Question-49950

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

Commented by ajfour last updated on 12/Dec/18

y = ∣ \frac{b}{2} + b \sin θ ∣, x = a \cos θ

F i n d a r e a e n c l o s e d b y t h e c u r v e .

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 = a b π + 4 \underset{0}{\int^{\frac{\sqrt{3}}{2} a}} (\frac{b}{2} - \frac{b}{a} \sqrt{a^{2} - x^{2}}) d x = (\frac{π}{3} + \frac{\sqrt{3}}{2}) a b

Commented by ajfour last updated on 12/Dec/18

T h a n k s S i r, v e r y p r e c i s e!

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