15.6.19 Find the mass and centroid

[karush]

Find the mass and centroid of the following thin plate assuming constant density
Sketch the region corresponding
to the plate and indicate the location
of the center is the mass
The region bounded by
$$y=ln x$$
$$x-axis$$
$$x=e$$
\begin{align}\displaystyle
\left(\overline{x},\overline{y}\right)
&=\left(\frac{my}{m}\frac{mx}{m}\right)\\
m&=\int_{a}^{b}\int_{a}^{b}xy \, dA\\
&=\int_{1}^{e}\int_{0}^{\ln{x}}xy \, dydx\\
&=\int_{1}^{e}x\left[\int_{1}^{\ln{x}} y \, dy\right]dx\\
&=\int_{1}^{e}x\left[ \frac{y^2}{2}\right]_0^{\ln{x}} \ dx\\
&=\int_{1}^{e}x\ln{x} \, dx\\
W|A &=\frac{1}{4}\left(1+e^2 \right)
\end{align}
so ?
the centroid is:
$$\left[\frac{e^2+1}{4},\frac{e}{2}-1\right]$$

 

[MarkFL]

Let's begin by sketching the region making up the lamina:

View attachment 7353

Okay, now for some constant mass density $\rho$ over this area $A$, we have by definition:

\(\displaystyle \rho=\frac{m}{A}\implies m=\rho A\)

To find $A$, I presume we are to use a double iterated integral. I would use vertical strips:

\(\displaystyle A=\int_{1}^{e}\int_{0}^{\ln(x)}\,dy\,dx=\int_{1}^{e}\ln(x)\,dx\)

Use IBP, where:

\(\displaystyle u=\ln(x)\implies du=\frac{1}{x}\,dx\)

\(\displaystyle dv=dx\implies v=x\)

Hence:

\(\displaystyle A=\left[x\ln(x)\right]_1^e-\int_1^e\,dx=\left(e-0\right)-(e-1)=1\)

And so we have:

\(\displaystyle m=\rho\)

To find the centroid $\left(\overline{x},\overline{y}\right)$, we use:

\(\displaystyle \overline{x}=\frac{1}{A}\int_1^e x\ln(x)\,dx\)

Use IBP where:

\(\displaystyle u=\ln(x)\implies du=\frac{1}{x}\,dx\)

\(\displaystyle dv=x\,dx\implies v=\frac{1}{2}x^2\)

And so we have (recalling $A=1$):

\(\displaystyle \overline{x}=\left[\frac{1}{2}x^2\ln(x)\right]_1^e-\frac{1}{2}\int_1^e x\,dx=\left(\frac{1}{2}e^2\right)-\left(\frac{1}{4}\left(e^2-1\right)\right)=\frac{e^2+1}{4}\)

Next, we use:

\(\displaystyle \overline{y}=\frac{1}{2}\int_1^e \ln^2(x)\,dx\)

Use IBP, where:

\(\displaystyle u=\ln^2(x)\implies du=\frac{2}{x}\ln(x)\,dx\)

\(\displaystyle dv=dx\implies v=x\)

Thus:

\(\displaystyle \overline{y}=\frac{1}{2}\left(\left[x\ln^2(x)\right]_1^e-2\int_1^e \ln(x)\,dx\right)=\frac{1}{2}\left(e-2\right)=\frac{e-2}{2}\)

And so the centroid is:

\(\displaystyle \left(\overline{x},\overline{y}\right)=\left(\frac{e^2+1}{4},\frac{e-2}{2}\right)\quad\checkmark\)

 

[karush]

mahalo

Awsome help as usual

 

[MarkFL]

I forgot to indicate where the centroid is on a graph:

View attachment 7354

 

[HallsofIvy]

Find the mass and centroid of the following thin plate assuming constant density
Sketch the region corresponding
to the plate and indicate the location
of the center is the mass
The region bounded by
$$y=ln x$$
$$x-axis$$
$$x=e$$
\begin{align}\displaystyle
\left(\overline{x},\overline{y}\right)
&=\left(\frac{my}{m}\frac{mx}{m}\right)\\
m&=\int_{a}^{b}\int_{a}^{b}xy \, dA\\

Where did "xy" come from? You are told that the density is constant, say C. The mass is $\int\int C dA$

&=\int_{1}^{e}\int_{0}^{\ln{x}}xy \, dydx\\
&=\int_{1}^{e}x\left[\int_{1}^{\ln{x}} y \, dy\right]dx\\
&=\int_{1}^{e}x\left[ \frac{y^2}{2}\right]_0^{\ln{x}} \ dx\\
&=\int_{1}^{e}x\ln{x} \, dx\\
W|A &=\frac{1}{4}\left(1+e^2 \right)
\end{align}
so ?
the centroid is:
$$\left[\frac{e^2+1}{4},\frac{e}{2}-1\right]$$