MvCalc33

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We consider the portion above the X-axis for calculating the area. Therefore,

Area A=2\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{{r=a}}^{{a(1+\cos \theta )}}rdrd\theta \,

=2\int _{{0}}^{{{\frac  {\pi }{2}}}}[{\frac  {r^{2}}{2}}]_{{a}}^{{a(1+\cos \theta )}}d\theta \,

=a^{2}\int _{{0}}^{{{\frac  {\pi }{2}}}}[(1+\cos \theta )^{2}-1]d\theta \,

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

=a^{2}\int _{{0}}^{{{\frac  {\pi }{2}}}}[2\cos \theta +{\frac  {1+\cos 2\theta }{2}}]d\theta \,

=a^{2}[2\sin \theta +{\frac  {1}{2}}(\theta +{\frac  {\sin 2\theta }{2}})]_{{0}}^{{{\frac  {\pi }{2}}}}\,

=a^{2}[2+{\frac  {1}{2}}({\frac  {\pi }{2}}+0)]=a^{2}(2+{\frac  {\pi }{4}})\,

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