Question The original question is in spanish ive included the english translation. Use the image included to answer. Centro de masa de una lámina plana Calcule el centro de masa de la lámina con densidad uniforme p que queda acotada entre las curvas y=x² & y=x³. Ver ilustración 2do gráfico. Calculate the center of mass of the sheet with uniform density rho that is bounded between the curves y=x^2 and y=x^3 y = x² y = x³ + Soto, M (2015) Gráficas y = x² y y = x³

Solution: Given that the curvesy=x^{2} \text { and } y=x^{3} \text {. }We know that to find the coordinates of the center of mass (\bar{x}, \bar{y}) of the sheet, we need to find the moment Mx of the sheet about the x-atis and the moment my about the y-axis. we also need to find the mass of the sheet.Ther\dot{x}=\frac{m_{y}}{m} \text { and } \bar{y}=\frac{m_{x}}{m} \text {. }whene\begin{array}{l}m=\iint_{R} \rho d A ; \rho \text { is the density } \\m_{x}=\iint_{R} y \rho d A \text { and } m_{y}=\iint_{R} x \rho d A\end{array}The region bounded by the wraves y=x^{2} and y=x^{3} has the points of intervals, whene\begin{aligned}& x^{2}=x^{3} \\\Rightarrow & x\left(x-x^{2}\right)=0 \\\Rightarrow & x^{2}(x-1)=0 \\\Rightarrow & x=0,1\end{aligned}So, the range of x is x : 0 to 1 the range of y is y : x^{3} to x^{2}Now\begin{aligned}m & =\iint_{R} \rho d A \\& =\rho \int_{x=0}^{1} \int_{y=x^{3}}^{x^{2}} d y d x \\& =\rho \int_{0}^{1}\left\{\int_{x^{3}}^{x^{2}} d y\right\} d x \\& \left.=\rho \int_{0}^{1}\{y]_{x^{3}}^{x^{2}}\right\} d x\end{aligned}\begin{aligned}& =\rho \int_{0}^{1}\left(x^{2}-x^{3}\right) d x . \\& =\rho\left[\frac{x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{1} \\& =\rho\left[\frac{1}{3}-\frac{1}{4}\right]=\frac{\rho}{12} \\\Rightarrow m & =\frac{\rho}{12}\end{aligned}Now,\begin{aligned}M_{x} & =\iint_{R} y \rho d A \\& =\rho \int_{0}^{1} \int_{x^{3}}^{x^{2}} y d y d x \\& =\rho \int_{0}^{1}\left\{\int_{x^{3}}^{x^{2}} y d y\right\} d x \\& =\rho \int_{0}^{1}\left\{\left[\frac{y^{2}}{2}\right]_{x^{3}}^{x^{2}}\right\} d x \\& =\rho \int_{0}^{1}\left(\frac{x^{4}}{2}-\frac{x^{6}}{2}\right) d x \\& =\frac{\rho}{2}\left\{\int_{0}^{1}\left(x^{4}-x^{6}\right) d x\right\} \\& =\frac{\rho}{2}\left[\frac{x^{5}}{5}-\frac{x^{7}}{7}\right]_{0}^{1} \\& =\frac{\rho}{2}\left(\frac{1}{5}-\frac{1}{7}\right) \\& =\frac{\rho}{2} \times \frac{2}{35} \\& =\frac{\rho}{35}\end{aligned}\Rightarrow m_{x}=\frac{\rho}{35}Now\text { Now } \begin{aligned}m_{y} & =\iint_{R} \rho x d \theta \\& =\rho \int_{x=0}^{1} \int_{y=x^{3}}^{x^{2}} x d y d x \\& =\rho \int_{0}^{1}\left\{\int_{x^{3}}^{x^{2}} x d y\right\} d x \\& =\rho \int_{0}^{1}\left\{x[y]_{x^{3}}^{x^{2}}\right\} d x \\& =\rho \int_{0}^{1} x\left(x^{2}-x^{3}\right) d x \\& =\rho \int_{0}^{1}\left(x^{3}-x^{4}\right) d x \\& =\rho\left[\frac{x^{4}}{4}-\frac{x^{5}}{5}\right]_{0}^{1} \\& =\rho\left(\frac{1}{4}-\frac{1}{5}\right) \\& =\frac{\rho}{20} \\\Rightarrow m y & \frac{\rho}{20}\end{aligned}\therefore the x-coordinate of the centre of mass\text { is } \bar{x}=\frac{M y}{m}=\frac{\rho / 20}{\rho / 12}=\frac{3}{5}\therefore the y-coordinate of the center of m ass\text { is } \bar{y}=\frac{M_{x}}{m}=\frac{\rho / 35}{s / 12}=\frac{12}{35}Hence the coordinate of the center of mass is given by(\bar{x}, \bar{y})=\left(\frac{3}{5}, \frac{12}{35}\right) \text { (Ans) } ...