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

[MarkFL]

Here is the question:

Find the area between those curves: x=y^2, and x=4-y^2?

Find the area between those curves: x=y^2, and x=4-y^2

the answer should be (16 sqrt2 ) /3

Here is a link to the question:

Find the area between those curves: x=y^2, and x=4-y^2? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Lazi,

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:

\(\displaystyle A=4\int_0^{\sqrt{2}}2-y^2\,dy=\frac{4}{3}\left[6y-y^3 \right]_0^{\sqrt{2}}=\frac{4}{3}\left(6\sqrt{2}-2\sqrt{2} \right)=\frac{16\sqrt{2}}{3}\)