Question Number 49950 by ajfour last updated on 12/Dec/18
Commented by ajfour last updated on 12/Dec/18
$$\:\:{y}\:=\:\mid\frac{{b}}{\mathrm{2}}+{b}\mathrm{sin}\:\theta\mid\:\:,\:{x}\:=\:{a}\mathrm{cos}\:\theta \\ $$$${Find}\:{area}\:{enclosed}\:{by}\:{the}\:{curve}. \\ $$
Answered by MJS last updated on 12/Dec/18
$${y}=\frac{{b}}{\mathrm{2}}+\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${y}=\mathrm{0}\:\Rightarrow\:{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a} \\ $$$${A}={ab}\pi+\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a}} {\int}}\left(\frac{{b}}{\mathrm{2}}−\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right){dx}=\left(\frac{\pi}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){ab} \\ $$
Commented by ajfour last updated on 12/Dec/18
$${Thanks}\:{Sir},\:{very}\:{precise}! \\ $$