# Thread: 244.15.6.1 center of mass

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

2.

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!

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

4. 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:

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

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

6. Originally Posted by karush
$\textsf{Find the center of mass of a thin plate of density}$
$\textsf{$\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.