# Centroid of the Region bounded by $y=\sqrt x$ and $y=x/2.$

Find the centroid of the region bounded by $$y=\sqrt x$$ and $$y=x/2.$$

So I did this problem by first:

Calculating the area using the formula: $$A=\int_0^4 \sqrt{x-\frac x2} ~\mathrm dx$$ and ended up with $$A=4/3.$$

Then I calculated $$M_x$$ using the formula: $$M_x=\frac12\int_0^4 \left(\left(\sqrt x\right)^2-\left(\frac x2\right)^2\right)~\mathrm dx$$ and got $$M_x=4/3.$$

I then calculated $$M_y$$ using the formula: $$M_y=\int_0^4\left(x\left(\sqrt{x}-\frac x2\right)\right)~\mathrm dx$$ and got $$M_y=32/15.$$

Lastly I found the coordinates of the centroid using: $$X=M_y/A$$ and $$Y=M_x/A$$ which resulted in $$X=8/5$$ and $$Y=1.$$

I was wondering if anyone would be willing to look over what I have said here to see if you get the same answers and if you get a different answer for one or all of the parts I will post the actual work of each part if you wouldn't mind looking it over to see what I did wrong.
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