Answer to Question #180961 in Calculus for John

Question

verify green's theorem by the force field F(x,y)= x^2 i - xy j , in moving a particle along the circle with in the first quadrent (x^2 + y-2)^2 = 4

Answer 1

Let's find work of the field directly:

$W = \int\limits_C {{x^2}} dx - xydy$

Parametric equation of a circle:

$x = 2\cos t \Rightarrow dx = - 2\sin tdt,\,\,y = 2\sin t + 2 \Rightarrow dy = 2\cos tdt,\,\, - \frac{\pi }{2} < t < \frac{\pi }{2}$

Then

$W = \int\limits_C {{x^2}} dx - xydy = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {4{{\cos }^2}t( - 2\sin t)dt - 4\cos t(\sin t + 1) \cdot 2\cos tdt} = 8\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^2}td\cos t + 8} \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^2}td\cos t} - 8\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^2}tdt} = \frac{{16}}{3}{\cos ^3}\left. t \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} - 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {(1 + \cos 2t)d2t} = 0 - 2\left( {2\left. t \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} + \sin 2\left. t \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}} \right) = - 4\pi$

Let's find

$\frac{{\partial P}}{{\partial y}} = \frac{{\partial {x^2}}}{{\partial y}} = 0;\,\,\frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( { - xy} \right)}}{{\partial x}} = - y$

Then, by Green's theorem, work is

$W = \int {\int\limits_D {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)} } dxdy = - \int {\int\limits_D y } dxdy$

Let's move on to polar coordinates:

$x = r\cos \varphi ,\,\,y = r\sin \varphi ,\,\,0 < \varphi < \frac{\pi }{2}$

${x^2} + {(y - 2)^2} = 4 \Rightarrow {r^2}{\cos ^2}\varphi + {r^2}{\sin ^2}\varphi - 4r\sin \varphi + 4 = 4 \Rightarrow {r^2} = 4r\sin \varphi \Rightarrow r = 4\sin \varphi$

Then

$0 < r < 4\sin \varphi$

So

$W = - \int {\int\limits_D y } dxdy = - \int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^{4\sin \varphi } {r\sin \varphi \cdot rdr} = - \frac{1}{3}\int\limits_0^{\frac{\pi }{2}} {\sin \varphi \left. { \cdot {r^3}} \right|_0^{4\sin \varphi }d\varphi } = - \frac{1}{3}\int\limits_0^{\frac{\pi }{2}} {64{{\sin }^4}\varphi d\varphi } = - \frac{{64}}{3}\int\limits_0^{\frac{\pi }{2}} {{{\left( {\frac{{1 - \cos 2\varphi }}{2}} \right)}^2}} d\varphi = - \frac{{16}}{3}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - 2\cos 2\varphi + {{\cos }^2}2\varphi } \right)d\varphi } = - \frac{8}{3}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - 2\cos 2\varphi } \right)} d2\varphi - \frac{8}{3}\int\limits_0^{\frac{\pi }{2}} {(1 + \cos 4\varphi )d\varphi = - \frac{8}{3}\left( {2\left. \varphi \right|_0^{\frac{\pi }{2}} - 2\sin 2\left. \varphi \right|_0^{\frac{\pi }{2}}} \right)} - \frac{2}{3}\left( {4\left. \varphi \right|_0^{\frac{\pi }{2}} + \sin 4\left. \varphi \right|_0^{\frac{\pi }{2}}} \right) = - \frac{8}{3}\pi - \frac{2}{3} \cdot 2\pi = - \frac{{12}}{3}\pi = - 4\pi$

Since in both cases we obtained the same value of the work, Green's theorem holds.