Question

 Verify Stokes’ theorem for 𝐴 = 2𝑥 − 𝑦 𝑖 − 𝑦𝑧² 𝑗 − 𝑦²𝑧𝑘, where S is the upper half of the sphere 𝑥²+ 𝑦² + 𝑧² = 1 and C is its boundary.

Answer 1

$\int_C\vec F\cdot \vec{dr}=\int\int_S curl \vec F\cdot \hat ndS\ \ (Stoke's\ Theorem)$

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

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

$r=x\vec i+y\vec j\ where\ \ 0\leq \theta\leq 2\pi\\\vec{dr}=dx\vec i+dy\vec j$

where,

$\vec F=(2x-y)\vec i-yz^2\vec j-y^2z\vec k\\\therefore\vec F\cdot \vec{dr}=(2x-y)dx\ (since\ z=0 )$

LHS $= \int_C \vec F\cdot \vec{dr}=\int_C(2x-y)dx$

$=\int_0^{2\pi}(2cost-sint)(-sint)dt\\\ \\=\int_0^{2\pi}(sin^2t-2costsint)dt\\\ \\=\int_0^{2\pi}\{\frac{1}{2}(1-cos2t)-sin2t\}dt\\\ \\=[\dfrac{t}{2}-\dfrac{sin2t}{4}-\dfrac{cos2t}{2}]_o^{2\pi}\\\ \\=(\dfrac{1}{2}-\dfrac{1}{2})+(\pi-0)\\\ \\=\pi$

Hence, $\vec F\cdot \vec{dr}=\pi\ \ \ .....(1)$

Also,

$curl \ \ \vec F=\triangledown \times \vec F$

$= \begin{vmatrix} \vec i & \vec j&\vec k \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y}&\frac{\delta}{\delta z}\\ 2x-y&-yz^2&-y^2z \end{vmatrix}\\\ \\=\vec i(-2yz+2yz)-\vec j (0)+\vec k (0+1)\\=\vec k$

$\therefore \vec {dS}=\hat n dS\\=dydz\ i+dzdx\ j+dxdy\ k$

Hence,

$RHS\ =\ \int\int_S\ curl\ \vec F\cdot \hat {n}dS=\int\int dxdy=\pi\ \ .....(2)$

$Since,\ \ \int\int dxdy$ represent the area of the circle $x^2+y^2=1$ which is $\pi$

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