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Question Number 106654 by mohammad17 last updated on 06/Aug/20

Answered by bemath last updated on 06/Aug/20

$$\mathrm{total}\mathrm{area}=2\underset{0}{\stackrel{2}{\int }}(4-{\mathrm{x}}^2)\mathrm{dx}$$$$=2\mid (4\mathrm{x}-\frac{1}{3}{\mathrm{x}}^3){\mid }_0^2=2(8-\frac{8}{3})=\frac{32}{3}$$$$\mathrm{Area}\mathrm{I}=2\underset{0}{\stackrel{\sqrt{\mathrm{a}}}{\int }}(\mathrm{a}-{\mathrm{x}}^2)\mathrm{dx}$$$$=2\mid (\mathrm{ax}-\frac{1}{3}{\mathrm{x}}^3){\mid }_0^{\sqrt{\mathrm{a}}}$$$$=2(\mathrm{a}\sqrt{\mathrm{a}}-\frac{\mathrm{a}\sqrt{\mathrm{a}}}{3})=\frac{4\mathrm{a}\sqrt{\mathrm{a}}}{3}$$$$\mathrm{totall}\mathrm{area}=2\times \mathrm{area}\mathrm{I}$$$$\frac{32}{3}=\frac{8\mathrm{a}\sqrt{\mathrm{a}}}{3}\Rightarrow 4={\mathrm{a}}^{3/2};\mathrm{a}=4^{2/3}$$$$\therefore \mathrm{a}=\sqrt[3]{16}.@\mathrm{bemath}@$$

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