1. Find the centroid of the region in the first quadrant bounded by the x-axis, the parabola y^2=x, and the line x+y=6.
I would like th confirm my answer since i can't check. the answer i get is (26/11,8/11)
Are you using something like
$\displaystyle
\bar x = \frac{{\iint\limits_R {x\,dA}}}
{{\iint\limits_R {dA}}}
$
and
$\displaystyle
\bar y = \frac{{\iint\limits_R {y\,dA}}}
{{\iint\limits_R {dA}}}
$
?
Then,
$\displaystyle
\bar x = \frac{\int_0^4 x^{\frac{3}{2}} dx + \int_4^6 (6x - x^2) dx}{\int_0^4 x^{\frac{1}{2}} dx + \int_4^6 (6 - x) dx}
$
and
$\displaystyle
\bar y = \frac{\int_0^2 (y^2 + 6y - y^3) dy}{\int_0^2 (y + 6 - y^2) dy}
$
... no?
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