Q 45. - Calculus

Question

$\mathcal{C}=\left\{(x,y)\mid x^2+y^2\le 1\right\}$

Find the centroid of $C$.

Step-by-Step Solution

Answer

The centroid is $\overline{x}=0,\overline{y}=0$.

Step 1: Given Information

It is given that $\mathcal{C}=\left\{(x,y)\mid x^2+y^2\le 1\right\}$

Step 2: Calculation of x -

The formula is

$\overline{x}=\frac{{\iint }_{\Omega }x\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}\text{ and }\overline{y}=\frac{{\iint }_{\Omega }y\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}$

As density is uniform $\rho (x,y)=1$

$\overline{x}=\frac{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}xdydx}{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dydx}=\frac{{\int }_{-1}^{1}x[y{]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}{{\int }_{-1}^1[y{]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}$

$\overline{x}=\frac{{\int }_{-1}^{1}x\left[\sqrt{1-x^2}-\left(-\sqrt{1-x^2}\right)\right]}{{\int }_{-1}^1\left[\sqrt{1-x^2}-\left(-\sqrt{1-x^2}\right)\right]dx}=\frac{2{\int }_{-1}^{1}x\sqrt{1-x^2}}{2{\int }_{-1}^1\sqrt{1-x^2}dx}$

Region is symmetric about $y$ axis

$\overline{x}=\frac{0}{\frac{\pi }{2}}\left[{\int }_{-a}^{a}f\left(x\right)dx=0,\text{ if }f\left(x\right)\text{ is odd }\right]$

$\overline{x}=0$

Step 3: Calculation of y -

Similarly $\overline{y}=\frac{{\iint }_{\Omega }y\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}$

$\overline{y}=\frac{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydydx}{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dydx}$

$\overline{y}=\frac{{\int }_{-1}^1\left[0\right]dx}{{\int }_{-1}^1[y{]}_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx}$

$\overline{y}=0$

The centroid is $\overline{x}=0,\overline{y}=0$