Answer to Question #197816 in Calculus for Rocky Valmores

Question #197816

 Verify Stokes’ theorem for 𝐴 = 2π‘₯ βˆ’ 𝑦 𝑖 βˆ’ 𝑦𝑧2 𝑗 βˆ’ 𝑦2π‘§π‘˜, where S is the upper half of the sphere π‘₯2+ 𝑦2 + 𝑧2 = 1 and C is its boundary.


1
Expert's answer
2021-05-27T18:07:45-0400

∫CFβƒ—β‹…drβƒ—=∫∫ScurlFβƒ—β‹…n^dS  (Stokeβ€²s Theorem)\int_C\vec F\cdot \vec{dr}=\int\int_S curl \vec F\cdot \hat ndS\ \ (Stoke's\ Theorem)


where C is the circle: x2+y2=1, z=0  (xyβˆ’plane)x^2+y^2=1,\ z=0\ \ (xy-plane)


i.e. x=cost, y=sint, z=0x=cost,\ y=sint,\ z=0

r=xiβƒ—+yjβƒ— where  0≀θ≀2Ο€drβƒ—=dxiβƒ—+dyjβƒ—r=x\vec i+y\vec j\ where\ \ 0\leq \theta\leq 2\pi\\\vec{dr}=dx\vec i+dy\vec j


where,

Fβƒ—=(2xβˆ’y)iβƒ—βˆ’yz2jβƒ—βˆ’y2zkβƒ—βˆ΄Fβƒ—β‹…drβƒ—=(2xβˆ’y)dx (since 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 )


LHS =∫CFβƒ—β‹…drβƒ—=∫C(2xβˆ’y)dx= \int_C \vec F\cdot \vec{dr}=\int_C(2x-y)dx


=∫02Ο€(2costβˆ’sint)(βˆ’sint)dt =∫02Ο€(sin2tβˆ’2costsint)dt =∫02Ο€{12(1βˆ’cos2t)βˆ’sin2t}dt =[t2βˆ’sin2t4βˆ’cos2t2]o2Ο€ =(12βˆ’12)+(Ο€βˆ’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


Hence, Fβƒ—β‹…drβƒ—=Ο€   .....(1)\vec F\cdot \vec{dr}=\pi\ \ \ .....(1)


Also,

curl  Fβƒ—=β–½Γ—Fβƒ—curl \ \ \vec F=\triangledown \times \vec F

=∣iβƒ—jβƒ—k⃗δδxδδyδδz2xβˆ’yβˆ’yz2βˆ’y2z∣ =iβƒ—(βˆ’2yz+2yz)βˆ’jβƒ—(0)+kβƒ—(0+1)=kβƒ—= \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


∴dSβƒ—=n^dS=dydz i+dzdx j+dxdy k\therefore \vec {dS}=\hat n dS\\=dydz\ i+dzdx\ j+dxdy\ k


Hence,

RHS = βˆ«βˆ«S curl Fβƒ—β‹…n^dS=∫∫dxdy=Ο€  .....(2)RHS\ =\ \int\int_S\ curl\ \vec F\cdot \hat {n}dS=\int\int dxdy=\pi\ \ .....(2)


Since,  βˆ«βˆ«dxdySince,\ \ \int\int dxdy represent the area of the circle x2+y2=1x^2+y^2=1 which is Ο€\pi


Thus, from (1) and (2) we conclude that the theorem is verified.


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