Question

Use a computer algebra system to find the mass, center of mass, and moments of inertia of the lamina that occupies the region \(D\) and has the given density function.

Short answer

The mass is of given lamina is \(m \approx 0.0304\)

The centre of mass of given lamina:

\(\begin{aligned}{l}\bar x \approx 1.4205\\\bar y = 0.2480\end{aligned}\)

The moment of inertia are following:

\(\begin{aligned}{l}{I_x} \approx 0.00201\\{I_y} \approx 0.06556\\{I_o} \approx 0.06757\end{aligned}\)

Step-by-step solution

Calculate the mass of the given lamina

Formula for calculating mass of the lamina:

Given that\({\rm{ }}\rho (x,y) = {x^2}{y^2}\)

\(\begin{aligned}{l} = \int_0^2 {\int_0^{x{e^{ - x}}} {{x^2}} } {y^2}dydx = \frac{1}{{729}}\left( {40 - \frac{{7192}}{{{e^6}}}} \right)\\ \approx 0.0304\end{aligned}\)

(Using a calculator to approximate these Integrals, because the problem asks us to use a calculator).

Calculate the centre of mass of the given lamina 

Remember that:

\(\bar x = \frac{{{M_y}}}{m}\)

And

Therefore

\(\begin{aligned}{l}\bar x = \frac{{{M_y}}}{m}\\ = \frac{{16\left( {5{e^6} - 1223} \right)}}{{40{e^6} - 7192}}\\ \approx 1.4205\end{aligned}\)

Remember that:

\(\bar y = \frac{{{M_x}}}{m}\)

And

Therefore

\(\begin{aligned}{l}\bar y = \frac{{{M_x}}}{m}\\ = \frac{{0.0075}}{{0.0304}}\\ \approx 0.2480\end{aligned}\)

Calculate the moment of inertia of the given lamina

Obtain the moment of inertia as follows:

\(\begin{aligned}{l} = \int_0^2 {\int_0^{x{e^{ - x}}} {{y^2}} } \left( {{x^2}{y^2}} \right)dydx\\ = \int_0^2 {\int_0^{x{e^{ - x}}} {{x^2}} } {y^4}dydx\\ \approx 0.00201\end{aligned}\)

\(\begin{aligned}{l} = \int_0^2 {\int_0^{x{e^{ - x}}} {{x^2}} } \left( {{x^2}{y^2}} \right)dydx\\ = \int_0^2 {\int_0^{x{e^{ - x}}} {{x^4}} } {y^2}dydx\\ = \frac{{80\left( {7{e^6} - 2101} \right)}}{{2187{e^6}}}\\ \approx 0.06556\end{aligned}\)

\({I_0} = {I_x} + {I_y}\)

\(\begin{aligned}{l} \approx 0.00201 + 0.06556\\ = 0.06757\end{aligned}\)

Thus, the mass is of given lamina is \(m \approx 0.0304\)

The centre of mass of given lamina:

\(\begin{aligned}{l}\bar x \approx 1.4205\\\bar y = 0.2480\end{aligned}\)

The moments of inertia are following:

\(\begin{aligned}{l}{I_x} \approx 0.00201\\{I_y} \approx 0.06556\\{I_o} \approx 0.06757\end{aligned}\)