presentation on calculus
TRANSCRIPT
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Welcome to Our
Mathematics Presentation
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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
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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. Greenโs 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
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Distance Between
Points
Plane and Line
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Theorem: The distance between a plane and point Q
(Not in plane) is, D= ๐๐๐๐ ๐๐ = ๐๐.๐
๐, 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
(aโ 0), let y=0 and z=0, then from the equation,
ax+d=0, we can conclude that the point
( -๐
๐, 0,0 ) lies in the plane and n is the normal to the
plane.
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Ex: Find the distance the point Q=( 1, 5, -5 ) and the plane
given by 3xโy+2z=6.
soln: We know that, ๐=< 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,
๐๐=< 1-2, 5-0, -4-0 >
=< -1, 5, -4 >
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Using the distance formula produces,
D=๐๐ .๐
๐
=<โ1, 5,โ4> .< 3,โ1, 2>
32+(โ1)2+22
=โ1 โ5 โ8
9+1+4
=โ16
14
=16
14
=4.276 (Answer)
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Ex: Find the distance the point Q=(1, 2, 3) and plane given by 2x-y+z=4.
Soln: We know that ๐=<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,
๐๐=<1-2, 2-0, 3-0>
=<-1, 2, 3>
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Using the distance formula produce
D= ๐๐ . ๐
๐
= <โ1, 2, 3> .<2,โ1, 1>
22+(โ1)2+12
= โ2โ2+3
4+1+1
= โ1
6
=1
6
=0.408 (Answer)
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Distance Between Two Parallel Planes
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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 = ๐๐ฅ0+๐๐ฆ0+๐๐ง0+๐
๐2+๐2+๐2
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
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Ex: Find the distance between the two parallel planes
given by 3x โ y + 2z โ 6 = 0 and 6x โ 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
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Distance between two parallel plane is
D=๐๐ฅ0+๐๐ฆ0+๐๐ง0
๐2+๐2+๐2
D=6โ2 + โ2โ0 + 4โ0 +4
62+(โ2)2+42
=12+4
36+4+16
=16
56
=16
56
=2.138 (Answer)
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Ex: Find the distance between the two parallel planes
given by x โ 3y + 4z =10 , x โ 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
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Distance between two parallel
plane
D=๐๐ฅ0+๐๐ฆ0+๐๐ง0
๐2+๐2+๐2
D=0โ1 + โ3โ0 + 4โ0 +(โ6)
12+(โ3)2+42
=10โ6
1+9+16
=4
26
=4
26
=0.7845 (Answer)
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Distance Between a Point a Line in Space
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Theorem: The distance between a point Q and a
line in space is given by D=๐๐ร๐
๐,where ๐ is
the directional vector for line and P is a point on the line.
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Ex : Find the distance between the point P= (3,โ1,4) and the line given by x =โ2+3t , y = โ2t, z = 1+4t.
๐๐๐๐: Using the direction number 3,โ2, 4
We know that the direction vector for the line is
๐=< 3, โ2, 4 >
To find a point on the line , let t= 0 and we obtain P = (โ2 , 0, 1)
point on the line. Then ๐๐=< 3โ (โ2) , โ1โ0 , 4โ1 >
=< 5, โ1 ,3 > and we can form
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๐๐ ร ๐ = ๐ ๐ ๐5 โ1 33 โ2 4
= ๐( โ4 + 6) โ ๐(20โ9) + ๐(โ10 + 3)
= 2 ๐ โ 11 ๐ โ 7๐
=< 2, โ11,โ7 >
D =๐๐ร๐
๐
=22+(โ11)2+(โ7)2
9+4+16
=4+121+49
29
=174
29
=2.45 (Answer)
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Vector Value Function
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A function of the form
r(t)= f(t) ๐ +g(t) ๐
and
r(t)= f(t) ๐ +g(t) ๐ + h(t)๐ is a vector valued function,
where the component of f, g, h are real valued
function of the parameter t,
r(t)= ๐ ๐ก , ๐(๐ก)
or
r(t)= ๐ ๐ก , ๐ ๐ก , โ(๐ก)
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Differention of a vector valued function:
Defn: The derivative of a vector valued function r is
defined by
rยด(t)= limโ๐กโ0
๐ ๐ก+โ๐ก โ๐(๐ก)
โ๐ก
for all t for which the limit exists. If rยด(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.
