The original question is in spanish ive included the english translation. Use the image included to answer.

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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) } ...