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For some reason I keep getting 15, and I've tried this question multiple times.

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Please, edit your title. –  Sigur Sep 26 '12 at 16:20
    
Did you mean $y=\frac{1}{3}, y=\sqrt{x}$? –  B. S. Sep 26 '12 at 16:22
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This does not make a closed region, so there is no answer without a range of $x$. –  Ross Millikan Sep 26 '12 at 16:29
    
What about between $0$ and $\frac19$? Still, I'd like to see what work Noah has done, especially since this is marked homework. –  Mike Sep 26 '12 at 16:34
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2 Answers 2

If you reflect on the $x=y$ line (i.e. exchange $x$ and $y$), you get the same picture with the same area. There $x$ goes (I guess from 0, but you didn't mention) to $\displaystyle\frac13$, and $y=\sqrt x$ as (the upper part of $y^2=x$) becomes $x^2=y$ when exchanging. So, you are about to calculate $$\int_0^{\frac13}x^2dx$$

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Else,

it is the area of the orthogonal parallelogram with vertices (0,0),$\left(\frac{1}{9},0\right)$,$\left(\frac{1}{9},\frac{1}{3}\right)$,$\left(0,\frac{1}{3}\right)$ if we substract the area under the plot of the function $y=\sqrt{x}$.

So $\large\frac{1}{27}-\int\limits_{0}^{\frac{1}{9}}\sqrt{x}\;\mathbb{d}x=\frac{1}{27}-\frac{2}{81}=\frac{1}{81}$.

we are looking for the orange colored area.

Note: we are looking for the orange colored area.

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