MvCalc45

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If (x1,y1) are the coordinates of the centroid,then

x_{1}={\frac  {\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kxydxdy(x)}{\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kxydxdy}}\,

y_{1}={\frac  {\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kxydxdy(y)}{\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kxydxdy}}\,

The numerator in the first one is

\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kx^{2}ydxdy\,

=k\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kx^{2}ydxdy\,

=k\int _{0}^{a}x^{2}{\frac  {b^{2}}{2}}[1-{\frac  {x^{2}}{a^{2}}}]dx\,

={\frac  {k}{2}}{\frac  {b^{2}}{a^{2}}}\int _{0}^{a}x^{2}(a^{2}-x^{2})dx\,

={\frac  {kb^{2}}{2a^{2}}}[a^{2}{\frac  {x^{3}}{3}}-{\frac  {x^{5}}{5}}]_{0}^{a}\,

={\frac  {kb^{2}}{2a^{2}}}[{\frac  {a^{5}}{3}}-{\frac  {a^{5}}{5}}]={\frac  {k}{15}}b^{2}a^{3}\,

Denominator=\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}kxydxdy\,

=k\int _{0}^{a}x[{\frac  {y^{2}}{2}}]_{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}dx\,

={\frac  {k}{2}}\int _{0}^{a}xb^{2}[1-{\frac  {x^{2}}{a^{2}}}]dx\,

={\frac  {kb^{2}}{2}}[{\frac  {x^{2}}{2}}-{\frac  {1}{a^{2}}}{\frac  {x^{4}}{4}}]_{0}^{a}\,

={\frac  {kb^{2}}{2}}[{\frac  {a^{2}}{2}}-{\frac  {a^{2}}{4}}]={\frac  {k}{8}}b^{2}a^{2}\,

Hence x_{1}={\frac  {8}{15}}a\,

By symmetry,y_{1}={\frac  {8}{15}}b\,

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