Question

The region between $f\left(x\right)=x^2\sqrt{3-x}$and the x-axis on [a, b] = [0, 3], with n = 6.

Short answer

$Centreofmass:(1,2)$

Step-by-step solution

Step1. Given Information

region,

$f\left(x\right)=x^2\sqrt{3-x}$

Step2. Calculate

C.O.M

$$\begin{gathered} f\left(x\right)=6-2x[0,3] \\ 0=6-2x \\ 2x=6 \\ x=\frac{6}{2} \\ x=3 \end{gathered}$$

Step3. Continue

$$\begin{gathered} M_x=\frac{1}{2}P{\int }_0^3{(6-2x)}^{2}dx{M}_x=P{\int }_0^{3}18-6x+2x^{2}dx \\ {M}_x=P{{[18x-\frac{6x^2}{2}+\frac{2x^3}{3}]}_0}^3{M}_x=P[54-54+\frac{54}{3}]=18P \end{gathered}$$

Step4. Continue

$$\begin{gathered} M_y=P{\int }_0^{3}x(6-2x)dx \\ {M}_y=P{\int }_0^{3}6x-2x^{2}dx \\ {M}_y=P{{[\frac{6x^2}{2}-\frac{2x^3}{3}]}_0}^3{M}_y=P[27-\frac{54}{3}] \\ {M}_y=P[27-18]=9P \end{gathered}$$

Step4. Continue

$$\begin{gathered} M=P{\int }_0^3(6-2x)dx=6x-x^2{{|}_0}^3 \\ =P[18-9] \\ =9P \end{gathered}$$

Step6. Continue

$$\begin{gathered} M=9P,M_y=9P,M_x=18P \\ C.o.M=(X_{cm},Y_{cm})=(\frac{M_y}{M},\frac{M_X}{M}) \\ =(\frac{9P}{9P},\frac{18P}{9P}) \\ =(1,2) \end{gathered}$$