# Find the centroid of the region in the first quadrant

Find the centroid of the region in the first quadrant bounded by the given curves.
$y={x}^{3},x={y}^{3}$
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Michele Grimsley
Step 1
Given
$y={x}^{3},x={y}^{3}$
Step 2
$f\left(x\right)={x}^{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(x\right)=y={x}^{\frac{1}{3}}$
${M}_{y}={\int }_{0}^{1}x\left[f\left(x\right)-g\left(x\right)\right]dx$
$={\int }_{0}^{1}x\left[{x}^{3}-{x}^{\frac{1}{3}}\right]dx$
$={\int }_{0}^{1}\left[{x}^{4}-{x}^{\frac{4}{3}}\right]dx$
$=\frac{1}{5}{\left[{x}^{5}\right]}_{0}^{1}-\frac{3}{7}{\left[{x}^{\frac{7}{3}}\right]}_{0}^{1}$
$=\frac{1}{5}-\frac{3}{7}=\frac{-8}{35}$
${M}_{x}=\frac{1}{2}{\int }_{0}^{1}\left[{\left(x\right)}^{6}-{\left(x\right)}^{\frac{2}{3}}\right]dx$
$=\frac{1}{2}{\left[\frac{1}{7}{\left[x\right]}^{7}-\frac{3}{5}{\left[x\right]}^{\frac{5}{3}}\right]}_{0}^{1}$
$=\frac{1}{2}\left[\frac{1}{7}-\frac{3}{5}\right]$
$=\frac{-8}{35}$
$M={\int }_{0}^{1}\left[f\left(x\right)-g\left(x\right)\right]dx$
$={\int }_{0}^{1}\left[{\left(x\right)}^{3}-{\left(x\right)}^{\frac{1}{3}}\right]dx$
$=\frac{1}{4}{\left[{x}^{4}\right]}_{0}^{1}-\frac{3}{4}{\left[{x}^{\frac{4}{3}}\right]}_{0}^{1}$