Check Stokes’ theorem using the function v = ay ˆx + bx ˆy (a and b are constants) and the circular path of radius R, centered at the origin in the xy plane. [Answer: π R^2(b − a)]

Question

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

Accepted Answer

[latex]∇ imes 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} ight | = \hat{z} (b − a). So \int{} (∇ imes v) · da = (b − a)\pi R^2[/latex] .

[latex]v · 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[/latex],

so [latex]dy = −(x/y) dx. So v · dl = ay dx + bx(−x/y) dx =\frac{1}{y}\left(ay^2 − bx^2 ight)dx [/latex] .

For the “upper” semicircle, [latex]y =\sqrt{R^2-x^{2} }, so v · dl =\frac{a\left(R^2−x^2 ight) −bx^2}{\sqrt{R^2−x^2} }dx [/latex].

[latex]\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} ight)− (a + b)\left[-\frac{x}{2}\sqrt{R^2 − x^2} +\frac{R^2}{2}\sin ^{-1}\left(\frac{x}{R} ight) ight] ight\}| ^{+R}_{-R}[/latex]

=[latex]\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) ight)=\frac{1}{2}R^2(a − b)\left(-\frac{\pi }{2}-\frac{\pi }{2} ight) [/latex]

=[latex]\frac{1}{2}\pi R^2 (b − a)[/latex].

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

[latex]\oint{}v · dl = \pi R^2(b − a)[/latex] .

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

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