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Question Number 167576 by mkam last updated on 19/Mar/22

$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{b}\mathit{y}\mathit{u}\mathit{s}\mathit{e}\mathit{i}\mathit{n}\mathit{g}\mathit{t}\mathit{h}\mathit{e}\mathit{p}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{c}\mathit{o}\mathit{r}\mathit{d}\mathit{i}\mathit{n}\mathit{a}\mathit{i}\mathit{t}\mathit{e}$$$$$$$${\int }_0^{\frac{\mathit{a}}{2}}{\int }_{\mathit{y}}^{\sqrt{{\mathit{a}}^2-{\mathit{y}}^2}}\mathit{x}\mathit{d}\mathit{x}\mathit{d}\mathit{y}=\frac{5{\mathit{a}}^3}{24}$$

Commented by mkam last updated on 20/Mar/22

$$?????$$

Answered by ArielVyny last updated on 19/Mar/22

$${\int }_0^{\frac{a}{2}}{\left[\frac{1}{2}{x}^2\right]}_y^{\sqrt{a^2-y^2}}dy={\int }_0^{\frac{a}{2}}\frac{1}{2}{a}^2-y^2$$$$=\frac{1}{2}{a}^2\left[\frac{a}{2}\right]-\frac{1}{3}\left[\frac{a^3}{8}\right]=\frac{6a^3}{24}-\frac{a^3}{24}=\frac{5a^3}{24}$$$$$$

Commented by ArielVyny last updated on 19/Mar/22

$$whyusingpolarcoordinates$$

Commented by mkam last updated on 19/Mar/22

$$becausemyteacherwantedthismethod$$

Commented by JDamian last updated on 19/Mar/22

Oh, what kind of you for doing his homework

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