MvCalc31

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Because of symmetry,we consider the area in the first quadrant,hence

A=4\int _{0}^{b}\int _{{0}}^{{a{\sqrt  {1-{\frac  {y^{2}}{b^{2}}}}}}}dxdy\,

=4\int _{0}^{b}a{\sqrt  {1-{\frac  {y^{2}}{b^{2}}}}}dy\,

=4a\int _{0}^{b}{\sqrt  {1-{\frac  {y^{2}}{b^{2}}}}}dy\,

Setting y=b\sin \theta \, we obtain dy=b\cos \theta d\theta \,,then

A=4a\int _{{0}}^{{{\frac  {\pi }{2}}}}\cos \theta (b\cos \theta )d\theta \,

=4ab\int _{{0}}^{{{\frac  {\pi }{2}}}}{\frac  {1+\cos 2\theta }{2}}d\theta \,

=2ab\times {\frac  {\pi }{2}}=\pi ab\,

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