MvCalc14

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\int _{0}^{a}[\int _{{0}}^{{{\sqrt  {a^{2}-x^{2}}}}}xydy]dx\,

Now \int _{{0}}^{{{\sqrt  {a^{2}-x^{2}}}}}xydy\,

=[{\frac  {xy^{2}}{2}}]_{{0}}^{{{\sqrt  {a^{2}-x^{2}}}}}\,

={\frac  {x(a^{2}-x^{2})}{2}}\,

={\frac  {1}{2}}[a^{2}x-x^{2}]\,

Therefore,I=\int _{0}^{a}[a^{2}x-x^{2}]dx\,

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

={\frac  {a^{4}}{2}}-{\frac  {a^{3}}{3}}\,

={\frac  {a^{3}}{6}}[3a-2]\,

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