Answer
$0$
Work Step by Step
The parametric equations are:
$ x=2 \cos t , y=2 \sin t$
Here, we have
$ dx= -2 \sin t dt, dy=2 \cos t dt$
Plug the above values in the given integral.
$\int_{0}^{2 \pi}( -2 \sin t )^2 (-2 \sin t dt)+(2 \cos t)^2 (2 \cos t dt)= \int_{0}^{2 \pi} (-8\sin^2 t+8 \cos^3 t) dt$
$\int_{0}^{2 \pi} (-8\sin t(1-\cos^2 t)+8 \cos t(1-\sin^2 t) ) dt=\int_{0}^{2 \pi} -8 sin t +8 \sin t\cos^2 t +8 \cos t -8 \cos t \sin^2 t$
Thus, we have
$[8\cos t-\dfrac{8}{3}\cos^3 t+8 \sin t-\dfrac{8}{3}\sin^3 t]_{0}^{2 \pi} =(8\cos (2 \pi)-\dfrac{8}{3}\cos^3 (2 \pi)+8 \sin (2 \pi)-\dfrac{8}{3}\sin^3 (2 \pi))-(8\cos (0)-\dfrac{8}{3}\cos^3 (0)+8 \sin (0)-\dfrac{8}{3}\sin^3 (0))= 0$
