# Lazi's question at Yahoo! Answers regarding the area bounded by two functions

Staff member

#### MarkFL

I would orient the coordinate axes such that the $x$-axis is vertical, and the $y$-axis is horizontal. To further simplify matters, I would vertical shift both functions down two units, then use symmetry so that we wish to find 4 times the following shaded area:
Hence the desired area $A$ is:
$$\displaystyle A=4\int_0^{\sqrt{2}}2-y^2\,dy=\frac{4}{3}\left[6y-y^3 \right]_0^{\sqrt{2}}=\frac{4}{3}\left(6\sqrt{2}-2\sqrt{2} \right)=\frac{16\sqrt{2}}{3}$$