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Ex: Find the derivative of each of the
following vector-valued function
(i) r(t)= t2 ๐ 4 ๐
Soln:
๐๐
๐๐ก/ rยด(t)=2t ๐
(ii) r(t)= 1
๐ก ๐ + lnt ๐+โฎ2t ๐
Soln: ๐๐
๐๐ก=
1
๐ก2 ๐ +
1
๐ก ๐+ 2โฎ2t ๐
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Velocity and Acceleration
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Defn: If x and y are twice differentiable functions of t
and r is a vector-valued function given by r(t)= x(t) ๐ +
y(t) ๐, then the velocity vector , acceleration vector , and
speed at time t are follows
Velocity = ๐ฃ ๐ก = rยด(t)= xยด(t) ๐ + yยด(t) ๐
Acceleration n= a(t)= rยดยด(t)= xยดยด(t) ๐ + yยดยด(t) ๐
Speed = ๐ฃ ๐ก = rยด(t) = [xยด(t)]2+[yยด(t)]2
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Example: Find the velocity and acceleration vector when t=0 and t=3 for the vector valued functionr(t)= <๐๐ก ๐๐๐ ๐ก, ๐๐ก๐ ๐๐๐ก, ๐๐ก >.
r(t) = ๐๐ก ๐๐๐ ๐ก ๐ + ๐๐ก๐ ๐๐๐ก ๐ + ๐๐ก ๐
Velocity = ๐๐ก๐
๐๐ฅ๐๐๐ ๐ก + ๐๐๐ ๐ก
๐
๐๐ฅ๐๐ก ๐ + ๐๐ก ๐
= (โ๐๐ก๐ ๐๐๐ก + ๐๐ก ๐๐๐ ๐ก ) ๐ +(๐๐ก ๐๐๐ ๐ก + ๐๐ก๐ ๐๐๐ก) ๐ + ๐๐ก ๐
= < ๐๐ก ๐๐๐ ๐ก โ ๐ ๐๐๐ก , ๐๐ก (๐ ๐๐๐ก + ๐๐๐ ๐ก) , ๐๐ก >
(Answer)
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Speed = rห(t)
= ๐๐ก๐๐๐ ๐ก โ ๐๐ก ๐ ๐๐๐ก 2 + (๐๐ก๐ ๐๐๐ก + ๐๐ก๐๐๐ ๐ก)2+(๐๐ก)2
= ๐๐ก 2 ๐๐๐ ๐ก โ ๐ ๐๐๐ก 2 + ๐๐ก 2 ๐ ๐๐๐ก + ๐๐๐ ๐ก 2 + ๐๐ก 2
= ๐๐ก 1 โ 2๐ ๐๐๐ก ๐๐๐ ๐ก + ๐ + 2๐ ๐๐๐ก ๐๐๐ ๐ก + 1
= ๐๐ก 3 (Answer)
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Tangent vector
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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)=๐ ฬ(๐ก)
๐ ฬ(๐ก), ๐ ฬ(๐ก) โ 0
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Ex: Find the unit tangent vector to the curve given by
r(t)=t ๐ +๐ก2 ๐ when t=1
Slon : Given
r(t)=t ๐ +๐ก2 ๐๐ ฬ(๐ก)= ๐ + 2t ๐
๐ ฬ(๐ก) = 12+(2๐ก)2
= 1 + 4๐ก2
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The tangen vector is
T(t)=๐ ฬ(๐ก)
๐ ฬ(๐ก)
=1
1+4๐ก2( ๐ + 2t ๐)
The unit tangent vector at t=1 is
T(1) =1
5( ๐ + 2 ๐)
(Answer)
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Arc Length and
Curvature
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Definition of Arc Length:
The arc length of a smooth plane curve c by the
parametric equations x=x(t) and y=(t), a โค t โค b is
S= ๐๐
[๐ฅโฒ(๐ก)]2+[๐ฆโฒ(๐ก)]2 dt . Where c is given by
r(t)= x(t)รฎ + y(t)ฤต . We can rewrite this equation for Arc
Length as, s= ๐๐๐โฒ ๐ก ๐๐ก
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Ex: Find the arc length of the curve given by
r(t)=cost รฎ + sint ฤต from t=0 to t=2ฯ
Solution:
Here, x(t)=cost y(t)= sint
Xโ(t)=-sint yโ(t)=cost
We have
S= ๐๐
[๐ฅโฒ(๐ก)]2+[๐ฆโฒ(๐ก)]2 dt
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S= 02๐
๐ ๐๐2๐ก + ๐๐๐ 2๐ก dt
S= 02๐
1dt
S= 02๐1dt
S=[๐ก]02๐
S=2ฯ (Answer)
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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=||๐๐
๐๐ ||
K=||Tโ(s)||
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Ex: Find the curvature of the line given by r(s)= (3-3
5s) รฎ +
4
5s ฤต
Solution:
r(s) = (3-3
5s) รฎ +
4
5s ฤต
rโ(s) = ( 0-3
5)รฎ +
4
5ฤต
||rโ(s)|| = (โ3
5)2+(
4
5)2
=9
25+
16
25
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=25
25
=1
T(s) = ๐โฒ(๐ )
||๐โฒ(๐ )||
= โ3
5รฎ +
4
5ฤต
Tโ(s) = 0.รฎ+0.ฤต
||Tโ(S)|| = 0
K = ||Tโ(S)|| = 0 (Answer)
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The Gradient of a
Function of Two
Variables
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Definition :
Let Z=f(x,y) be a function of x and y such that ๐๐ฅand ๐๐ฆ exit. Then the gradient of f, denoted by
๐ป๐( ๐ฅ, ๐ฆ ) ,is the vector of
๐ป๐ ๐ฅ, ๐ฆ = ๐๐ฅ( x,y ) ๐ + ๐๐ฆ ( x,y) ๐
We read ๐ป๐ ๐๐ โฒโฒ๐๐๐๐โฒโฒ. Another notation for the
gradient is grad ๐ ๐ฅ, ๐ฆ .
