Question Number 27500 by abdo imad last updated on 07/Jan/18
$${find}\:\int\int_{\Delta} \sqrt{\mathrm{4}\:−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:}\:\:{dxdy}\:{with} \\ $$$$\Delta=\left\{\left({x},{y}\right)\:\in\mathbb{R}^{\mathrm{2}} /\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{2}{x}\right\} \\ $$
Commented by abdo imad last updated on 10/Jan/18
![let use the changement x=rcosθ and y=rsinθ x^2 +y^2 ≤2x ⇔ r^2 ≤ 2r cosθ ⇔ 0<r≤ 2cosθ I= ∫∫_(−_ (π/2)<θ<(π/2) and 0<r≤2cosθ) (√( 4−r^2 )) rdrdθ I =∫_(−(π/2)) ^(π/2) ( ∫_0 ^(2cosθ) r(√( 4−r^2 )) dr)dθ but ∫_0 ^(2cosθ) r(√(4−r^2 )) dr = [−(1/3)(4−r^2 )^(3/2) ]_0 ^(2cosθ) = −(1/3)((4−4cos^2 θ)^(3/2) −4^(3/2) =−(1/3)(4^(3/2) (sin^2 )^(3/2) −8)= −(1/3)(8 sin^3 θ −8 ) I= −(8/3) ∫_(−(π/2)) ^(π/2) (sin^3 θ −1)dθ = ((8π)/3) −(8/3) ∫_(−(π/2)) ^(π/2) sin^3 θdθ we find the value of I by linearisation of sin^3 θ....](Q27575.png)
$${let}\:{use}\:{the}\:{changement}\:{x}={rcos}\theta\:{and}\:{y}={rsin}\theta \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}{x}\:\Leftrightarrow\:{r}^{\mathrm{2}} \leqslant\:\mathrm{2}{r}\:{cos}\theta\:\:\Leftrightarrow\:\:\mathrm{0}<{r}\leqslant\:\mathrm{2}{cos}\theta \\ $$$${I}=\:\int\int_{−_{} \frac{\pi}{\mathrm{2}}<\theta<\frac{\pi}{\mathrm{2}}\:{and}\:\mathrm{0}<{r}\leqslant\mathrm{2}{cos}\theta} \:\:\sqrt{\:\mathrm{4}−{r}^{\mathrm{2}} }\:{rdrd}\theta \\ $$$${I}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\:\:\int_{\mathrm{0}} ^{\mathrm{2}{cos}\theta} {r}\sqrt{\:\mathrm{4}−{r}^{\mathrm{2}} }\:{dr}\right){d}\theta\:\:{but}\:\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{cos}\theta} {r}\sqrt{\mathrm{4}−{r}^{\mathrm{2}} }\:{dr}\:\:=\:\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}−{r}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\right]_{\mathrm{0}} ^{\mathrm{2}{cos}\theta} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\left(\mathrm{4}−\mathrm{4}{cos}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \left({sin}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{8}\right)=\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{8}\:{sin}^{\mathrm{3}} \theta\:−\mathrm{8}\:\right) \\ $$$${I}=\:−\frac{\mathrm{8}}{\mathrm{3}}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\left({sin}^{\mathrm{3}} \theta\:−\mathrm{1}\right){d}\theta\:\:\:=\:\frac{\mathrm{8}\pi}{\mathrm{3}}\:−\frac{\mathrm{8}}{\mathrm{3}}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}^{\mathrm{3}} \theta{d}\theta\:\:\: \\ $$$${we}\:{find}\:{the}\:{value}\:{of}\:{I}\:{by}\:{linearisation}\:{of}\:{sin}^{\mathrm{3}} \theta.... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by abdo imad last updated on 10/Jan/18
$${the}\:\:{fonction}\:{is}\:{impar}\:{so}\:\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{3}} {dx}=\mathrm{0}\:\:{and}\:{I}=\:\frac{\mathrm{8}\pi}{\mathrm{3}}\:. \\ $$