Compute the curveintegral $$I=\int_\gamma x(4x^2+4y+1) \ dx+16y \ dy,$$
where $\gamma$ is the part of the curve $\left(4x^2+4y^2\right)^2=-x^2+16$ positioned in the right hemisphere going from the point $[0,-1]$ to $[0,1].$
So I have that $(P,Q)=(4x^3+4xy+x,16y),$ parametrizing by
$$\left\{ \begin{array}{rcr} x & = & \cos{t} \\ y & = & \sin{t} \\ \end{array} \right. \implies\left\{ \begin{array}{rcr} dx & = & -\sin{t} \\ dy & = & \cos{t} \\ \end{array} \right., \quad t\in[-\pi/2,\pi/2]$$
so using $I=\int_\gamma Pdx+Qdy$ I get
$$I=\int_{-\pi/2}^{\pi/2}(-4\cos^3{t}\sin{t}-4\sin^2{t}\cos{t}-\cos{t}\sin{t}+16\sin{t}\cos{t}) \ dt = 0-\frac{8}{3}+0=-\frac{8}{3}.$$
Is this correct?