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

Green's Theorem:

$\oint_C (Ldx+Mdy)=\iint_D(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y})dxdy$

$\oint_C((x − 2y^2) dx + (y^4 + 2xy) dy)=$$\iint_D(2y+4y)dxdy=\iint_D6ydxdy$

Let's move to polar coordinates:

$x=r\cos \phi$

$y=r\sin\phi$

$2\leq r\leq 4$

$\frac{\pi}{2}\leq\phi\leq \arctan(\frac{2\sqrt 3}{-2})=\frac{2\pi}{3}$

$\displaystyle\iint_D6ydxdy=\int_{\frac{\pi}{2}}^{\frac{2\pi}{3}}d\phi\int_2^46r\sin\phi \cdot rdr=$$\int_{\frac{\pi}{2}}^{\frac{2\pi}{3}}\sin\phi d\phi\int_2^46r^2dr=-(\cos\phi|_{\frac{\pi}{2}}^{\frac{2\pi}{3}})\cdot(2r^3|_2^4)=\\$$=-(0.5-0)\cdot2(4^3-2^3)=-56$

Answer: $\oint_C((x − 2y^2) dx + (y^4 + 2xy) dy)=-56$ .