Multiply either side by $r$, use $r^2=x^2+y^2, r\sin t=y,r\cos t=x$
But we don't need $ab\ne 0$
If $a=0,r=b\cos t\implies r^2=b r\cos t\implies x^2+y^2-bx=0$
$\implies (x-\frac b 2)^2+y^2=(\frac b 2)^2$ is a circle.
If both $a,b$ are $0, r=0\implies x^2+y^2=0$ which is point-circle.
To integrate this region in polar coordinates,
it is advisable to break up the integral into two parts,
as shown in the figures below:
The two parts of the integral are divided by the diagonal line through
the upper right corner of the rectangle.
Since the sides of the rectangle are $a$ and $b$,
this diagonal line is at the angle $\arctan \frac ba.$
For $0 \leq \theta \leq \arctan \frac ba,$
you would integrate over $0 \leq r \leq a \sec\theta,$
and for $\arctan \frac ba \leq \theta \leq \frac\pi2,$
you would integrate over $0 \leq r \leq b \csc\theta.$
If you actually try this, I think you'll find that it is no easier than
doing the integration in rectangular coordinates.
It may even be worse.
An alternative approach, rather than combining $x^2+y^2$ into $r^2$,
is to integrate the terms separately:
$$\begin{eqnarray}
m &=& \int_0^a\int_0^b (1+x^2+y^2)\,dy\,dx \\
&=& \int_0^a\int_0^b dy\,dx
+\int_0^a\int_0^b x^2 \,dy\,dx
+\int_0^a\int_0^b y^2 \,dy\,dx
\end{eqnarray}$$
$$\begin{eqnarray}
m\bar{x} &=& \int_0^a\int_0^b x(1+x^2+y^2)\,dy\,dx \\
&=& \int_0^a\int_0^b x \,dy\,dx
+\int_0^a\int_0^b x^3 \,dy\,dx
+\int_0^a\int_0^b xy^2 \,dy\,dx
\end{eqnarray}$$
$$\begin{eqnarray}
m\bar{y} &=& \int_0^a\int_0^b y(1+x^2+y^2)\,dy\,dx \\
&=& \int_0^a\int_0^b y \,dy\,dx
+\int_0^a\int_0^b x^2y \,dy\,dx
+\int_0^a\int_0^b y^3 \,dy\,dx
\end{eqnarray}$$
Now you have nine integrals to solve, but they're all quite simple.
Best Answer
I must say that your differentiation went wrong near the beginning of the problem, and thus this problem is a little over complicated from that. The y derivative of the first term in the vector field is $ 1 + \sin y$, and the x derivative of the second term is $ \sin y $ the difference of these two is not $\sin y + 1$.