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Question Number 204500 by DeArtist last updated on 19/Feb/24

$$\mathrm{Given}\mathrm{that}I=\int {\int }_R{(x^2+y^2)}^{\frac{5}{2}}dxdy\mathrm{where}R$$$$\mathrm{is}\mathrm{the}\mathrm{region}{x}^2+y^2⩽a^2$$$$\mathrm{use}\mathrm{a}\mathrm{suitable}\mathrm{transformation}\mathrm{to}\mathrm{evaluate}I$$

Answered by witcher3 last updated on 19/Feb/24

$${\mathrm{x}}^2+{\mathrm{y}}^2⩽{\mathrm{a}}^2\Rightarrow (\mathrm{x}=\mathrm{rcos}\left(\mathrm{t}\right);\mathrm{y}=\mathrm{rsin}\left(\mathrm{t}\right))$$$$\mathrm{r}\in [0,\mid \mathrm{a}\mid ]\mathrm{t}\in [0,2\pi ]$$$$\mathrm{I}={\int }_0^{2\pi }{\int }_0^{\mid \mathrm{a}\mid }{\mathrm{r}}^5.\mathrm{rdrdt}$$$$={\int }_0^{2\pi }{\int }_0^{\mid \mathrm{a}\mid }{\mathrm{r}}^6\mathrm{dr}.\mathrm{dt}={\int }_0^{2\pi }\mathrm{dt}.{\int }_0^{\mid \mathrm{a}\mid }{\mathrm{r}}^6\mathrm{dr}$$$$=\frac{2\pi }{7}.\mid \mathrm{a}{\mid }^7$$$$$$

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