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Ex:
Find the gradient of ๐ ๐ฅ, ๐ฆ = ๐ฆ ๐๐ ๐ฅ + ๐ฅ๐ฆ2 at
the point (1,2).
๐บ๐๐๐:
Grad f or ๐ป๐ = ๐๐ฅ ( x,y ) ๐ + ๐๐ฆ ( x,y)
๐โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ(i)
Now,
๐๐ฅ=๐ฆ
๐ฅ+ ๐ฆ2 , ๐๐ฆ = ๐๐ ๐ฅ + 2๐ฅ๐ฆ
โด ๐๐ฅ( 1,2)= 2 + 22 = 6
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โด ๐๐ฆ ( 1,2) = ln 1 + 2 โ 1 โ 2 = 4
From (i) we get,
๐ป๐ = 6 ๐ + 4 ๐ (Ans:)
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Tangent Plane and Normal Line
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Definition: Let F be differentiable at the point P= (๐ฅ0 ,๐ฆ0 ,๐ง0 ) u the surface s
given by F(x, y, z) = 0 such that ๐ป๐น ๐ฅ0 ,๐ฆ0 ,๐ง0 โ 0.
1). The plane through P that is normal to ๐ป๐น ๐ฅ0 ,๐ฆ0 ,๐ง0 is called the tangent
plane to S at P.
2). The line through P having the direction of ๐ป๐น ๐ฅ0 ,๐ฆ0 ,๐ง0 is called the normal
line to S at P.
If F is differentiable at ๐ฅ0 ,๐ฆ0 ,๐ง0 then an equation of the tangent plane to the
surface given by
๐น ๐ฅ, ๐ฆ, ๐ง = 0 at ๐ฅ0 ,๐ฆ0 ,๐ง0 is
๐น๐ฅ ๐ฅ0 ,๐ฆ0 ,๐ง0 (xโ๐ฅ0)+ ๐น๐ฆ ๐ฅ0 ,๐ฆ0 ,๐ง0 (yโ๐ฆ0) +๐น๐ง ๐ฅ0 ,๐ฆ0 ,๐ง0 (zโ๐ง0)= 0.
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Ex :
Find an equation of the tangent plane to the hyperboloid given by
at ๐ง2โ2๐ฅ2 โ
2๐ฆ2 = 12 the point (1,โ1, 4) .
๐บ๐๐๐:
An the equation of the tangent plane to the Given hyperboloid
can be rewrite
as ๐น ๐ฅ, ๐ฆ, ๐ง = 12 โ ๐ง2 +2๐ฅ2 + 2๐ฆ2
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๐น๐ฅ ๐ฅ, ๐ฆ, ๐ง =4x โด ๐น๐ฅ 1,โ1, 4 =4
๐น๐ฆ ๐ฅ, ๐ฆ, ๐ง =4y โด ๐น๐ฆ 1,โ1, 4
= โ4
๐น๐ง ๐ฅ, ๐ฆ, ๐ง = โ2๐ง โด ๐น๐ง 1,โ1, 4 =
โ8
The tangent plane is ,
4(xโ1) + โ4 (y+1) + ( โ8)(zโ4) = 0
โ4xโ4โ4yโ4โ8z+32 = 0
โ4xโ4yโ8z+24 = 0 (Answer)
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Curl of a vector field
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Curl of a vector field:
The curl of the F(x,y,z)= M ๐+N ๐+p ๐ is
Curl F(x,y,z)= ๐ป x F(x,y,z)
= (๐๐
๐๐ฆ-๐๐
๐๐ง) ๐ -(
๐๐
๐๐ฅ-๐๐
๐๐ง) ๐+(
๐๐
๐๐ฅ-๐๐
๐๐ฆ) ๐
If curl F=0, then F is irrational.
