β« C F β β
d r β = β« β« S c u r l F β β
n ^ d S ( S t o k e β² s T h e o r e m ) \int_C\vec F\cdot \vec{dr}=\int\int_S curl \vec F\cdot \hat ndS\ \ (Stoke's\ Theorem) β« C β F β
d r = β« β« S β c u r l F β
n ^ d S ( St o k e β² s T h eore m )
where C is the circle: x 2 + y 2 = 1 , z = 0 ( x y β p l a n e ) x^2+y^2=1,\ z=0\ \ (xy-plane) x 2 + y 2 = 1 , z = 0 ( x y β pl an e )
i.e. x = c o s t , y = s i n t , z = 0 x=cost,\ y=sint,\ z=0 x = cos t , y = s in t , z = 0
r = x i β + y j β w h e r e 0 β€ ΞΈ β€ 2 Ο d r β = d x i β + d y j β r=x\vec i+y\vec j\ where\ \ 0\leq \theta\leq 2\pi\\\vec{dr}=dx\vec i+dy\vec j r = x i + y j β w h ere 0 β€ ΞΈ β€ 2 Ο d r = d x i + d y j β
where,
F β = ( 2 x β y ) i β β y z 2 j β β y 2 z k β β΄ F β β
d r β = ( 2 x β y ) d x ( s i n c e z = 0 ) \vec F=(2x-y)\vec i-yz^2\vec j-y^2z\vec k\\\therefore\vec F\cdot \vec{dr}=(2x-y)dx\ (since\ z=0 ) F = ( 2 x β y ) i β y z 2 j β β y 2 z k β΄ F β
d r = ( 2 x β y ) d x ( s in ce z = 0 )
LHS = β« C F β β
d r β = β« C ( 2 x β y ) d x = \int_C \vec F\cdot \vec{dr}=\int_C(2x-y)dx = β« C β F β
d r = β« C β ( 2 x β y ) d x
= β« 0 2 Ο ( 2 c o s t β s i n t ) ( β s i n t ) d t = β« 0 2 Ο ( s i n 2 t β 2 c o s t s i n t ) d t = β« 0 2 Ο { 1 2 ( 1 β c o s 2 t ) β s i n 2 t } d t = [ t 2 β s i n 2 t 4 β c o s 2 t 2 ] o 2 Ο = ( 1 2 β 1 2 ) + ( Ο β 0 ) = Ο =\int_0^{2\pi}(2cost-sint)(-sint)dt\\\ \\=\int_0^{2\pi}(sin^2t-2costsint)dt\\\ \\=\int_0^{2\pi}\{\frac{1}{2}(1-cos2t)-sin2t\}dt\\\ \\=[\dfrac{t}{2}-\dfrac{sin2t}{4}-\dfrac{cos2t}{2}]_o^{2\pi}\\\ \\=(\dfrac{1}{2}-\dfrac{1}{2})+(\pi-0)\\\ \\=\pi = β« 0 2 Ο β ( 2 cos t β s in t ) ( β s in t ) d t = β« 0 2 Ο β ( s i n 2 t β 2 cos t s in t ) d t = β« 0 2 Ο β { 2 1 β ( 1 β cos 2 t ) β s in 2 t } d t = [ 2 t β β 4 s in 2 t β β 2 cos 2 t β ] o 2 Ο β = ( 2 1 β β 2 1 β ) + ( Ο β 0 ) = Ο
Hence, F β β
d r β = Ο . . . . . ( 1 ) \vec F\cdot \vec{dr}=\pi\ \ \ .....(1) F β
d r = Ο ..... ( 1 )
Also,
c u r l F β = β½ Γ F β curl \ \ \vec F=\triangledown \times \vec F c u r l F = β½ Γ F
= β£ i β j β k β Ξ΄ Ξ΄ x Ξ΄ Ξ΄ y Ξ΄ Ξ΄ z 2 x β y β y z 2 β y 2 z β£ = i β ( β 2 y z + 2 y z ) β j β ( 0 ) + k β ( 0 + 1 ) = k β = \begin{vmatrix}
\vec i & \vec j&\vec k \\
\frac{\delta}{\delta x} & \frac{\delta}{\delta y}&\frac{\delta}{\delta z}\\ 2x-y&-yz^2&-y^2z
\end{vmatrix}\\\ \\=\vec i(-2yz+2yz)-\vec j (0)+\vec k (0+1)\\=\vec k = β£ β£ β i Ξ΄ x Ξ΄ β 2 x β y β j β Ξ΄y Ξ΄ β β y z 2 β k Ξ΄z Ξ΄ β β y 2 z β β£ β£ β = i ( β 2 yz + 2 yz ) β j β ( 0 ) + k ( 0 + 1 ) = k
β΄ d S β = n ^ d S = d y d z i + d z d x j + d x d y k \therefore \vec {dS}=\hat n dS\\=dydz\ i+dzdx\ j+dxdy\ k β΄ d S = n ^ d S = d y d z i + d z d x j + d x d y k
Hence,
R H S = β« β« S c u r l F β β
n ^ d S = β« β« d x d y = Ο . . . . . ( 2 ) RHS\ =\ \int\int_S\ curl\ \vec F\cdot \hat {n}dS=\int\int dxdy=\pi\ \ .....(2) R H S = β« β« S β c u r l F β
n ^ d S = β«β« d x d y = Ο ..... ( 2 )
S i n c e , β« β« d x d y Since,\ \ \int\int dxdy S in ce , β«β« d x d y represent the area of the circle x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 which is Ο \pi Ο
Thus, from (1) and (2) we conclude that the theorem is verified.
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