Question 1.55: Check Stokes’ theorem using the function v = ay ˆx + bx ˆy (...

Check Stokes’ theorem using the function v=ayx^+bxy^v=ay\hat{x}+bx\hat{y} (a and b are constants) and the circular path of radius R, centered at the origin in the xy plane. [Answer: πR2(ba)]π R^2(b − a)]

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×v=x^y^z^xyzaybx0=z^(ba). So (×v)da=(ba)πR2∇\times v=\left | \begin{matrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ay & bx & 0 \end{matrix} \right | = \hat{z} (b − a).   So  \int{} (∇\times v) · da = (b − a)\pi R^2 .

vdl=(ayx^+bxy^)(dxx^+dyy^+dzz^)=aydx+bxdy;x2+y2=R22xdx+2ydy=0v · dl = (ay \hat{x} + bx \hat{y} ) · (dx \hat{x} + dy \hat{y} + dz \hat{z} ) = ay dx + bx dy; x^2 + y^2 = R^2 \Rightarrow 2xdx + 2y dy = 0,

so dy=(x/y)dx. So vdl=aydx+bx(x/y)dx=1y(ay2bx2)dxdy = −(x/y) dx.   So   v · dl = ay dx + bx(−x/y) dx =\frac{1}{y}\left(ay^2 − bx^2\right)dx .

For the “upper” semicircle, y=R2x2, so  vdl=a(R2x2)bx2R2x2dxy =\sqrt{R^2-x^{2} },   so    v · dl =\frac{a\left(R^2−x^2\right) −bx^2}{\sqrt{R^2−x^2} }dx .

vdl=RRaR2(a+b)x2R2x2dx={aR2sin1(xR)(a+b)[x2R2x2+R22sin1(xR)]}R+R\int{}v · dl =\int\limits_{R}^{-R}{} \frac{aR^2 − (a + b)x^2}{\sqrt{R^2 − x^2} }dx =\left\{aR^2\sin ^{-1}\left(\frac{x}{R} \right)− (a + b)\left[-\frac{x}{2}\sqrt{R^2 − x^2} +\frac{R^2}{2}\sin ^{-1}\left(\frac{x}{R} \right) \right] \right\}| ^{+R}_{-R}

=12R2(ab)sin1(x/R)+RR=12R2(ab)(sin1(1)sin1(+1))=12R2(ab)(π2π2)\frac{1}{2}R^2(a − b)\sin ^{-1}(x/R)\mid ^{-R}_{+R}=\frac{1}{2}R^2(a − b)\left(\sin ^{-1}(−1) − \sin ^{-1}(+1) \right)=\frac{1}{2}R^2(a − b)\left(-\frac{\pi }{2}-\frac{\pi }{2} \right)

=12πR2(ba)\frac{1}{2}\pi R^2 (b − a).

And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so

vdl=πR2(ba)\oint{}v · dl = \pi R^2(b − a) .

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