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For the following exercises, calculate the center of mass for the collection of masses given.
${m}_{1}=2$ at ${x}_{1}=1$ and ${m}_{2}=4$ at ${x}_{2}=2$
${m}_{1}=1$ at ${x}_{1}=\mathrm{-1}$ and ${m}_{2}=3$ at ${x}_{2}=2$
$\frac{5}{4}$
$m=3$ at $x=0,1,2,6$
Unit masses at $(x,y)=(1,0),(0,1),(1,1)$
$\left(\frac{2}{3},\frac{2}{3}\right)$
${m}_{1}=1$ at $(1,0)$ and ${m}_{2}=4$ at $(0,1)$
${m}_{1}=1$ at $(1,0)$ and ${m}_{2}=3$ at $(2,2)$
$\left(\frac{7}{4},\frac{3}{2}\right)$
For the following exercises, compute the center of mass $\stackrel{\u2013}{x}.$
$\rho =1$ for $x\in (\mathrm{-1},3)$
$\rho =1$ for $x\in (0,1)$ and $\rho =2$ for $x\in (1,2)$
$\rho =\text{sin}\phantom{\rule{0.2em}{0ex}}x$ for $x\in (0,\pi )$
$\frac{\pi}{2}$
$\rho =\text{cos}\phantom{\rule{0.2em}{0ex}}x$ for $x\in \left(0,\frac{\pi}{2}\right)$
$\rho ={e}^{x}$ for $x\in \left(0,2\right)$
$\frac{{e}^{2}+1}{{e}^{2}-1}$
$\rho ={x}^{3}+x{e}^{\text{\u2212}x}$ for $x\in (0,1)$
$\rho =x\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}x$ for $x\in (0,\pi )$
$\frac{{\pi}^{2}-4}{\pi}$
$\rho =\sqrt{x}$ for $x\in \left(1,4\right)$
$\rho =\text{ln}\phantom{\rule{0.2em}{0ex}}x$ for $x\in \left(1,e\right)$
$\frac{1}{4}\left(1+{e}^{2}\right)$
For the following exercises, compute the center of mass $\left(\stackrel{\u2013}{x},\stackrel{\u2013}{y}\right).$ Use symmetry to help locate the center of mass whenever possible.
$\rho =7$ in the square $0\le x\le 1,$ $0\le y\le 1$
$\rho =3$ in the triangle with vertices $(0,0),$ $(a,0),$ and $(0,b)$
$\left(\frac{a}{3},\frac{b}{3}\right)$
$\rho =2$ for the region bounded by $y=\text{cos}(x),$ $y=\text{\u2212}\text{cos}(x),$ $x=-\frac{\pi}{2},$ and $x=\frac{\pi}{2}$
For the following exercises, use a calculator to draw the region, then compute the center of mass $\left(\stackrel{\u2013}{x},\stackrel{\u2013}{y}\right).$ Use symmetry to help locate the center of mass whenever possible.
[T] The region bounded by $y=\text{cos}(2x),$ $x=-\frac{\pi}{4},$ and $x=\frac{\pi}{4}$
$\left(0,\frac{\pi}{8}\right)$
[T] The region between $y=2{x}^{2},$ $y=0,$ $x=0,$ and $x=1$
[T] The region between $y=\frac{5}{4}{x}^{2}$ and $y=5$
$(0,3)$
[T] Region between $y=\sqrt{x},$ $y=\text{ln}(x),$ $x=1,$ and $x=4$
[T] The region bounded by $y=0,$ $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$
$\left(0,\frac{4}{\pi}\right)$
[T] The region bounded by $y=0,$ $x=0,$ and $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$
[T] The region bounded by $y={x}^{2}$ and $y={x}^{4}$ in the first quadrant
$\left(\frac{5}{8},\frac{1}{3}\right)$
For the following exercises, use the theorem of Pappus to determine the volume of the shape.
Rotating $y=mx$ around the $x$ -axis between $x=0$ and $x=1$
Rotating $y=mx$ around the $y$ -axis between $x=0$ and $x=1$
$\frac{m\pi}{3}$
A general cone created by rotating a triangle with vertices $(0,0),$ $(a,0),$ and $(0,b)$ around the $y$ -axis. Does your answer agree with the volume of a cone?
A general cylinder created by rotating a rectangle with vertices $(0,0),$ $(a,0),(0,b),$ and $(a,b)$ around the $y$ -axis. Does your answer agree with the volume of a cylinder?
$\pi {a}^{2}b$
A sphere created by rotating a semicircle with radius $a$ around the $y$ -axis. Does your answer agree with the volume of a sphere?
For the following exercises, use a calculator to draw the region enclosed by the curve. Find the area $M$ and the centroid $\left(\stackrel{\u2013}{x},\stackrel{\u2013}{y}\right)$ for the given shapes. Use symmetry to help locate the center of mass whenever possible.
[T] Quarter-circle: $y=\sqrt{1-{x}^{2}},$ $y=0,$ and $x=0$
$\left(\frac{4}{3\pi},\frac{4}{3\pi}\right)$
[T] Triangle: $y=x,$ $y=2-x,$ and $y=0$
[T] Lens: $y={x}^{2}$ and $y=x$
$\left(\frac{1}{2},\frac{2}{5}\right)$
[T] Ring: ${y}^{2}+{x}^{2}=1$ and ${y}^{2}+{x}^{2}=4$
[T] Half-ring: ${y}^{2}+{x}^{2}=1,$ ${y}^{2}+{x}^{2}=4,$ and $y=0$
$\left(0,\frac{28}{9\pi}\right)$
Find the generalized center of mass in the sliver between $y={x}^{a}$ and $y={x}^{b}$ with $a>b.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the y -axis.
Find the generalized center of mass between $y={a}^{2}-{x}^{2},$ $x=0,$ and $y=0.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the y -axis.
Center of mass: $\left(\frac{a}{6},\frac{4{a}^{2}}{5}\right),$ volume: $\frac{2\pi {a}^{4}}{9}$
Find the generalized center of mass between $y=b\phantom{\rule{0.2em}{0ex}}\text{sin}(ax),$ $x=0,$ and $x=\frac{\pi}{a}.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the y -axis.
Use the theorem of Pappus to find the volume of a torus (pictured here). Assume that a disk of radius $a$ is positioned with the left end of the circle at $x=b,$ $b>0,$ and is rotated around the y -axis.
Volume: $2{\pi}^{2}{a}^{2}\left(b+a\right)$
Find the center of mass $\left(\stackrel{\u2013}{x},\stackrel{\u2013}{y}\right)$ for a thin wire along the semicircle $y=\sqrt{1-{x}^{2}}$ with unit mass. ( Hint: Use the theorem of Pappus.)
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