Answer
$0$
Work Step by Step
$\textbf{F}(x,y)=3x\textbf{i}+4y\textbf{j}$
$C: \textbf{r}(t)=t\textbf{i}+\sqrt{4-t^2}\textbf{j}, -2\leq t\leq2$
$\textbf{r}(t)=(t,\sqrt{4-t^2})=(x(t),y(t)).$
Curve $C$ is the upper half of the circle with its center at $(0,0)$ and radius $R=2.$
We have
\begin{eqnarray*}\int_{C}\textbf{F}\cdot d\textbf{r}&=&\int_{-2}^{2}(\textbf{F}\circ\textbf{r})(t)\cdot(x'(t),y'(t))\,dt\\
&=&\int_{-2}^{2}(3x(t)x'(t)+4y(t)y'(t))\,dt\\
&=&\int_{-2}^{2}(3t(t')+4\sqrt{4-t^2}(\sqrt{4-t^2})')\,dt\\
&=&\int_{-2}^{2}\left(3t+4\sqrt{4-t^2}\frac{-2t}{2\sqrt{4-t^2}}\right)\,dt\\
&=&\int_{-2}^{2}(3t-4t)\,dt\\
&=&-\int_{-2}^{2}t\,dt=0
\end{eqnarray*}
