Question Number 28427 by abdo imad last updated on 25/Jan/18
$${find}\:\int\int_{{D}} \sqrt{\mathrm{2}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }\:\:{dxdy}\:\:{with} \\ $$$${D}=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\sqrt{\mathrm{2}}\:\right\} \\ $$
Commented by abdo imad last updated on 27/Jan/18
![let put I= ∫∫_D (√(2−x^2 −y^2 )) let use the ch. x=rcosθ and y=rsinθ x^2 +y^2 ≤(√2) ⇒ 0< r^2 ≤(√2)⇒ 0<r≤^4 (√2) I= ∫∫_(0<r≤^4 (√2) and 0≤θ≤2π) (√(2−r^2 )) rdrdθ I= (∫_0 ^4_(√2) r(√(2−r^2 )) dr) ( ∫_0 ^(2π) θ) =2π ∫_0 ^4_(√2) r(√(2−r^2 )) dr =−((2π)/3)∫_0 ^4_(√2) 3 (−r)(2−r^2 )^(1/2) dr =−((2π)/3) [ (2−r^2 )^(3/2) ]_0 ^4_(√2) = ((−2π)/3)((2 −(^4 (√(2))) )^2 )^(3/2) −2^(3/2) ) =((−2π)/3) ( (2−(√(2)))^(3/2) −2(√2)) = ((−2π)/3)( (2−(√(2))) (√(2−(√2))) −2(√2))) .](https://www.tinkutara.com/question/Q28570.png)
$${let}\:{put}\:{I}=\:\int\int_{{D}} \sqrt{\mathrm{2}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }\:\:{let}\:{use}\:{the}\:{ch}.\:{x}={rcos}\theta\:{and} \\ $$$${y}={rsin}\theta\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\sqrt{\mathrm{2}}\:\:\Rightarrow\:\mathrm{0}<\:{r}^{\mathrm{2}} \leqslant\sqrt{\mathrm{2}}\Rightarrow\:\mathrm{0}<{r}\leqslant\:^{\mathrm{4}} \sqrt{\mathrm{2}} \\ $$$${I}=\:\int\int_{\mathrm{0}<{r}\leqslant\:^{\mathrm{4}} \sqrt{\mathrm{2}}\:\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi} \sqrt{\mathrm{2}−{r}^{\mathrm{2}} \:}\:{rdrd}\theta \\ $$$${I}=\:\left(\int_{\mathrm{0}} ^{\mathrm{4}_{\sqrt{\mathrm{2}}} } \:\:{r}\sqrt{\mathrm{2}−{r}^{\mathrm{2}} }\:{dr}\right)\:\left(\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \theta\right)\:=\mathrm{2}\pi\:\int_{\mathrm{0}} ^{\mathrm{4}_{\sqrt{\mathrm{2}}} } \:\:{r}\sqrt{\mathrm{2}−{r}^{\mathrm{2}} }\:{dr} \\ $$$$=−\frac{\mathrm{2}\pi}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{4}_{\sqrt{\mathrm{2}}} } \:\mathrm{3}\:\left(−{r}\right)\left(\mathrm{2}−{r}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dr}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=−\frac{\mathrm{2}\pi}{\mathrm{3}}\:\left[\:\left(\mathrm{2}−{r}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{4}_{\sqrt{\mathrm{2}}} } \:=\:\frac{−\mathrm{2}\pi}{\mathrm{3}}\left(\left(\mathrm{2}\:−\left(^{\mathrm{4}} \sqrt{\left.\mathrm{2}\right)}\:\right)^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \right) \\ $$$$=\frac{−\mathrm{2}\pi}{\mathrm{3}}\:\left(\:\left(\mathrm{2}−\sqrt{\left.\mathrm{2}\right)}\:^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\mathrm{2}\sqrt{\mathrm{2}}\right)\right. \\ $$$$=\:\frac{−\mathrm{2}\pi}{\mathrm{3}}\left(\:\left(\mathrm{2}−\sqrt{\left.\mathrm{2}\right)}\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\:\:−\mathrm{2}\sqrt{\mathrm{2}}\right)\right)\:\:. \\ $$
Answered by ajfour last updated on 27/Jan/18
![I=∫_0 ^( 2π) [(1/2)∫_0 ^( (√2)) (√(2−(r^2 ))) d(r^2 )]dθ =[((2π(2−r^2 )(√(2−r^2 )))/3)]∣_(r^2 =(√2)) ^0 I = ((2π)/3)[2(√2)−(2−(√2))(√(2−(√2))) ] .](https://www.tinkutara.com/question/Q28571.png)
$${I}=\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \left[\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\sqrt{\mathrm{2}}} \sqrt{\mathrm{2}−\left({r}^{\mathrm{2}} \right)}\:{d}\left({r}^{\mathrm{2}} \right)\right]{d}\theta \\ $$$$=\left[\frac{\mathrm{2}\pi\left(\mathrm{2}−{r}^{\mathrm{2}} \right)\sqrt{\mathrm{2}−{r}^{\mathrm{2}} }}{\mathrm{3}}\right]\mid_{{r}^{\mathrm{2}} =\sqrt{\mathrm{2}}} ^{\mathrm{0}} \\ $$$${I}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\left[\mathrm{2}\sqrt{\mathrm{2}}−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\:\right]\:. \\ $$