Show that $\int_C A\cdot dr=\frac{5}{3}$, where $A=(x^2-3y^2)i + (y^2-2x^2)j$ and $C$ is a closed curve in the $xy$-plane Show that $\int_C \mathbf{A}\cdot d\mathbf{r}=\frac{5}{3}$, where $\mathbf{A}=(x^2-3y^2)\mathbf{i}+(y^2-2x^2)\mathbf{j}$ and $C$ is a closed curve in the $xy$-plane such that
$$x=3\cos t, ~~ y=2\sin t, ~~~ 0\leq t\leq 2\pi.$$

\begin{align*}
\int_C \mathbf{A}\cdot d\mathbf{r}&amp;=\int_C(x^2-3y^2) dx + (y^2-2x^2) dy\\&amp;=\int_C \left[-3\, \left( 9\,  \cos^{2} \left( t \right)   -12\,
  \sin^{2} \left( t \right)  \right)   \sin \left( t
 \right) +2\, \left( -18\, \cos^{2} \left( t \right)   +
4\,  \sin^{2} \left( t \right) \right) \cos \left( t
 \right)\right] dt\\&amp;=\int_0^{2\pi}(-63\,\sin \left( t \right)  \cos^{2} \left( t \right)  
-44\, \cos^{3} \left( t \right)   +36\,\sin \left( t
 \right) +8\,\cos \left( t \right))dt\\&amp;= 0
\end{align*}

Don't know where is the mistake. Please help me to solve.
Hint: Use Green's theorem to get the solution easily.
Hint: Use Green's theorem to get the solution easily.