Find the centroid of the region in the first quadrant bounded by the given curves.

$y={x}^{3},x={y}^{3}$

vetrila10
2021-12-01
Answered

Find the centroid of the region in the first quadrant bounded by the given curves.

$y={x}^{3},x={y}^{3}$

You can still ask an expert for help

Michele Grimsley

Answered 2021-12-02
Author has **19** answers

Step 1

Given

$y={x}^{3},x={y}^{3}$

Step 2

$f\left(x\right)={x}^{3}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}g\left(x\right)=y={x}^{\frac{1}{3}}$

${M}_{y}={\int}_{0}^{1}x[f\left(x\right)-g\left(x\right)]dx$

$={\int}_{0}^{1}x[{x}^{3}-{x}^{\frac{1}{3}}]dx$

$={\int}_{0}^{1}[{x}^{4}-{x}^{\frac{4}{3}}]dx$

$=\frac{1}{5}{\left[{x}^{5}\right]}_{0}^{1}-\frac{3}{7}{\left[{x}^{\frac{7}{3}}\right]}_{0}^{1}$

$=\frac{1}{5}-\frac{3}{7}=\frac{-8}{35}$

${M}_{x}=\frac{1}{2}{\int}_{0}^{1}[{\left(x\right)}^{6}-{\left(x\right)}^{\frac{2}{3}}]dx$

$=\frac{1}{2}{[\frac{1}{7}{\left[x\right]}^{7}-\frac{3}{5}{\left[x\right]}^{\frac{5}{3}}]}_{0}^{1}$

$=\frac{1}{2}[\frac{1}{7}-\frac{3}{5}]$

$=\frac{-8}{35}$

$M={\int}_{0}^{1}[f\left(x\right)-g\left(x\right)]dx$

$={\int}_{0}^{1}[{\left(x\right)}^{3}-{\left(x\right)}^{\frac{1}{3}}]dx$

$=\frac{1}{4}{\left[{x}^{4}\right]}_{0}^{1}-\frac{3}{4}{\left[{x}^{\frac{4}{3}}\right]}_{0}^{1}$

Given

Step 2

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