Problem: Compute ∫C2ydx−3dy, where C is an ellipse parametrized by r(θ)=(5+4cosθ,3+2sinθ).
Solution: First calculate that x′(t)=−4sinθ and y′(t)=2cosθ. Plug everything into the integrand to obtain:
(5+4cosθ,3+2cosθ)=(2(3+2sinθ)(−4sinθ)−3(2cosθ))dθ.
Simplify this and evaluate:
∫02π(2(3+2sinθ)(−4sinθ)−3(2cosθ))dθ=−∫02π24sinθ+16sin2θ+6cosθ=−16π.