
MHB Master
#1
February 8th, 2018,
20:55
$\textsf{Find the center of mass of a thin plate of density}$
$\textsf{ $\delta=3$ bounded by the lines $x=0, y=x$, and the parabola
$y=2x^2$ 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

February 8th, 2018 20:55
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MHB Master
#2
February 12th, 2018,
16:28
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
$y=2x^2$ in $Q1$}\\$
\begin{align*}\displaystyle
M&=\int_{0}^{\sqrt{2}}\int_{2x^2}^{1}3 \, dy \, dx\\
&=3\int_{0}^{\sqrt{2}}\biggr[y\biggr]_{y=2x^2}^{y=1}\, dx
=3\int_{0}^{\sqrt{2}}(x^21) \, dx
=3\biggr[\frac{x^3}{3}x\biggr]_0^{\sqrt{2}}=\sqrt{2}\\
M_y&=\int_{0}^{\sqrt{2}}\int_{2x^2}^{1} \, dy \, dx
=\int_{0}^{\sqrt{2}}\biggr[y\biggr]_{y=2x^2}^{y=1}\, dx\\
&=\int_{0}^{\sqrt{2}}(x^21) \, 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.

Pessimist Singularitarian
#3
February 12th, 2018,
22:18
The first thing I would do is sketch the bounded area:
Now, let's compute the mass (noting that the curves $y=x$ and $y=2x^2$ intersect at $x=1$ in QI):
$ \displaystyle m=\rho A=3\int_{0}^{1}\int_{x}^{2x^2}\,dy\,dx=3\int_{0}^{1}2xx^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}^{2x^2}y\,dy\,dx=\frac{3}{2}\int_{0}^{1}\left(2x^2\right)^2x^2\,dx=\frac{3}{2}\int_{0}^{1}x^45x^2+4\,dx=\frac{19}{5}$
$ \displaystyle M_y=3\int_{0}^{1}x\int_{x}^{2x^2}\,dy\,dx=3\int_{0}^{1}2xx^2x^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}$

MHB Master
#4
February 12th, 2018,
23:38
Thread Author
ok i graphed this but
it not it
why

MHB Craftsman
#5
February 12th, 2018,
23:54
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
$y=2x^2$ in the $Q1$}$
$\color{red}{x=0}$ is the yaxis as shown in Mark’s sketch, not the xaxis as shown in your sketch.