๐ป x F(x,y,z)=
i ๐ ๐๐
๐๐ฅ
๐
๐๐ฆ
๐
๐๐ง
๐ ๐ ๐
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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
๐ป x F(x,y,z)=
i ๐ ๐๐
๐๐ฅ
๐
๐๐ฆ
๐
๐๐ง
2xy (x2 + z2) 2zy
= i{๐
๐๐ฆ. 2zy-
๐
๐๐ง(x2+z2)} โ j(
๐
๐๐ฅ.2zy -
๐
๐๐ง.2xy) +k{
๐
๐๐ฅ(x2+z2) -
๐
๐๐ฆ.2xy}
=i(2z-2z) โj(0-0) +k(2x-2x)
=0i + 0j + 0k
=0
If curl F=0, then the vector field F is irrational.
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Curl and Divergence
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Curl;
Curl ๐น = ๐ป ร ๐น ๐ฅ, ๐ฆ, ๐ง
=๐
๐๐ฅ+
๐
๐๐ฆ+
๐
๐๐งร ๐ + ๐ + ๐
If ๐๐ข๐๐ ๐น = 0 ๐โ๐๐ ๐น is irrational.
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Problem:
Find the curl F of the followings
F(x, y, z) = ๐โ๐ฅ๐ฆ๐ง( ๐ + ๐ + ๐) and point (3,2,0)
Solution:
F(x, y, z) = ๐โ๐ฅ๐ฆ๐ง( ๐ + ๐ + ๐)
F x, y, z = ๐โ๐ฅ๐ฆ๐ง ๐ + ๐โ๐ฅ๐ฆ๐ง ๐ + ๐โ๐ฅ๐ฆ๐ง ๐
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๐๐ข๐๐ ๐น(๐ฅ, ๐ฆ, ๐ง) = ๐ป ร ๐น(๐ฅ, ๐ฆ, ๐ง)
=
๐ ๐ ๐๐
๐๐ฅ
๐
๐๐ฆ
๐
๐๐ง
๐โ๐ฅ๐ฆ๐ง ๐โ๐ฅ๐ฆ๐ง ๐โ๐ฅ๐ฆ๐ง
= (โ๐ฅ๐ง๐โ๐ฅ๐ฆ๐ง + ๐ฅ๐ฆ๐โ๐ฅ๐ฆ๐ง) ๐ โ โ๐ฆ๐ง๐โ๐ฅ๐ฆ๐ง + ๐ฅ๐ฆ๐โ๐ฅ๐ฆ๐ง ๐ +
(โ๐ฆ๐ง๐โ๐ฅ๐ฆ๐ง + ๐ฅ๐ง๐โ๐ฅ๐ฆ๐ง) ๐
At point (3, 2, 0)
Curl F= (0+6๐0) ๐ โ 0 + 6๐0 ๐ + (0 + 0) ๐
=6 ๐ + 6 ๐ (Ans)
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Divergence:
div F = ๐ป. ๐น(๐ฅ, ๐ฆ, ๐ง)
div F=๐๐
๐๐ฅ+
๐๐
๐๐ฆ+
๐๐
๐๐ง
where F(x,y,z) = M ๐ + ๐ ๐ + ๐ ๐
If div F=0, then it is said to be divergence free.
Relation between curl & divergence is div(curlF)=0.
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Problem:
Find the divergence of ๐น ๐ฅ, ๐ฆ, ๐ง = ๐ฅ๐ฆ๐ง ๐ โ ๐ฆ ๐ + ๐ง ๐ and also find the
relation between curl & divergence.
