The area bounded by the curve y=|x^2-9| and the line y=3 is:

Question

The area bounded by the curve $y=|x^2-9|$ and the line $y=3$ is:

Accepted Answer

$9-x^2=3\Rightarrow x=\pm\sqrt6$; $x^2-9=3\Rightarrow x=\pm2\sqrt3$. The enclosed loop (for $x\ge0$) runs from $\sqrt6$ to $2\sqrt3$, split at $x=3$:

$$\int_{\sqrt6}^3(x^2-6)\,dx+\int_3^{2\sqrt3}(12-x^2)\,dx=(4\sqrt6-9)+(16\sqrt3-27)=16\sqrt3+4\sqrt6-36.$$

Doubling for the symmetric loop on $x<0$:
$$\text{Area}=2(16\sqrt3+4\sqrt6-36)=8(4\sqrt3+\sqrt6-9).$$

Solutions · 1