MvCalc34

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By symmetry,the required area is A=4\int _{{0}}^{{{\frac  {\pi }{4}}}}\int _{{0}}^{{a{\sqrt  {\cos 2\theta }}}}rdrd\theta \,

=4\int _{{0}}^{{{\frac  {\pi }{4}}}}[{\frac  {r^{2}}{2}}]_{{0}}^{{a{\sqrt  {\cos 2\theta }}}}\,

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

=2a^{2}[{\frac  {\sin 2\theta }{2}}]_{{0}}^{{{\frac  {\pi }{4}}}}\,

=a^{2}\,

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