Finding Center of Northeast Quadrant - Radius 6cm

[blumfeld0]

Hi. this is probably a really dumb question but here goes.
I know the radius of a circle, say 6 cm.
i need to know what is the (x,y) coordinate of the CENTER of the Northeast quadrant (first quadrant).
is it just 3,3? or do i need to use some angle and trig to figure this out.

thank you in advance

 

[HallsofIvy]

What you are calling the "center" is technically the "centroid"- the point at which the "center of mass" would be if the figure were a thin plane of constant density. From formulas for center of mass, you want determine that the x and y coordinates of the centroid $(\overline{x},\overline{y})$ are given by:
$Area*\overline{x}= \int\int x dA$ and
$Area*\overline{y}= \int\int y dA$ where "dA" is the differential of area for the figure. For a quadrant of a circle of radius A, we would have
$$(1/4)\pi R^2 \overline{x}= \int_{r=0}^R\int_{\theta=0}^{\pi/2}(r cos \theta)(r dr d\theta)$$\
$$= \left(\int_{r=0}^R r^2 dr\right)\left(\int_{\theta= 0}^{\pi/2} cos\theta d\theta\right)$$
$$= \left((1/3)R^3\right)(1)$$
so
[tex]\overline{x}= \frac{4}{3\pi}R[/itex]<br /> By symmetry, $\overline{y}= \overline{x}$[/tex]

 

[blumfeld0]

wow. thank you!