Find the centroid of the region in the first quadrant

Maria Huey 2021-12-17 Answered
Find the centroid of the region in the first quadrant bounded by the given curves.
\(\displaystyle{y}={x}^{{{3}}}\), \(\displaystyle{x}={y}^{{{3}}}\)

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Expert Answer

alexandrebaud43
Answered 2021-12-18 Author has 132 answers
Step 1
Given
\(\displaystyle{y}={x}^{{{3}}}\), \(\displaystyle{x}={y}^{{{3}}}\)
Step 2
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}\) and \(\displaystyle{g{{\left({x}\right)}}}={y}={x}^{{{\frac{{{1}}}{{{3}}}}}}\)
\(\displaystyle{M}_{{{y}}}={\int_{{{0}}}^{{{1}}}}{x}{\left[{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{0}}}^{{{1}}}}{x}{\left[{x}^{{{3}}}-{x}^{{{\frac{{{1}}}{{{3}}}}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{0}}}^{{{1}}}}{\left[{x}^{{{4}}}-{x}^{{{\frac{{{4}}}{{{3}}}}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{5}}}}{{\left[{x}^{{{5}}}\right]}_{{{0}}}^{{{1}}}}-{\frac{{{3}}}{{{7}}}}{{\left[{x}^{{{\frac{{{7}}}{{{3}}}}}}\right]}_{{{0}}}^{{{1}}}}\)
\(\displaystyle={\frac{{{1}}}{{{5}}}}-{\frac{{{3}}}{{{7}}}}={\frac{{-{8}}}{{{35}}}}\)
\(\displaystyle{M}_{{{x}}}={\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{1}}}}{\left[{\left({x}\right)}^{{{6}}}-{\left({x}\right)}^{{{\frac{{{2}}}{{{3}}}}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{{\left[{\frac{{{1}}}{{{7}}}}{\left[{x}\right]}^{{{7}}}-{\frac{{{3}}}{{{5}}}}{\left[{x}\right]}^{{{\frac{{{5}}}{{{3}}}}}}\right]}_{{{0}}}^{{{1}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{1}}}{{{7}}}}-{\frac{{{3}}}{{{5}}}}\right]}\)
\(\displaystyle={\frac{{-{8}}}{{{35}}}}\)
\(\displaystyle{M}={\int_{{{0}}}^{{{1}}}}{\left[{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{0}}}^{{{1}}}}{\left[{\left({3}\right)}^{{{x}}}-{\left({x}\right)}^{{{\frac{{{1}}}{{{3}}}}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{{\left[{x}^{{{4}}}\right]}_{{{0}}}^{{{1}}}}-{\frac{{{3}}}{{{4}}}}{{\left[{x}^{{{\frac{{{4}}}{{{3}}}}}}\right]}_{{{0}}}^{{{1}}}}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}-{\frac{{{3}}}{{{4}}}}={\frac{{-{1}}}{{{2}}}}\)
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Barbara Meeker
Answered 2021-12-19 Author has 6441 answers
Good afternoon, your help in learning math helps me and my friends a lot, thank you for that.
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