Centroid

Find the centroid of the region bounded by the following curves $\displaystyle y = x^2, y = x+3$

Centroid : (___, ___)

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Originally Posted by viet

Find the centroid of the region bounded by the following curves $\displaystyle y = x^2, y = x+3$

Centroid : (___, ___)

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I will assume it is a homogenous density.

In that case fist find the mass.
In this case simply the area.

$\displaystyle \int_D \int dA$
To express the region we need the intersection points between the line and parabola.
$\displaystyle x+3=x^2$
$\displaystyle x^2-x-3=0$
$\displaystyle x=.5\pm .5\sqrt{13}$
The upper curve is,
$\displaystyle x+3$ and lower is $\displaystyle x^2$.
Thus,
$\displaystyle m=\int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}} \int_{x^2}^{x+3} dy\, dx$
Thus,
$\displaystyle \int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}}x+3-x^2dx\approx 5.67$.

Now we find the first moments.

$\displaystyle M_x=\int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}} \int_{x^2}^{x+3}y dy\, dx$
Thus,
$\displaystyle \int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}}\frac{1}{2}(x+3)^2-\frac{1}{2}(x^2)^2 dx\approx 11.84$

$\displaystyle M_y=\int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}} \int_{x^2}^{x+3} xdy\, dx$
Thus,
$\displaystyle \int_{-.5-.5\sqrt{13}}^{.5+.5\sqrt{13}}x(x+3)-x(x^2)dx=8.14$

Thus,
$\displaystyle \bar{x}= \frac{M_x}{M}=\frac{11.84}{5.67}\approx 2.08$

$\displaystyle \bar{y}=\frac{M_y}{M}=\frac{8.14}{5.67}\approx 1.43$

the y coordinate $\displaystyle y_c$ is given by
$\displaystyle \frac{1}{2A}\int_a^b(f(x)-g(x))(f(x)+g(x))dx$.
and the x coordinate $\displaystyle x_c$ is given by
$\displaystyle \frac{1}{A}\int_a^bx(f(x)-g(x))dx$

$\displaystyle A$ is the area of the region between the two graphs and $\displaystyle a,b$ are the points where they intersect.

i got it thanks