Solution:
Given that
๐น ๐ฅ, ๐ฆ, ๐ง = ๐ฅ๐ฆ๐ง ๐ โ ๐ฆ ๐ + ๐ง ๐
๐๐ข๐๐ ๐น ๐ฅ, ๐ฆ, ๐ง = ๐ป ร ๐น ๐ฅ, ๐ฆ, ๐ง
=
๐ ๐ ๐๐
๐๐ฅ
๐
๐๐ฆ
๐
๐๐ง
๐ฅ๐ฆ๐ง โ๐ฆ ๐ง
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= (0-0) ๐ โ 0 โ ๐ฅ๐ฆ ๐ + (0 โ ๐ฅ๐ง) ๐
Curl F(๐ฅ, ๐ฆ, ๐ง) = ๐ฅ๐ฆ ๐ โ ๐ฅ๐ง ๐
We know that
div F=๐๐
๐๐ฅ+
๐๐
๐๐ฆ+
๐๐
๐๐ง
Here, m=xyz n= -y p= z
๐๐
๐๐ฅ= ๐ฆ๐ง
๐๐
๐๐ฅ= โ1
๐๐
๐๐ฅ= 1
๐๐๐ฃ๐น ๐ฅ, ๐ฆ, ๐ง = ๐ฆ๐ง โ 1 + 1 = ๐ฆ๐ง
At the point (1,2,1)
๐๐๐ฃ ๐น ๐ฅ, ๐ฆ, ๐ง = 2 ร 1 = 2 (Ans)
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Now
div (curl F) = ๐๐
๐๐ฅ+
๐๐
๐๐ฆ+
๐๐
๐๐ง
Here, ๐ = 0 ๐ = ๐ฅ๐ฆ ๐ = โ๐ฅ๐ง
๐๐
๐๐ฅ= 0
๐๐
๐๐ฆ= ๐ฅ
๐๐
๐๐ง= โ๐ฅ
Relation between curl & divergence
div (curl F)=0 + ๐ฅ โ ๐ฅ = 0
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Line Integral
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De๐๐ 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
๐
๐ ๐ฅ, ๐ฆ ๐๐ = limโ โ0
๐=1
๐
๐ ๐ฅ๐, ๐ฆ๐ โ๐ ๐ ๐๐๐๐๐ ๐๐
๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ = limโ โ0
๐=1
๐
๐ ๐ฅ๐, ๐ฆ๐, ๐ง๐ โ๐ ๐ ๐๐๐๐๐
๐ ๐๐๐๐, ๐๐๐๐ฃ๐๐๐๐ ๐กโ๐ ๐๐๐๐๐ก ๐๐ฅ๐๐ ๐ก๐
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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= ๐โฒ(๐ก) ๐๐ก
= ๐ฅโฒ(๐ก) 2 + ๐ฆโฒ(๐ก) 2 dt
A similar formula holds for a space curve as indicated in
the following theorem.
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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.
๐
๐ ๐ฅ, ๐ฆ ๐๐ = ๐
๐
๐(๐ฅ ๐ก , ๐ฆ ๐ก ) (๐ฅโฒ ๐ก + ๐ฆ ๐ก )2
If c is given by
r(t)=x(t)i + y(t)j + z(t)k where aโค ๐ก โค ๐ ๐กโ๐๐,
๐ ๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ = ๐๐๐(๐ฅ, ๐ฆ, ๐ง) ((๐ฅโฒ ๐ก )2+(๐ฆโฒ ๐ก )2 + (๐งโฒ(๐ก))2
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Ex-Evaluate ๐ ๐ฅ2 โ ๐ฆ + 3๐ง ๐๐ ,where c is the line segment show
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Sol-To find the parametric from of equation of a line
X=t,
Y=2t,
Z=t,
Again
Xโ(t)=1,
Yโ(t)=2,
Zโ(t)=1,
Hear
12 + 22 + 12
= 6
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Thus line integral take the fowling form ๐ ๐ฅ2 โ ๐ฆ + 3๐ง ๐๐ ,where c is a line
segment
= 01๐ก2 โ ๐ก + 3๐ก 6 ๐๐ก
= 6๐ก3
3โ
2๐ก2
๐ก+
3๐ก2
2 0
1
= 61
3โ
2
2+
3
2โ (0 โ 0 + 0)
= 6(2โ6+9
6)
=6โ5
6
=5
6(Ans)
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Greenโs Theorem
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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
๐ ๐๐๐ฅ + ๐๐๐ฆ = ๐ (๐๐
๐๐ฅโ
๐๐
๐๐ฆ)๐๐ด
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Ex : Evaluate ๐ (๐ก๐๐โ1๐ฅ + ๐ฆ2)๐๐ฅ + (๐๐ฆ โ ๐ฅ2)๐๐ฆ
Where c is the pat enclosing the annular region.
