Question

Use Green’s Theorem to evaluate

∮C(x − 2y2) dx + (y4 + 2xy) dy where C consists of the line segment

from (0, 2) to (0, 4), followed by the curve with parametric equations x = 4 cos t, y = 4 sin t from (0, 4) to (−2, 2√3), then the line segment from (−2, 2√3) to (−1, √3), and finally the curve with parametric equations x = 2 sin t, y = 2 cos t from (−1, √3) to (0, 2).

Answer 1

ANSWER $\oint_{C} (x-2y^{2}) dx+(y^{4}+2xy) dy = \frac{28}{3}$

EXPLANATION

$C$ is the positively oriented , piecewise smooth , simple closed curve . Let $D$ is the region bounded by $C.$ Denote $L(x,y)=x-2y^{2}, M(x,y)=y^{4}+2xy$ , then $\frac{\partial L}{\partial y}=-4y, \frac{\partial M}{\partial x}=2y$ . Since all functions are continuous in $\R^{2}$ , by Green's theorem we have $\oint_{C} \left ( Ldx+Mdy \right )=\iint_{D} \left ( \frac{\partial M}{\partial x}-\frac{\partial L }{\partial y} \right )dxdy$ , where the path of integration along $C$ is counterclockwise. Therefore,

$\oint_{C} (x-2y^{2}) dx+(y^{4}+2xy) dy =\iint_{D} ( 2y+4y)dxdy=6\iint_{D} y dxdy$ .

To calculate the integral we replace the Cartesian coordinates with polar coordinates: $x=r\cos t , y=r\sin t.$ So,$\iint_{D} y dxdy=\iint_{Δ} r\cdot r \sin t dr dt$ , where

$Δ= \left\{ \left( t,r \right) :\frac { \pi }{ 2 } \le t\le \frac { 2\pi }{ 3 } ,\quad 2\le r\le 4 \right\}$ .

$\iint_{Δ} r\cdot r \sin t dr dt= \left( \int _{ \frac { \pi }{ 2 } }^{ \frac { 2\pi }{ 3 } }{ \sin { t } dt } \right) \cdot \left( \int _{ 2 }^{ 4 }{ { r }^{ 2 }dr } \right) =$$= \left [-\cos t \right ]_{\frac{\pi}{2}}^{\frac{2\pi}{3}} \cdot\left [ \frac{r^{3} }{3} \right ]_{2}^{4}=\left ( -\cos \frac{2\pi}{3} \right )\left ( \frac{64-8}{3} \right )=\frac{1}{2}\cdot\frac{56}{3}=\frac{28}{3}$