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Question Number 57821 by behi83417@gmail.com last updated on 12/Apr/19

$$\mathbf{find}\mathbf{the}\mathbf{common}\mathbf{area}\mathbf{of}:$$$$\{\begin{array}{l}\frac{{\mathbf{x}}^2}{3}+{\mathbf{y}}^2=1\\ {\mathbf{x}}^2+\frac{{\mathbf{y}}^2}{3}=1\end{array}$$

Answered by MJS last updated on 13/Apr/19

$$4(\underset{0}{\stackrel{\frac{\sqrt{3}}{2}}{\int }}\sqrt{1-\frac{x^2}{3}}dx+\underset{\frac{\sqrt{3}}{2}}{\stackrel{1}{\int }}\sqrt{3-3x^2}dx)=\frac{2\sqrt{3}}{3}\pi$$

Answered by mr W last updated on 13/Apr/19

$$r^2=\frac{a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$$$$A_{common}=8{\int }_0^{\frac{\pi }{4}}\frac{r^{2}d\theta }{2}=4a^2{b}^2{\int }_0^{\frac{\pi }{4}}\frac{d\theta }{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$$$$=4ab{\left[{\tan }^{-1}\left(\frac{a}{b}\tan \theta \right)\right]}_0^{\frac{\pi }{4}}$$$$=4ab{\tan }^{-1}\left(\frac{a}{b}\right)$$$$witha=1,b=\sqrt{3}$$$$A_{common}=4\sqrt{3}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{4\sqrt{3}\pi }{6}=\frac{2\sqrt{3}\pi }{3}$$

Commented by behi83417@gmail.com last updated on 13/Apr/19

$$thankyousomuchsirMJSandsirMrW.$$

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