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Question Number 49950 by ajfour last updated on 12/Dec/18

Commented by ajfour last updated on 12/Dec/18

$$y=\mid \frac{b}{2}+b\sin \theta \mid ,x=a\cos \theta$$$$Findareaenclosedbythecurve.$$

Answered by MJS last updated on 12/Dec/18

$$y=\frac{b}{2}+\frac{b}{a}\sqrt{a^2-x^2}$$$$y=0\Rightarrow x=\pm \frac{\sqrt{3}}{2}a$$$$A=ab\pi +4\underset{0}{\stackrel{\frac{\sqrt{3}}{2}a}{\int }}(\frac{b}{2}-\frac{b}{a}\sqrt{a^2-x^2})dx=(\frac{\pi }{3}+\frac{\sqrt{3}}{2})ab$$

Commented by ajfour last updated on 12/Dec/18

$$ThanksSir,veryprecise!$$

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