Resolved: "Locate the centroid x of the area."

sputneyoh

Answered question

2021-12-18

Locate the centroid x of the area.

usaho4w

Beginner2021-12-19Added 39 answers

Area

d A = y d x

d A = 4 - (4 - \frac{x^{2}}{16} d x)

d A = \frac{x^{2}}{16} d x

A = \int_{0}^{8} 4 - (4 - \frac{x^{2}}{16} d x)

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

= 10.667 m^{2}

Cleveland Walters

Beginner2021-12-20Added 40 answers

Centroid

\overset{―}{x} = \frac{\int \overset{―}{x} d A}{\int d A}

= \frac{\int_{0}^{8} x (\frac{x^{3}}{16} d x)}{10.667}

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

= \frac{64}{10.667}

\overset{―}{x} = 6 m

\overset{―}{y} = \frac{\int \overset{―}{x} d A}{\int d A}

= \int_{0}^{8} \frac{\frac{y}{2} (\frac{x^{2}}{16}) d x}{10.667}

= \int_{0}^{8} \frac{\frac{1}{2} [4 + (4 - \frac{x^{2}}{16})] (\frac{x^{2}}{16}) d x}{10.667}

= \int_{0}^{8} \frac{(\frac{128 - x^{2}}{16}) \frac{x^{2}}{16} d x}{21.334}

= \frac{\int_{0}^{8} (12 x^{2} - x^{4}) d x}{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}

\overset{―}{y} = 2.8 m

nick1337

Expert2021-12-28Added 777 answers

why is $d A = 4 - (4 - (x^{2} / 16)) d x$ when $d A = y d x$ and $y = 4 - (x^{2} / 16)$

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