sputneyoh

2021-12-18

Locate the centroid x of the area.

usaho4w

Area
$dA=ydx$
$dA=4-\left(4-\frac{{x}^{2}}{16}dx\right)$
$dA=\frac{{x}^{2}}{16}dx$
$A={\int }_{0}^{8}4-\left(4-\frac{{x}^{2}}{16}dx\right)$
$={\left(\frac{{x}^{3}}{48}\right)}_{0}^{8}$

Cleveland Walters

Centroid
$\stackrel{―}{x}=\frac{\int \stackrel{―}{x}dA}{\int dA}$
$=\frac{{\int }_{0}^{8}x\left(\frac{{x}^{3}}{16}dx\right)}{10.667}$
$=\frac{{\left(\frac{{x}^{4}}{64}\right)}_{0}^{8}}{10.667}$
$=\frac{64}{10.667}$

$\stackrel{―}{y}=\frac{\int \stackrel{―}{x}dA}{\int dA}$
$={\int }_{0}^{8}\frac{\frac{y}{2}\left(\frac{{x}^{2}}{16}\right)dx}{10.667}$
$={\int }_{0}^{8}\frac{\frac{1}{2}\left[4+\left(4-\frac{{x}^{2}}{16}\right)\right]\left(\frac{{x}^{2}}{16}\right)dx}{10.667}$
$={\int }_{0}^{8}\frac{\left(\frac{128-{x}^{2}}{16}\right)\frac{{x}^{2}}{16}dx}{21.334}$
$=\frac{{\int }_{0}^{8}\left(12{x}^{2}-{x}^{4}\right)dx}{256×21.334}$
$=\frac{{\left(\frac{128{x}^{3}}{3}-\frac{{x}^{5}}{5}\right)}_{0}^{8}}{256×21.334}$
$=\frac{15291.733}{256×21.334}$

nick1337

why is $dA=4-\left(4-\left({x}^{2}/16\right)\right)dx$ when $dA=ydx$ and $y=4-\left({x}^{2}/16\right)$