Question Number 61648 by maxmathsup by imad last updated on 05/Jun/19
$${calculate}\:\int\int_{{W}} \:\left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}}{dxdy}\:\:\:\:{with} \\ $$$${W}\:=\left\{\:\left({x},{y}\right)\:\in\:{R}^{\mathrm{2}} \:\:/\:\:\:\:\mathrm{1}\leqslant{x}\:\leqslant\sqrt{\mathrm{3}}\:\:{and}\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{2}{y}\:\leqslant\:\mathrm{2}\:\right\} \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/19
![we have x^2 +y^2 −2y ≤2 ⇒x^2 +y^2 −2y +1−1 ≤2 ⇒x^2 +(y−1)^2 ≤3 let use the diffeomorphism x=rcosθ and y−1 =rsinθ we have 1≤x^2 ≤3 ⇒1≤x^2 +(y−1)^2 ≤6 ⇒1≤r^2 ≤6 ⇒1≤r≤(√6) ∫∫_W (x^2 −2y^2 )(√(x^2 +y^2 +3))dxdy =∫∫_(1≤r≤(√6) and −(π/2)≤θ≤(π/2)) (r^2 cos^2 θ−2r^2 sin^2 θ)(√(r^2 +3))rdrdθ =∫_1 ^(√6) r^3 (√(3+r^2 ))dr ∫_(−(π/2)) ^(π/2) (cos^2 θ −2sin^2 θ)dθ ∫_(−(π/2)) ^(π/2) (cos^2 θ −2sin^2 θ)dθ =∫_(−(π/2)) ^(π/2) (((1+cos(2θ))/2) −(1−cos(2θ))dθ =(1/2) ∫_(−((.π)/2)) ^(π/2) (1+cos(2θ)−2 +2cos(2θ))dθ =(1/4) ∫_0 ^(π/2) {3cosθ −1}dθ = (3/4)[sinθ]_0 ^(π/2) −(π/8) =(3/4) −(π/8) . chang.(√(3+r^2 ))=t give 3+r^2 =t^2 ⇒rdr =tdt ∫_1 ^(√6) r^3 (√(3+r^2 ))dr = ∫_2 ^3 (t^2 −3)t dt = ∫_2 ^3 (t^3 −3t)dt =[(1/4)t^4 −(3/2)t^2 ]_2 ^3 =(3^4 /4) −(3^3 /2) −(2^4 /2) +(3/2).2^2 =((81)/4) −((27)/2) −8 +6 =((81−54)/4) −2 =((81−62)/4) =((18)/4) =(9/2) ⇒ ⇒ I = (9/2)((3/4)−(π/8)) =((27)/8) −((9π)/(16)) .](https://www.tinkutara.com/question/Q61690.png)
$${we}\:{have}\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{2}{y}\:\leqslant\mathrm{2}\:\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}\:+\mathrm{1}−\mathrm{1}\:\leqslant\mathrm{2}\:\Rightarrow{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} \:\leqslant\mathrm{3} \\ $$$${let}\:{use}\:{the}\:{diffeomorphism}\:\:\:{x}={rcos}\theta\:\:{and}\:{y}−\mathrm{1}\:={rsin}\theta \\ $$$${we}\:{have}\:\:\mathrm{1}\leqslant{x}^{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow\mathrm{1}\leqslant{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{6}\:\Rightarrow\mathrm{1}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{6}\:\:\Rightarrow\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{6}} \\ $$$$\int\int_{{W}} \left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+\mathrm{3}}{dxdy}\:=\int\int_{\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{6}}\:\:\:\:{and}\:\:\:−\frac{\pi}{\mathrm{2}}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \left({r}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−\mathrm{2}{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\sqrt{{r}^{\mathrm{2}} +\mathrm{3}}{rdrd}\theta \\ $$$$=\int_{\mathrm{1}} ^{\sqrt{\mathrm{6}}} {r}^{\mathrm{3}} \sqrt{\mathrm{3}+{r}^{\mathrm{2}} }{dr}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}^{\mathrm{2}} \theta\:−\mathrm{2}{sin}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}^{\mathrm{2}} \theta\:−\mathrm{2}{sin}^{\mathrm{2}} \theta\right){d}\theta\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:−\left(\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\frac{.\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)−\mathrm{2}\:+\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{3}{cos}\theta\:−\mathrm{1}\right\}{d}\theta \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}}\left[{sin}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\frac{\pi}{\mathrm{8}}\:=\frac{\mathrm{3}}{\mathrm{4}}\:−\frac{\pi}{\mathrm{8}}\:.\:{chang}.\sqrt{\mathrm{3}+{r}^{\mathrm{2}} }={t}\:{give}\:\mathrm{3}+{r}^{\mathrm{2}} ={t}^{\mathrm{2}} \:\Rightarrow{rdr}\:={tdt} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{6}}} {r}^{\mathrm{3}} \sqrt{\mathrm{3}+{r}^{\mathrm{2}} }{dr}\:=\:\int_{\mathrm{2}} ^{\mathrm{3}} \:\left({t}^{\mathrm{2}} −\mathrm{3}\right){t}\:{dt}\:=\:\int_{\mathrm{2}} ^{\mathrm{3}} \left({t}^{\mathrm{3}} −\mathrm{3}{t}\right){dt}\:=\left[\frac{\mathrm{1}}{\mathrm{4}}{t}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} \right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$=\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{4}}\:−\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}}\:−\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}}.\mathrm{2}^{\mathrm{2}} \:\:=\frac{\mathrm{81}}{\mathrm{4}}\:−\frac{\mathrm{27}}{\mathrm{2}}\:−\mathrm{8}\:+\mathrm{6}\:=\frac{\mathrm{81}−\mathrm{54}}{\mathrm{4}}\:−\mathrm{2}\:=\frac{\mathrm{81}−\mathrm{62}}{\mathrm{4}}\:=\frac{\mathrm{18}}{\mathrm{4}}\:=\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:{I}\:=\:\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{4}}−\frac{\pi}{\mathrm{8}}\right)\:=\frac{\mathrm{27}}{\mathrm{8}}\:−\frac{\mathrm{9}\pi}{\mathrm{16}}\:. \\ $$$$ \\ $$