Question

Find the centroid of the quarter of the unit circle lying in the third quadrant.

Answer 1

As you know that  Centroid is the point of any curve ,Starting from a line segment it's centroid is it's mid point,  want to say that centroid means the point where the curve is balanced such that if at that point if you put anything it will remain balanced in all directions, whether it is a one dimensional figure,two dimensional figure,or three dimensional figure.For a circle it's center is the Centroid.

Now  coming to your question

Centroid of the quarter of the unit circle lying in the third quadrant.

Circle being a two dimensional figure. The equation of circle is

$x^2+y^2=1$

Applying the rule of integration i.e from (-1,0) to (0,-1). i.e from $x_{1}=-1 \text{ to } x_{2} =0 \text{ and } y_{1}=0\text{ to } y_{2} =-1$-----[∵ we have to find centroid of a circle lying in the third quadrant.]

$x^2+ y^2=1\\y^2=1-x^2\\y=\sqrt{1-x^{2}}$

Applying the formula of centroid i.e

$\bar{x}=\frac{1}{A}\int_{a}^{b}xy\text{ dx }\\\bar{y}=\frac{1}{2A}\int_{a}^{b}y^{2}\text{ dx }\\ \text{ or }\\\bar{y}=\frac{1}{A}\int_{a}^{b}yx\text{ dy }$

where A is the area of circle which is $\pi r^{2}$.

Area of unit circle=π

Area of Quadrant of a circle=π/4

$\bar{x}=\frac{4}{\pi}\int_{-1}^{0}x\sqrt{1-x^2}\text{ dx }$

put, (1-x²)=t

differentiating both sides

-2 x dx = dt

x dx= - (dt/2)

Also when , x= -1, t=0 and when x=0 then t=1

$\bar{x}=-\frac{4}{3\pi}\int_{0}^{1}\sqrt{t}\text{ dt }}$

$\bar{x}=-\frac{4}{3\pi}{ t ^ \frac{3}{2}}_{0}^{1}$

x coordinate=  $\bar{x} =-\frac{4}{3\pi}$

Now finding the y -cordinate by using the above formula

$\bar{y}=\frac{2}{\pi }\int_{0}^{-1}y^{2}\text{ dx }$

$\bar{y}=\frac{2}{\pi}\int\limits^{-1}_{0} {1-x^2} \, dx$

$\bar{y}=\frac{2}{\pi }[x-\frac{x^3}{3}]_{0}^{-1}$

$\bar{y}=-\frac{4}{3\pi}\\centroid(-\frac{4}{3\pi},-\frac{4}{3\pi})$