๐บ๐๐๐: In polar coordinate
dA=rdrd๐
1โคrโค3 0โค ๐ โค ๐
Here, M = (๐ก๐๐โ1๐ฅ + ๐ฆ2 ) , N = ๐๐ฆ โ ๐ฅ2
๐๐
๐๐ฆ= 2๐ฆ
๐๐
๐๐ฅ=โ2x
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x =rcos๐ , y =rsin๐
๐๐
๐๐ฅโ
๐๐
๐๐ฆ= โ2xโ2y
= โ2(x+y)
= โ2(rcos๐ + rsin๐)
= โ2r(cos๐ + sin๐)
According to the green theorem,
๐ (๐ก๐๐โ1๐ฅ + ๐ฆ2)๐๐ฅ + (๐๐ฆ โ ๐ฅ2)๐๐ฆ = ๐ (๐๐
๐๐ฅโ
๐๐
๐๐ฆ)๐๐ด
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= โ 0
๐
1
3
2๐2 (cos๐ + sin๐) ๐๐๐๐
= 0
๐
โ2๐3
3
3
(cos๐ + sin๐)๐๐
= 0
๐ โ2 ร 33
3+2 ร 13
3(cos๐ + sin๐)๐๐
= 0
๐
โ18 +2
3(cos๐ + sin๐)๐๐
= 0
๐
โ52
3(cos๐ + sin๐)๐๐
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= โ52
3sin๐ โ cos๐ 0
๐
= โ52
3sin๐ โ cos๐ โ sin0 + cos0
= โ52
30 โ 1 โ 0 + 1
= โ52
3ร 2
= โ104
3
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Surface Integral
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Theorem: Let, S be a surface equation z = g(x,y) and R its projection on the xy -
plane. If g , ๐๐ฅand ๐๐ฆ are continuous on R and f is continuous on S, then the surface
integral of f over S
is ๐๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ = ๐
๐ ๐ฅ, ๐ฆ, ๐ ๐ฅ, ๐ฆ . 1 + [๐๐ฅ(๐ฅ, ๐ฆ)]2 + [๐๐ฆ(๐ฅ, ๐ฆ)]
2 ๐๐ด
If, S is the graph of y = g(x,z) and R is its projection on the xz - plane, then
๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ =
๐
๐ ๐ฅ, ๐ง, ๐ ๐ฅ, ๐ง . 1 + [๐๐ฅ(๐ฅ, ๐ง)]2 + [๐๐ง(๐ฅ, ๐ง)]
2 ๐๐ด
If, S is the graph of x = g(y,z) and R is its projection on the yz - plane, then
๐
๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ =
๐
๐ ๐ฆ, ๐ง, ๐ ๐ฆ, ๐ง . 1 + [๐๐ฆ(๐ฆ, ๐ง)]2 + [๐๐ง(๐ฆ, ๐ง)]
2 ๐๐ด
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Example: Evaluate the surface integral,
๐(๐ฆ2 + 2๐ฆ๐ง)๐๐ , where S is the first-octant portion of the
2x + y + 2z = 6 .
Solution:
We can re-write as,
2z = 6 โ 2๐ฅ โ ๐ฆ or, z = 1
26 โ 2๐ฅ โ ๐ฆ or, g(x,y) =
1
2(6 โ 2๐ฅ โ ๐ฆ)
โด ๐๐ฅ ๐ฅ, ๐ฆ = โ1
๐๐ฆ ๐ฅ, ๐ฆ = โ1
2
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Now, 1 + [๐๐ฅ(๐ฅ, ๐ฆ)]2 + [๐๐ฆ(๐ฅ, ๐ฆ)]
2
= 1 + (โ1)2+(โ1
2)2
= 1 + 1 +1
4
=4+4+1
4
=9
4
=3
2
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For limit,
z = 1
2(6 โ 2๐ฅ โ ๐ฆ)
or, 0 = 1
2(6 โ 2๐ฅ โ ๐ฆ)
or, 0 = 6 โ 2๐ฅ โ ๐ฆ
or, y = 6 โ 2๐ฅ
or, y = 2(3 โ ๐ฅ)
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Again,
y = 2 3 โ ๐ฅ
or, o = 2(3 โ ๐ฅ)
or, o = 3 โ ๐ฅ
or, x = 3
So, the limit is,
0 โค ๐ฆ โค 2(3 โ ๐ฅ) and 0 โค ๐ฅ โค 3
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โด
๐
(๐ฆ2 + 2๐ฆ๐ง)๐๐ =3
2
๐
๐ฆ2 + 6๐ฆ โ 2๐ฅ๐ฆ โ ๐ฆ2 ๐๐ด
= 3
2 ๐3 02(3โ๐ฅ)
(6๐ฆ โ 2๐ฅ๐ฆ) ๐๐ฆ ๐๐ฅ
= 3 03[3๐ฆ2
2โ
๐ฅ๐ฆ2
2]0
2(3โ๐ฅ)
๐๐ฅ
= 3 03[3
22 3 โ ๐ฅ 2 โ
๐ฅ
2{2 3 โ ๐ฅ }2] ๐๐ฅ
= 3 03[6(9 โ 6๐ฅ + ๐ฅ2) โ 2๐ฅ(9 โ 6๐ฅ + ๐ฅ2)] ๐๐ฅ
= 3 03[54 โ 36๐ฅ + 6๐ฅ2 โ 18๐ฅ + 12๐ฅ2 โ 2๐ฅ3] ๐๐ฅ
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= 3 03[โ2๐ฅ3 + 18๐ฅ2 โ 54๐ฅ + 54] ๐๐ฅ
= 3 [โ2๐ฅ4
4+
18๐ฅ3
3โ
54๐ฅ2
2+ 54]0
3๐๐ฅ
= 3 [โ81
2+ 162 โ 243 + 162]
= 121.5 (Answer)
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Divergence Theorem
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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
๐
๐น.๐๐๐ =
๐
๐๐๐ฃ ๐น. ๐๐ฃ
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Example :
Let, Q be the solid region bounded by the co-ordinate planes and
the plane 2x + 2y + z = 6 and let, x i +๐ฆ2 ๐ + ๐ ๐ . Find
๐
๐น .๐๐๐ .
