jmc

Completing the square in $x$ and $y,$ we get $$(x+3{)}^2+4(y-1{)}^2=4.$$Then $$\frac{(x+3{)}^2}{4}+\frac{(y-1{)}^2}{1}=1,$$so the semi-major axis is 2, the semi-minor axis is 1, and the area is then $\boxed{2\pi }.$