Let $x_1$ be the red curve and $x_2$ be the blue curve. I've tried integrating $x_1-x_2$ as well as $x_2-x_1$ from $y=0$ to $y=9/2$ and both answers were marked wrong.
What do I do here? I'm thinking it might be better to integrate with respect to $x$ and break it down into intervals, but I'm not sure how to go about doing this.
$\endgroup$ 11 Answer
$\begingroup$In that region, the smallest $y$-coordinate of a point is $0$ (occuring at the point $(0,0)$) and the greatest $x$-coordinate is $\frac92$ (occuring at the point $\left(-\frac{27}4,\frac92\right)$).
So, you should consider the integral$$\int_0^{9/2}(3y-y^2)-(y^2-6y)\,\mathrm dy=\int_0^{9/2}9y-2y^2\,\mathrm dy=\frac{243}8.$$
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