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Question Number 72175 by mr W last updated on 26/Oct/19

$${({\mathit{x}}^2+{\mathit{y}}^2-1)}^3={\mathit{x}}^2{\mathit{y}}^3$$$$$$$$howlarge\left(area\right)isyourheart\mathrm{♡}?$$

Commented by mr W last updated on 26/Oct/19

Commented by MJS last updated on 26/Oct/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{for}x;\mathrm{we}\mathrm{get}3\mathrm{solutions}$$$$\mathrm{for}{x}^2\Rightarrow 6\mathrm{solutions}\pm \sqrt{x^2}\mathrm{but}\mathrm{then}\mathrm{the}\mathrm{exact}$$$$\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{upper}\mathrm{border}\mathrm{is}\mathrm{a}\mathrm{problem}$$$$$$$$\mathrm{I}\mathrm{also}\mathrm{tried}\mathrm{polar}\mathrm{coordinates}\mathrm{but}\mathrm{cannot}\mathrm{solve}$$$$\mathrm{for}r$$

Commented by mr W last updated on 26/Oct/19

$$thankssir!itseemstobeimpossibleto$$$$gettheexactvalueofthearea.$$

Answered by ajfour last updated on 26/Oct/19

$${(r^2-1)}^2=r^5\cos {}^2\theta \sin {}^3\theta$$$$letr=t\sec \varphi$$$$\Rightarrow t^4\tan {}^4\varphi =t^5\sec {}^5\varphi \cos {}^2\theta \sin {}^3\theta$$$$\Rightarrow t=\frac{\sin {}^4\varphi \cos \varphi }{\sin {}^3\theta \cos {}^2\theta }$$$$A={\int }_0^{\pi }{r}^{2}d\theta ={\int }_0^{\pi }{\left(\frac{\sin {}^4\varphi }{\sin {}^3\theta \cos {}^2\theta }\right)}^{2}d\theta$$$$...$$

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