Check Stokes’ theorem using the function v=ayx^+bxy^ (a and b are constants) and the circular path of radius R, centered at the origin in the xy plane. [Answer: πR2(b−a)]
Check Stokes’ theorem using the function v=ayx^+bxy^ (a and b are constants) and the circular path of radius R, centered at the origin in the xy plane. [Answer: πR2(b−a)]
∇×v=∣∣∣∣∣∣∣x^∂x∂ayy^∂y∂bxz^∂z∂0∣∣∣∣∣∣∣=z^(b−a). So ∫(∇×v)⋅da=(b−a)πR2 .
v⋅dl=(ayx^+bxy^)⋅(dxx^+dyy^+dzz^)=aydx+bxdy;x2+y2=R2⇒2xdx+2ydy=0,
so dy=−(x/y)dx. So v⋅dl=aydx+bx(−x/y)dx=y1(ay2−bx2)dx .
For the “upper” semicircle, y=R2−x2, so v⋅dl=R2−x2a(R2−x2)−bx2dx.
∫v⋅dl=R∫−RR2−x2aR2−(a+b)x2dx={aR2sin−1(Rx)−(a+b)[−2xR2−x2+2R2sin−1(Rx)]}∣−R+R=21R2(a−b)sin−1(x/R)∣+R−R=21R2(a−b)(sin−1(−1)−sin−1(+1))=21R2(a−b)(−2π−2π)
=21πR2(b−a).
And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so
∮v⋅dl=πR2(b−a) .