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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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I think this may be a matter of opinion. –  David Mitra Dec 9 '12 at 19:04
    
@DavidMitra Yeah I realise that. 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? - Why is 16 the right number? –  Flavius Dec 9 '12 at 19:06
    
$g(x)=x^2$ touches the $x$-axis even without any translation, doesn't it? –  Hagen von Eitzen Dec 9 '12 at 19:07
    
@HagenvonEitzen Yes, but the question is to find the area in an elegant way. –  Flavius Dec 9 '12 at 19:08
    
The plots do not match your function expressions. So how should we know what the significance of 16 should be? Moreover, your second plot is only translated up by $4$ for the red line, not $16$, and then things are somehow scaled!? –  Hagen von Eitzen Dec 9 '12 at 19:09

1 Answer 1

up vote 2 down vote accepted

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 $4$.

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