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Concept:
Green’s Theorem:
If M(x, y), N(x, y), My and Nx be continuous in a region E of the xy-plane bounded by a closed curve C, then
∫C(Mdx+Ndy)=∫∫E(∂N∂x−∂M∂y)dxdy
Calculation:
The given function is in the form of
(M dx + N dy)
M = 4y2x, N = 8x2y
∂M∂y=8xy,∂N∂x=16xy
∫C(4y2xdx+8x2ydy)=∫∫S(∂N∂x−∂M∂y)dxdy
=∫∫S(16xy−8xy)dxdy
=∫∫S8xydxdy
=∫01∫018xydxdy
=∫01[4xy2]01dx
=∫014xdx
=[2x2]01=2
y = x2 – 4x + 4
dy = 2x dx – 4 dx = (2x - 4) dx
dr→=dxi^+dyj^
∮C(yi^−3xj^)⋅(dxi^+dyj^)
=∮Cydx−3xdy
=∮C(x2−4x+4)dx−3x(2x−4)dx
=∮C(−5x2+8x+4)dx
∫x=02(−5x2+8x+4)dx
=[−5x33+4x2+4x]02
=−53(8)+4(4)+4(2)
−403+24=323
The value of the integral
∮sr→.n→ds
According to Gauss Divergence theorem:
∮A.ds=∭(∇.A)dv
∮F.→n^ds=∭(∇.F)dv
Given that S is a closed surface:
∮r.→n^ds=∭(∇.r)dv
r→=xi^+yj^+zk^
∇.r=∂∂x(x)+∂∂y(y)+∂∂z(z)=3
∮r.→n^ds=∭(∇.r)dv=∭3dv=3V
The value of ∫C(2x+3y)dx−(3x−4y)dy where c is the circle with radius as 1 and centre at origin.
C ≡ x2 + y2 = 1
Let x = r cos θ, y = r sin θ
⇒ dx = -sin θ dθ, dy = cos θ dθ
∫θ=02π[(2cosθ+3sinθ)(−sinθ)dθ−(3cosθ−4sinθ)(cosθ)dθ]=∫02π[−2sinθcosθ−3sin2θ−3cos2θ+4sinθcosθ]dθ=∫02π(2sinθcosθ−3)dθ=∫02π(sin2θ−3)dθ
= -3(2π) = -6π
By stokes theorem,
∮CF.dr=∫∫SCurlF.Nds
F→(x,y,z)=xi^+yj^+3(x2+y2)k^
∇×F→=|ijk∂∂x∂∂y∂∂zxy3(x2+y2)|
= i (6y) – j(6x) = 6y î - 6x ĵ
S: x2 + y2 + z2 – 64 = 0
ds=+2xi^+2yj^+2zk^
(∇×F→)⋅ds=(6yi^−6xj^)⋅(2xi^+2yj^+2zk^)
= 12xy – 12xy + 0 = 0
⇒∫CF→⋅dr→=0
If S is the surface of the sphere x2 + y2 + z2 = a2, then the value of
From Gauss divergence theorem,
∬SF.ds=∭V∇.F→dV
F→=(x+z)i+(y+z)j+(x+y)k
∇.F→=1+1+0=2
∭V∇.F→dV=∭2dV
2∭dV
The volume of the sphere =43πa3
⇒∭V∇.F→dV=2×43πa3=83πa3
If S be any closed surface, evaluate ∫SCurlF→.ds→
Explanation:
Cut open the surface S by any plane and Let S1, S2 denotes its upper and lower portions.
Let C be the common curve bounding both these portions.
∫SCurlF→.d→s=∫S1CurlF→.d→s+∫S2CurlF→.d→s
∫SCurlF→.d→s=∫CF→.d→r⇒∫SCurlF→.d→s=∫CF→.d→r−∫CF→.d→r=0
The second integral is negative because it is traversed in a direction opposite to that of the first.
The numerical value of work done (rounded off to the nearest integer) by a position-dependent force F→=xi^+5xyj^ (where i^ and j^ are unit vectors) along the path, y=x22, from (0, 0) to (2, 2) in the xy plane is ________.
F→=xi^+5xyj^
Work done is given by:
W=∫CF→.dr→
r→=xi^+yj^
W=∫CF→.dr→=∫C(xi^+5xyj^).d(xi^+yj^)=∫C(xdx+5xydy)
Substituting y=x22 and x is varying from 0 to 2
dy = x dx
∫CF→.dr→=∫C(xdx+5x(x22)xdx)
=∫02(x+52x4)dx
=[x22]02+[x52]02
F→=3xyi^−y2j^ and R→=xi^+yj^
∫CF→.dR→=∫C(3xyi^−y2j^).d(xi^+yj^)=∫C(3xydx−y2dy)
Substituting y = 2x2 and x is varying from 0 to 1
⇒ dy = 4x dx
∫CF→.dR→=∫C(3x(2x2)dx−(2x2)2(4xdx)
=∫01(6x3−16x5)dx
=[64x4]01−[166x6]01
=64−86=18−3212=−76=−1.17
Alternate method:
Let x = t
⇒ dx = dt
y = 2x2 = 2t2
⇒ dy = 4tdt
t varies from 0 to 1.
∫CF→.dR→=∫C(3(t)(2t2)dt−(2t2)2(4tdt))
=∫01(6t3−16t5)dt
=[64t4]01−[166t6]01
The value of the line integral 2π∮γ(−y3dx+x3dy), where γ is the circle x2 + y2 = 1 oriented counter clockwise, is ________.
2π∮γ(−y3dx+x3dy)
It is in the form of ∫C(Mdx+Ndy)
M = -y3 and N = x3
∂N∂x=3x2
∂M∂y=−3y2
2π∮γ(−y3dx+x3dy)=2π∫∫E(3x2+3y2)dxdy
Changing to polar coordinates (r, θ), r varies from 0 to 1 and θ varies from 0 to 2π.
=2π∫01∫02π(3(rcosθ)2+3(rsinθ)2)rdθdr
=2π∫01∫02π3r3dθdr
=2π∫01[3r3θ]02πdr
=6π∫012πr3dr
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