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

Answered by bemath last updated on 06/Aug/20

total area = 2∫_0 ^2 (4−x^2 )dx = 2∣(4x−(1/3)x^3 )∣_0 ^2 =2(8−(8/3))=((32)/3) Area I = 2∫_0 ^(√a) (a−x^2 )dx =2∣(ax−(1/3)x^3 )∣_0 ^(√a) =2(a(√a)−((a(√a))/3))=((4a(√a))/3) totall area = 2×area I ((32)/3) = ((8a(√a))/3) ⇒4 = a^(3/2) ; a = 4^(2/3) ∴ a = ((16))^(1/(3 )) . @bemath@

$$\mathrm{total}\:\mathrm{area}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\left(\mathrm{4}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\: \\ $$$$=\:\mathrm{2}\mid\left(\mathrm{4x}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \right)\mid_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{2}\left(\mathrm{8}−\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\mathrm{32}}{\mathrm{3}} \\ $$$$\mathrm{Area}\:\mathrm{I}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{a}}} {\int}}\left(\mathrm{a}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\: \\ $$$$=\mathrm{2}\mid\left(\mathrm{ax}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \right)\mid_{\mathrm{0}} ^{\sqrt{\mathrm{a}}} \\ $$$$=\mathrm{2}\left(\mathrm{a}\sqrt{\mathrm{a}}−\frac{\mathrm{a}\sqrt{\mathrm{a}}}{\mathrm{3}}\right)=\frac{\mathrm{4a}\sqrt{\mathrm{a}}}{\mathrm{3}} \\ $$$$\mathrm{totall}\:\mathrm{area}\:=\:\mathrm{2}×\mathrm{area}\:\mathrm{I} \\ $$$$\frac{\mathrm{32}}{\mathrm{3}}\:=\:\frac{\mathrm{8a}\sqrt{\mathrm{a}}}{\mathrm{3}}\:\Rightarrow\mathrm{4}\:=\:\mathrm{a}^{\mathrm{3}/\mathrm{2}} \:;\:\mathrm{a}\:=\:\mathrm{4}^{\mathrm{2}/\mathrm{3}} \\ $$$$\therefore\:\mathrm{a}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{16}}\:.\:\:@\mathrm{bemath}@ \\ $$

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