Solutions - 0110-02

Option 1: In terms of $x$.

(1) Mass of region:

$$\begin{array}{c}m=\rho {\int }_0^1{e}^x-1dx\\ \\ \gg \gg \rho (e^x-x){|}_0^1\gg \gg \rho (e-2)\end{array}$$

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(2) Integral formula for $M_y$:

$$\begin{array}{ccc}& M_y={\int }_a^b\rho x(f(x)-g(x))dx& \\ & & \\ & \gg \gg \rho {\int }_0^{1}x(e^x-1)dx& \\ & & \\ & \gg \gg \rho {(x{e}^x-e^x-\frac{x^2}{2})|}_0^1\gg \gg \frac{\rho }{2}& \text{(Int. by parts)}\end{array}$$

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(3) Integral formula for $M_x$:

$$\begin{array}{c}{M}_x={\int }_a^b\frac{\rho }{2}(f{(x)}^2-g{(x)}^2)dx\\ \\ \gg \gg \frac{\rho }{2}{\int }_0^1{e}^{2x}-1dx\\ \\ \gg \gg \frac{\rho }{2}{(\frac{1}{2}{e}^{2x}-x)|}_0^1\gg \gg \frac{\rho }{4}(e^2-3)\end{array}$$

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(4) Compute $(\bar{x},\bar{y})$:

$$\begin{array}{c}(\bar{x},\bar{y})=(\frac{M_y}{m},\frac{M_x}{m})\\ \\ \gg \gg (\frac{\rho /2}{\rho (e-2)},\frac{\frac{\rho }{4}(e^2-3)}{\rho (e-2)})\\ \\ \gg \gg (\frac{1}{2(e-2)},\frac{e^2-3}{4(e-2)})\end{array}$$

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Option 2: In terms of $y$.

(1) Mass:

$$\begin{array}{c}m/\rho \gg \gg {\int }_1^{e}1-\ln ydy\\ \\ \gg \gg y-(y\ln y-y){|}_1^e\\ \\ \gg \gg e-e+e-(1-(0-1))\\ \\ \gg \gg e-2\end{array}$$

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(2) $M_x$:

$$\begin{array}{ccc}& M_x/\rho \gg \gg {\int }_1^{e}y(1-\ln y)dy& \\ & & \\ & \gg \gg \frac{1}{4}(e^2-3)& \text{(Int. by parts)}\end{array}$$

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(3) $M_y$:

$$\begin{array}{ccc}& M_y/\rho \gg \gg {\int }_1^e\frac{1}{2}(1^2-(\ln y{)}^2)dy& \\ & & \\ & \gg \gg 1/2& \text{(}\text{I}\text{n}\text{t}\text{.}\text{b}\text{y}\text{p}\text{a}\text{r}\text{t}\text{s}\text{,}u=(\ln y{)}^2\text{)}\end{array}$$