sputneyoh

2021-12-18

Locate the centroid x of the area.

usaho4w

Beginner2021-12-19Added 39 answers

Area

$dA=ydx$

$dA=4-(4-\frac{{x}^{2}}{16}dx)$

$dA=\frac{{x}^{2}}{16}dx$

$A={\int}_{0}^{8}4-(4-\frac{{x}^{2}}{16}dx)$

$={\left(\frac{{x}^{3}}{48}\right)}_{0}^{8}$

$=10.667\text{}{m}^{2}$

Cleveland Walters

Beginner2021-12-20Added 40 answers

Centroid

$\stackrel{\u2015}{x}=\frac{\int \stackrel{\u2015}{x}dA}{\int dA}$

$=\frac{{\int}_{0}^{8}x\left(\frac{{x}^{3}}{16}dx\right)}{10.667}$

$=\frac{{\left(\frac{{x}^{4}}{64}\right)}_{0}^{8}}{10.667}$

$=\frac{64}{10.667}$

$\stackrel{\u2015}{x}=6\text{}m$

$\stackrel{\u2015}{y}=\frac{\int \stackrel{\u2015}{x}dA}{\int dA}$

$={\int}_{0}^{8}\frac{\frac{y}{2}\left(\frac{{x}^{2}}{16}\right)dx}{10.667}$

$={\int}_{0}^{8}\frac{\frac{1}{2}[4+(4-\frac{{x}^{2}}{16})]\left(\frac{{x}^{2}}{16}\right)dx}{10.667}$

$={\int}_{0}^{8}\frac{\left(\frac{128-{x}^{2}}{16}\right)\frac{{x}^{2}}{16}dx}{21.334}$

$=\frac{{\int}_{0}^{8}(12{x}^{2}-{x}^{4})dx}{256\times 21.334}$

$=\frac{{(\frac{128{x}^{3}}{3}-\frac{{x}^{5}}{5})}_{0}^{8}}{256\times 21.334}$

$=\frac{15291.733}{256\times 21.334}$

$\stackrel{\u2015}{y}=2.8\text{}m$

nick1337

Expert2021-12-28Added 777 answers

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