First we need to calculate intersection points. This we can determine from these two equations
Also since `x(t_1)=-2` and `x(t_2)=2` it's easy to see that area under `y=4` (this is simple to calculate because it is actually square) is `4cdot(2-(-2))=16.` If we subtract area under the other curve we will get area between the curves.
And since area under parametrically defined curve `x=f(t),` `y=g(t)` is given by formula
` ` `8-6ln3` <--Your solution
Area between the curves is equal to `8-6ln3.`