]
0
푑푥
3
[
= 3 0
3
2
2 3 − 푥 2 −
푥
2
{2 3 − 푥 }2] 푑푥
3
[6(9 − 6푥 + 푥2) − 2푥(9 − 6푥 + 푥2)] 푑푥
= 3 0
3
[54 − 36푥 + 6푥2 − 18푥 + 12푥2 − 2푥3] 푑푥
= 3 0

3
[−2푥3 + 18푥2 − 54푥 + 54] 푑푥
= 3 0
= 3 [−
2푥4
4
+
18푥3
3
−
54푥2
2
3푑푥
+ 54]0
= 3 [−
81
2
+ 162 − 243 + 162]
= 121.5 (Answer)

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
푆
퐹. 푁푑푠 =
푄
푑푖푣 퐹. 푑푣

Example :
Let, Q be the solid region bounded by the co-ordinate planes and
the plane 2x + 2y + z = 6 and let, x i +푦2 푗 + 푍 푘 . Find
푆
퐹 . 푁푑푠 .
Where, S is the surface of Q.

Solution :
Here, M = x ∴
휕푀
휕푥
= 1
N = 푦2 ∴
휕푁
휕푦
= 2푦
P = Z ∴
휕푃
휕푧
= 1
∴ 푑푖푣 퐹 =
휕푀
휕푥
+
휕푁
휕푦
+
휕푃
휕푧
= 1 + 2푦 + 1
= 2 + 2푦

For limit, 푍 = 6 − 2푥 − 2푦
or, 0 = 6 − 2푥 − 2푦
or, 2푦 = 6 − 2푥
or, 푦 = 3 − 푥
Again, 푦 = 3 − 푥
or, 0 = 3 − 푥
or, 푥 = 3
So, the limit is,
0 ≤ 푧 ≤ 6 − 2푥 − 2푦 , 0 ≤ 푦 ≤ 3 − 푥 푎푛푑 0 ≤ 푥 ≤ 3

∴
푆
퐹. 푁푑푠 =
푄
푑푖푣 퐹. 푑푣
3
= 0
3−푥
0
6−2푥−2푦
0
(2 + 2푦) 푑푧 푑푦 푑푥
3
= 0
3−푥
0
6−2푥−2푦푑푦 푑푥
[2푧 + 2푦푧]0
3
= 0
3−푥
0
[2 6 − 2푥 − 2푦 + 2푦 6 − 2푥 − 2푦 ] 푑푦 푑푥
3
= 0
3−푥
0
[12 − 4푥 − 4푦 + 12푦 − 4푥푦 − 4푦2] 푑푦 푑푥
3
= 0
3−푥
0
[−4푦2 − 4푥 + 8푦 − 4푥푦 + 12] 푑푦 푑푥

3
[−
= 0
4푦3
3
− 4푥푦 +
8푦2
2
−
4푥푦2
2
+ 12푦]03−푥푑푥
3
[−
= 0
4
3
3 − 푥 3 − 4푥 3 − 푥 + 4 3 − 푥 2 − 2푥 3 − 푥 2 + 12 3 − 푥 ] 푑푥
3
[−
= 0
4
3
27 − 27푥 + 9푥2 − 푥3 − 12푥 + 4푥2 + 4 9 − 6푥 + 푥2 −
2푥 9 − 6푥 + 푥2 + 36 − 12푥]
=
3
[−36 + 36푥 − 12푥2 +
0
4
3
푥3 − 12푥 + 4푥2 + 36 − 24푥 + 4푥2 − 18푥 +
3
[+
= 0
4
3
푥3 − 2푥3 + 8푥2 − 30푥 + 36] 푑푥

= [
4푥4
3.4
−
2푥4
4
+
8푥3
3
−
30푥2
2
+ 36푥 ]0 3
= [
푥4
3
−
푥4
2
+
8푥3
3
− 15푥2 + 30푥 ]0 3
= [
34
3
−
34
2
+
8∗33
3
− 15 ∗ 32 + 36 ∗ 3 ]
= 27 − 40.5 + 72 − 135 + 108
= 31.5 (Answer)

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,
퐹. 푁 푑푠 = 푑푖푣 퐹. 푑푣

Example: Let Q be a solid region bounded by the co-ordinate plane and
the plane 2x+2y+z=6 and let,
F= x푖 +푦2푗 +z푘 Find 퐹.푁 푑푠 where S is the surface of Q .
Solution:
M=x
휕푀
휕푥
= 1
N=푦2 휕푁
휕푦
= 2y
P=z
휕푃
휕푧
=1
div F = 1+2y+1
=2+2y

퐹. 푁 푑푠 = 푑푖푣 퐹. 푑푣
3
= 0
3−푥
0
6−2푥−2푦
0
(2 − 2푦) 푑푧푑푦푑푥
3
= 0
3−푥
0
6−2푦−2푧
2푧 + 2푦푧 0
3
=4 0
3−푥
0
(1 + 푥)(3 − 푥 − 푦)푑푦푑푥
3
=4 0
3−푥
0
3 − 푥 − 푦 + 푥푦 − 푦2 푑푦푑푥
3
3푦 − 푥푦 − 푦2 −
=4 0
푥
2
푦2 −
3−푥
푦3
3 0
푑푥
3
[3 3 − 푥 − 푥 3 − 푥 − 3 − 푥 2 −
=4 0
푥
2
3 − 푥 2 −
1
3
(3 − 푥)3]푑푥
3
[9 − 3푥 + 푥2 − 9 + 6푥 − 푥3 −
=4 0
9푥−6푥2+푥3
2
−
27−푥3−27푥+9푥2
3
]푑푥

3
[
=4 0
9푥
2
− 9 −
푥3
6
]푑푥
3 9푥2
=4 0
4
− 9푥 −
3
푥4
24 0
=4(
81
4
− 27 −
91
6
)
=81−108−
91
6
=−
253
6
=−42.16(Ans)

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
푐 퐹. 푑푟 = 푠 푐푢푟푙 퐹 . 푛. 푑푠

Example Let C be the oriented triangle ying in the plane 2x+3y+z=6 as
shown in evaluating 푐 퐹 푑푟 when
E(x,y,z) = -y2 i + z j +x k =2x+2y+
Curl F =
푖 푗 푘
푑
푑푥
푑
푑푦
푑
푑푧
−푦2 푧 푥
= i(-1) -j (1) -1k (0+2y)
Consider z= 6-2x-2y=g(x,y)

We have 푠 퐹. 푁푑푠 = 푅 퐹. −푔푥 푥, 푦 푖 − 푔푦 푥, 푦 푗 + 푘 푗퐴
For an upcurved normal vector tobe obtain here
푔푥= -2
푔푦 =-2
N=(2i+2j+k)
Then 푠 퐹. 푁푑푠 = 푅 ( −푖 − 푗 + 2푦푘 . (2푖 + 2푗 + 푘))푗퐴
3
= 0
3−푥
0
−2 − 2 + 2푦 푑푦푑푥

3
푦2 − 4푦
= 0
3
(3 − 푥2 − 4(3 − 푥) 푑푥
= 0
3
(9 − 6푥 + 푥2 − 12 + 4푥 dx
= 0
3
푥2 − 2푥 − 3 푑푥
= 0
=
푥3
3
− 푥2 − 3푥
33
3
=
− 32 − 3.3
= 9- 9- 9
=-9 (Answer)

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