## Calculus 8th Edition

$\int_Cydx+(x+y^{2})dy=0$
Rewrite the equation of the ellipse as $(\frac{x}{3})^{2}+(\frac{y}{2})^{2}=1$ We can parameterize the ellipse as $C:x=3cost,y=2sint$ Starting from $(3,0)$ when we complete on round of the ellipse , $t$ increases from $0$ to $2\pi$ Note that $dx=-3sintdt$ and $dy=2costdt$ Therefore, $\int_Cydx+(x+y^{2})dy=\int_{0}^{2\pi}2sint(-3sint)dt+(3cost+4sin^{2}t)(2cost)dt$ $=\int_{0}^{2\pi}6cos2t+8(1-cos^{2}t)(cost)dt$ $=\int_{0}^{2\pi}6cos2t+8cost-8cos^{3}tdt$ Use formula: $4cos^{3}\theta=3cos\theta-cos3\theta$ $=\int_{0}^{2\pi}6cos2t+8cost-6cost+2cos3tdt$ $=0$ Hence, $\int_Cydx+(x+y^{2})dy=0$