Where, S is the surface of Q.
![Page 83: Presentation on calculus](https://reader038.vdocuments.mx/reader038/viewer/2022110310/55a3a2281a28abf32d8b4808/html5/thumbnails/83.jpg)
Solution :
Here, M = x โด๐๐
๐๐ฅ= 1
N = ๐ฆ2 โด๐๐
๐๐ฆ= 2๐ฆ
P = Z โด๐๐
๐๐ง= 1
โด ๐๐๐ฃ ๐น =๐๐
๐๐ฅ+
๐๐
๐๐ฆ+
๐๐
๐๐ง
= 1 + 2๐ฆ + 1
= 2 + 2๐ฆ
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For limit, ๐ = 6 โ 2๐ฅ โ 2๐ฆ
or, 0 = 6 โ 2๐ฅ โ 2๐ฆ
or, 2๐ฆ = 6 โ 2๐ฅ
or, ๐ฆ = 3 โ ๐ฅ
Again, ๐ฆ = 3 โ ๐ฅ
or, 0 = 3 โ ๐ฅ
or, ๐ฅ = 3
So, the limit is,
0 โค ๐ง โค 6 โ 2๐ฅ โ 2๐ฆ , 0 โค ๐ฆ โค 3 โ ๐ฅ ๐๐๐ 0 โค ๐ฅ โค 3
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โด
๐
๐น.๐๐๐ =
๐
๐๐๐ฃ ๐น. ๐๐ฃ
= 03 03โ๐ฅ
06โ2๐ฅโ2๐ฆ
(2 + 2๐ฆ) ๐๐ง ๐๐ฆ ๐๐ฅ
= 03 03โ๐ฅ
[2๐ง + 2๐ฆ๐ง]06โ2๐ฅโ2๐ฆ
๐๐ฆ ๐๐ฅ
= 03 03โ๐ฅ
[2 6 โ 2๐ฅ โ 2๐ฆ + 2๐ฆ 6 โ 2๐ฅ โ 2๐ฆ ] ๐๐ฆ ๐๐ฅ
= 03 03โ๐ฅ
[12 โ 4๐ฅ โ 4๐ฆ + 12๐ฆ โ 4๐ฅ๐ฆ โ 4๐ฆ2] ๐๐ฆ ๐๐ฅ
= 03 03โ๐ฅ
[โ4๐ฆ2 โ 4๐ฅ + 8๐ฆ โ 4๐ฅ๐ฆ + 12] ๐๐ฆ ๐๐ฅ
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= 03[โ
4๐ฆ3
3โ 4๐ฅ๐ฆ +
8๐ฆ2
2โ
4๐ฅ๐ฆ2
2+ 12๐ฆ]0
3โ๐ฅ๐๐ฅ
= 03[โ
4
33 โ ๐ฅ 3 โ 4๐ฅ 3 โ ๐ฅ + 4 3 โ ๐ฅ 2 โ 2๐ฅ 3 โ ๐ฅ 2 + 12 3 โ ๐ฅ ] ๐๐ฅ
= 03[โ
4
327 โ 27๐ฅ + 9๐ฅ2 โ ๐ฅ3 โ 12๐ฅ + 4๐ฅ2 + 4 9 โ 6๐ฅ + ๐ฅ2 โ
2๐ฅ 9 โ 6๐ฅ + ๐ฅ2 + 36 โ 12๐ฅ]
=
03[โ36 + 36๐ฅ โ 12๐ฅ2 +
4
3๐ฅ3 โ 12๐ฅ + 4๐ฅ2 + 36 โ 24๐ฅ + 4๐ฅ2 โ 18๐ฅ +
= 03[+
4
3๐ฅ3 โ 2๐ฅ3 + 8๐ฅ2 โ 30๐ฅ + 36] ๐๐ฅ
![Page 87: Presentation on calculus](https://reader038.vdocuments.mx/reader038/viewer/2022110310/55a3a2281a28abf32d8b4808/html5/thumbnails/87.jpg)
= [4๐ฅ4
3.4โ
2๐ฅ4
4+
8๐ฅ3
3โ
30๐ฅ2
2+ 36๐ฅ ]0
3
= [๐ฅ4
3โ
๐ฅ4
2+
8๐ฅ3
3โ 15๐ฅ2 + 30๐ฅ ]0
3
= [34
3โ
34
2+
8โ33
3โ 15 โ 32 + 36 โ 3 ]
= 27 โ 40.5 + 72 โ 135 + 108
= 31.5 (Answer)
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Divergence Theorem
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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,
๐น.๐ ๐๐ = ๐๐๐ฃ ๐น. ๐๐ฃ
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Example: Let Q be a solid region bounded by the co-ordinate plane and
the plane 2x+2y+z=6 and let,
F= x ๐+๐ฆ2 ๐+z ๐ Find ๐น.๐ ๐๐ where S is the surface of Q .
