Let's find work of the field directly:
W = ∫ C x 2 d x − x y d y W = \int\limits_C {{x^2}} dx - xydy W = C ∫ x 2 d x − x y d y
Parametric equation of a circle:
x = 2 cos t ⇒ d x = − 2 sin t d t , y = 2 sin t + 2 ⇒ d y = 2 cos t d t , − π 2 < t < π 2 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} x = 2 cos t ⇒ d x = − 2 sin t d t , y = 2 sin t + 2 ⇒ d y = 2 cos t d t , − 2 π < t < 2 π
Then
W = ∫ C x 2 d x − x y d y = ∫ − π 2 π 2 4 cos 2 t ( − 2 sin t ) d t − 4 cos t ( sin t + 1 ) ⋅ 2 cos t d t = 8 ∫ − π 2 π 2 cos 2 t d cos t + 8 ∫ − π 2 π 2 cos 2 t d cos t − 8 ∫ − π 2 π 2 cos 2 t d t = 16 3 cos 3 t ∣ − π 2 π 2 − 2 ∫ − π 2 π 2 ( 1 + cos 2 t ) d 2 t = 0 − 2 ( 2 t ∣ − π 2 π 2 + sin 2 t ∣ − π 2 π 2 ) = − 4 π 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 W = C ∫ x 2 d x − x y d y = − 2 π ∫ 2 π 4 cos 2 t ( − 2 sin t ) d t − 4 cos t ( sin t + 1 ) ⋅ 2 cos t d t = 8 − 2 π ∫ 2 π cos 2 t d cos t + 8 − 2 π ∫ 2 π cos 2 t d cos t − 8 − 2 π ∫ 2 π cos 2 t d t = 3 16 cos 3 t ∣ − 2 π 2 π − 2 − 2 π ∫ 2 π ( 1 + cos 2 t ) d 2 t = 0 − 2 ( 2 t ∣ − 2 π 2 π + sin 2 t ∣ − 2 π 2 π ) = − 4 π
Let's find
∂ P ∂ y = ∂ x 2 ∂ y = 0 ; ∂ Q ∂ x = ∂ ( − x y ) ∂ x = − y \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 ∂ y ∂ P = ∂ y ∂ x 2 = 0 ; ∂ x ∂ Q = ∂ x ∂ ( − x y ) = − y
Then, by Green's theorem, work is
W = ∫ ∫ D ( ∂ Q ∂ x − ∂ P ∂ y ) d x d y = − ∫ ∫ D y d x d y W = \int {\int\limits_D {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)} } dxdy = - \int {\int\limits_D y } dxdy W = ∫ D ∫ ( ∂ x ∂ Q − ∂ y ∂ P ) d x d y = − ∫ D ∫ y d x d y
Let's move on to polar coordinates:
x = r cos φ , y = r sin φ , 0 < φ < π 2 x = r\cos \varphi ,\,\,y = r\sin \varphi ,\,\,0 < \varphi < \frac{\pi }{2} x = r cos φ , y = r sin φ , 0 < φ < 2 π
x 2 + ( y − 2 ) 2 = 4 ⇒ r 2 cos 2 φ + r 2 sin 2 φ − 4 r sin φ + 4 = 4 ⇒ r 2 = 4 r sin φ ⇒ r = 4 sin φ {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 x 2 + ( y − 2 ) 2 = 4 ⇒ r 2 cos 2 φ + r 2 sin 2 φ − 4 r sin φ + 4 = 4 ⇒ r 2 = 4 r sin φ ⇒ r = 4 sin φ
Then
0 < r < 4 sin φ 0 < r < 4\sin \varphi 0 < r < 4 sin φ
So
W = − ∫ ∫ D y d x d y = − ∫ 0 π 2 d φ ∫ 0 4 sin φ r sin φ ⋅ r d r = − 1 3 ∫ 0 π 2 sin φ ⋅ r 3 ∣ 0 4 sin φ d φ = − 1 3 ∫ 0 π 2 64 sin 4 φ d φ = − 64 3 ∫ 0 π 2 ( 1 − cos 2 φ 2 ) 2 d φ = − 16 3 ∫ 0 π 2 ( 1 − 2 cos 2 φ + cos 2 2 φ ) d φ = − 8 3 ∫ 0 π 2 ( 1 − 2 cos 2 φ ) d 2 φ − 8 3 ∫ 0 π 2 ( 1 + cos 4 φ ) d φ = − 8 3 ( 2 φ ∣ 0 π 2 − 2 sin 2 φ ∣ 0 π 2 ) − 2 3 ( 4 φ ∣ 0 π 2 + sin 4 φ ∣ 0 π 2 ) = − 8 3 π − 2 3 ⋅ 2 π = − 12 3 π = − 4 π 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 W = − ∫ D ∫ y d x d y = − 0 ∫ 2 π d φ 0 ∫ 4 s i n φ r sin φ ⋅ r d r = − 3 1 0 ∫ 2 π sin φ ⋅ r 3 ∣ ∣ 0 4 s i n φ d φ = − 3 1 0 ∫ 2 π 64 sin 4 φ d φ = − 3 64 0 ∫ 2 π ( 2 1 − c o s 2 φ ) 2 d φ = − 3 16 0 ∫ 2 π ( 1 − 2 cos 2 φ + cos 2 2 φ ) d φ = − 3 8 0 ∫ 2 π ( 1 − 2 cos 2 φ ) d 2 φ − 3 8 0 ∫ 2 π ( 1 + cos 4 φ ) d φ = − 3 8 ( 2 φ ∣ 0 2 π − 2 sin 2 φ ∣ 0 2 π ) − 3 2 ( 4 φ ∣ 0 2 π + sin 4 φ ∣ 0 2 π ) = − 3 8 π − 3 2 ⋅ 2 π = − 3 12 π = − 4 π
Since in both cases we obtained the same value of the work, Green's theorem holds.
#339914 on Dec 2023
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