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Having the two functions $f(x) = x^2 + \sqrt{16-x^2}$ and $g(x) = x^2 - \sqrt{16-x^2}$

plotted as

enter image description here

how to find out the area of the enclosed area which looks like a "mouth" in an elegant way?

My thoughts

I've thought about adding $16$ to both functions, which should have no effect on the integrals of the two functions, but is this the most elegant way?

I've come up with 16 by observing that the lower curve only touches the $x$ axis, anything else (like 15 or 17) makes it cut the $x$ axis, which complicates things.

But why is $16$ the right number?

With $+16$, the two functions look like this:

enter image description here

If $+16$ is the most elegant technique, how to proceed afterwards?

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    Shouldn't you add $4$? You essentially want to shift up, and $g(0)=-4$. (By the way, the inequality sign in your definition of $g$ is a typo, correct?)2012-12-09

1 Answers 1

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All you need is the difference $f(x)-g(x)$ and integrate between intersectiopn points (i.e. zeroes of the difference), so there is no need to translate the two functions beforehand.

Then the most elegant way (assuming $g(x)$ should rather be $x^2-\sqrt{16-x^2}$) is to observe that the term $x^2$ can be subtracted from both $f$ and $g$ and that the area is then simply a circular disc with radius $4$.