Question Number 204500 by DeArtist last updated on 19/Feb/24
$$\mathrm{Given}\:\mathrm{that}\:{I}\:=\:\int\int_{{R}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} {dxdy}\:\mathrm{where}\:{R} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{region}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:{a}^{\mathrm{2}} \\ $$$$\mathrm{use}\:\mathrm{a}\:\mathrm{suitable}\:\mathrm{transformation}\:\mathrm{to}\:\mathrm{evaluate}\:{I} \\ $$
Answered by witcher3 last updated on 19/Feb/24
![x^2 +y^2 ≤a^2 ⇒(x=rcos(t);y=rsin(t)) r∈[0,∣a∣] t∈[0,2π] I=∫_0 ^(2π) ∫_0 ^(∣a∣) r^5 .rdrdt =∫_0 ^(2π) ∫_0 ^(∣a∣) r^6 dr.dt=∫_0 ^(2π) dt.∫_0 ^(∣a∣) r^6 dr =((2π)/7).∣a∣^7](Q204501.png)
$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{a}^{\mathrm{2}} \Rightarrow\left(\mathrm{x}=\mathrm{rcos}\left(\mathrm{t}\right);\mathrm{y}=\mathrm{rsin}\left(\mathrm{t}\right)\right) \\ $$$$\mathrm{r}\in\left[\mathrm{0},\mid\mathrm{a}\mid\right]\:\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\mid\mathrm{a}\mid} \mathrm{r}^{\mathrm{5}} .\mathrm{rdrdt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\mid\mathrm{a}\mid} \mathrm{r}^{\mathrm{6}} \mathrm{dr}.\mathrm{dt}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{dt}.\int_{\mathrm{0}} ^{\mid\mathrm{a}\mid} \mathrm{r}^{\mathrm{6}} \mathrm{dr} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{7}}.\mid\mathrm{a}\mid^{\mathrm{7}} \\ $$$$ \\ $$