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.


I have checked your work and everything is correct. You have explained every step very clearly.

Do not be surprised if the centroid is not inside the region. That happens in many cases.

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