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  1. MHB Master
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    #1
    $\textsf{Find the center of mass of a thin plate of density}$
    $\textsf{ $\delta=3$ bounded by the lines $x=0, y=x$, and the parabola
    $y=2-x^2$ in the $Q1$}$
    $\begin{array}{llcr}\displaystyle
    &\textit{Mass}\\
    &&\displaystyle M=\iint\limits_{R}\delta \, dA\\
    &\textit{First Moments}\\
    &&\displaystyle M_y=\iint\limits_{R}x\delta \, dA
    &\displaystyle M_x=\iint\limits_{R}y\delta \, dA\\
    &\textit{Center of mass}\\
    &&\displaystyle\bar{x}=\displaystyle\frac{M_y}{M},
    \displaystyle\bar{y}=\displaystyle\frac{M_x}{M}\\
    \\
    &&\color{red}
    {\displaystyle \, \bar{x}=\frac{5}{14},
    \bar{y}=\displaystyle\frac{38}{35}}\\
    \end{array}$


    ok I just barely had to time to post this
    equations are just from reference
    red is answer

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  3. MHB Master
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    #2 Thread Author
    Ok I'm starting over on this a small step at a time...

    $\textsf{Find the center of mass of a thin plate of density}\\$
    $\textsf{$\delta=3$ bounded by the lines $x=0, y=x$, and the parabola
    $y=2-x^2$ in $Q1$}\\$
    \begin{align*}\displaystyle
    M&=\int_{0}^{\sqrt{2}}\int_{2-x^2}^{1}3 \, dy \, dx\\
    &=3\int_{0}^{\sqrt{2}}\biggr[y\biggr]_{y=2-x^2}^{y=1}\, dx
    =3\int_{0}^{\sqrt{2}}(x^2-1) \, dx
    =3\biggr[\frac{x^3}{3}-x\biggr]_0^{\sqrt{2}}=\sqrt{2}\\
    M_y&=\int_{0}^{\sqrt{2}}\int_{2-x^2}^{1} \, dy \, dx
    =\int_{0}^{\sqrt{2}}\biggr[y\biggr]_{y=2-x^2}^{y=1}\, dx\\
    &=\int_{0}^{\sqrt{2}}(x^2-1) \, dx = -\sqrt{2}\\
    \end{align*}

    something isn't happening right!


    $\textsf{the answer utimately is:}\\$
    $\color{red}{\, \bar{x}=\displaystyle\frac{5}{14},\bar{y}=\frac{38}{35}}$



    Last edited by karush; February 12th, 2018 at 20:14.

  4. Pessimist Singularitarian
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    #3
    The first thing I would do is sketch the bounded area:



    Now, let's compute the mass (noting that the curves $y=x$ and $y=2-x^2$ intersect at $x=1$ in QI):

    $ \displaystyle m=\rho A=3\int_{0}^{1}\int_{x}^{2-x^2}\,dy\,dx=3\int_{0}^{1}2-x-x^2\,dx=3\left(2-\frac{1}{2}-\frac{1}{3}\right)=\frac{7}{2}$

    Next, let's compute the moments of the lamina:

    $ \displaystyle M_x=3\int_{0}^{1}\int_{x}^{2-x^2}y\,dy\,dx=\frac{3}{2}\int_{0}^{1}\left(2-x^2\right)^2-x^2\,dx=\frac{3}{2}\int_{0}^{1}x^4-5x^2+4\,dx=\frac{19}{5}$

    $ \displaystyle M_y=3\int_{0}^{1}x\int_{x}^{2-x^2}\,dy\,dx=3\int_{0}^{1}2x-x^2-x^3\,dx=\frac{5}{4}$

    Hence:

    $ \displaystyle \overline{x}=\frac{M_y}{m}=\frac{\frac{5}{4}}{\frac{7}{2}}=\frac{5}{14}$

    $ \displaystyle \overline{y}=\frac{M_x}{m}=\frac{\frac{19}{5}}{\frac{7}{2}}=\frac{38}{35}$

  5. MHB Master
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    #5
    Quote Originally Posted by karush View Post
    $\textsf{Find the center of mass of a thin plate of density}$
    $\textsf{ $\delta=3$ bounded by the lines $\color{red}{x=0}$, $ y=x$, and the parabola
    $y=2-x^2$ in the $Q1$}$
    $\color{red}{x=0}$ is the y-axis as shown in Mark’s sketch, not the x-axis as shown in your sketch.

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