Solution:
M=x ๐๐
๐๐ฅ= 1
N=๐ฆ2๐๐
๐๐ฆ= 2y
P=z ๐๐
๐๐ง=1
div F = 1+2y+1
=2+2y
![Page 91: Presentation on calculus](https://reader038.vdocuments.mx/reader038/viewer/2022110310/55a3a2281a28abf32d8b4808/html5/thumbnails/91.jpg)
๐น.๐ ๐๐ = ๐๐๐ฃ ๐น. ๐๐ฃ
= 03 03โ๐ฅ
06โ2๐ฅโ2๐ฆ
(2 โ 2๐ฆ) ๐๐ง๐๐ฆ๐๐ฅ
= 03 03โ๐ฅ
2๐ง + 2๐ฆ๐ง 06โ2๐ฆโ2๐ง
=4 03 03โ๐ฅ
(1 + ๐ฅ)(3 โ ๐ฅ โ ๐ฆ)๐๐ฆ๐๐ฅ
=4 03 03โ๐ฅ
3 โ ๐ฅ โ ๐ฆ + ๐ฅ๐ฆ โ ๐ฆ2 ๐๐ฆ๐๐ฅ
=4 033๐ฆ โ ๐ฅ๐ฆ โ ๐ฆ2 โ
๐ฅ
2๐ฆ2 โ
๐ฆ3
3 0
3โ๐ฅ
๐๐ฅ
=4 03[3 3 โ ๐ฅ โ ๐ฅ 3 โ ๐ฅ โ 3 โ ๐ฅ 2 โ
๐ฅ
23 โ ๐ฅ 2 โ
1
3(3 โ ๐ฅ)3]๐๐ฅ
=4 03[9 โ 3๐ฅ + ๐ฅ2 โ 9 + 6๐ฅ โ ๐ฅ3 โ
9๐ฅโ6๐ฅ2+๐ฅ3
2โ
27โ๐ฅ3โ27๐ฅ+9๐ฅ2
3]๐๐ฅ
![Page 92: Presentation on calculus](https://reader038.vdocuments.mx/reader038/viewer/2022110310/55a3a2281a28abf32d8b4808/html5/thumbnails/92.jpg)
=4 03[9๐ฅ
2โ 9 โ
๐ฅ3
6]๐๐ฅ
=4 03 9๐ฅ2
4โ 9๐ฅ โ
๐ฅ4
24 0
3
=4(81
4โ 27 โ
91
6)
=81โ108โ91
6
=โ253
6
=โ42.16(Ans)
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STOKES THEOREM
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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
๐ ๐น. ๐๐ = ๐ ๐๐ข๐๐ ๐น . ๐. ๐๐
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Example Let C be the oriented triangle ying in the plane 2x+3y+z=6 as
shown in evaluating ๐ ๐น ๐๐ when
E(x,y,z) = -y2 i + z j +x k =2x+2y+
Curl F =
๐ ๐ ๐๐
๐๐ฅ
๐
๐๐ฆ
๐๐๐ง
โ๐ฆ2 ๐ง ๐ฅ
= i(-1) -j (1) -1k (0+2y)
Consider z= 6-2x-2y=g(x,y)
![Page 96: Presentation on calculus](https://reader038.vdocuments.mx/reader038/viewer/2022110310/55a3a2281a28abf32d8b4808/html5/thumbnails/96.jpg)
We have ๐ ๐น.๐๐๐ = ๐
๐น. โ๐๐ฅ ๐ฅ, ๐ฆ ๐ โ ๐๐ฆ ๐ฅ, ๐ฆ ๐ + ๐ ๐๐ด
For an upcurved normal vector tobe obtain here
๐๐ฅ= -2
๐๐ฆ =-2
N=(2i+2j+k)
Then ๐ ๐น.๐๐๐ = ๐
( โ๐ โ ๐ + 2๐ฆ๐ . (2๐ + 2๐ + ๐))๐๐ด
= 03 03โ๐ฅ
โ2 โ 2 + 2๐ฆ ๐๐ฆ๐๐ฅ
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= 03๐ฆ2 โ 4๐ฆ
= 03(3 โ ๐ฅ2 โ 4(3 โ ๐ฅ) ๐๐ฅ
= 03(9 โ 6๐ฅ + ๐ฅ2 โ 12 + 4๐ฅ dx
= 03๐ฅ2 โ 2๐ฅ โ 3 ๐๐ฅ
=๐ฅ3
3โ ๐ฅ2 โ 3๐ฅ
=33
3โ 32 โ 3.3
= 9- 9- 9
=-9 (Answer)
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