Centroid

**viet** · December 7th 2006, 05:40 PM

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

Centroid : (___, ___)

**ThePerfectHacker** · December 7th 2006, 07:34 PM

Originally Posted by viet

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

Centroid : (___, ___)

I will assume it is a homogenous density.

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

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

Now we find the first moments.

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

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

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

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

**putnam120** · December 7th 2006, 08:41 PM

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

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

**viet** · December 7th 2006, 08:46 PM

i got it thanks

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