Question Number 49942 by turbo msup by abdo last updated on 12/Dec/18
![calculate ∫∫_D (√(x^4 −y^4 ))dxdy with D =[0,1]×[0,1]](https://www.tinkutara.com/question/Q49942.png)
$${calculate}\:\int\int_{{D}} \sqrt{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }{dxdy} \\ $$$${with}\:{D}\:=\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},\mathrm{1}\right] \\ $$
Commented by Abdo msup. last updated on 16/Dec/18
![changement (r,θ)→(x,y)/ x =rcosθ and y =rsinθ we have 0≤x≤1 and 0≤y≤1 ⇒0≤x^2 +y^2 ≤2 ⇒0<r<(√2) but x≥0 and y≥0 ⇒ θ∈[0,(π/2)] ⇒ ∫∫_D (√(x^4 −y^4 ))dxdy =∫∫_D (√(x^2 −y^2 ))(√(x^2 +y^2 ))dxdy =∫∫_(0<r<(√(2 ))and 0<θ<(π/2)) (√(r^2 (cos^2 θ −sin^2 θ))r r drdθ =∫_0 ^(√2) r^3 dr ∫_0 ^(π/2) (√(cos^2 θ−sin^2 θ))dr dθ =[(r^4 /4)]_0 ^(√2) .∫_0 ^(π/2) (√(cos(2θ)))dθ =∫_0 ^(π/2) (√(cos(2θ)))dθ ....be continued...](https://www.tinkutara.com/question/Q50335.png)
$${changement}\:\left({r},\theta\right)\rightarrow\left({x},{y}\right)/\:{x}\:={rcos}\theta\:\:{and}\:{y}\:={rsin}\theta \\ $$$${we}\:{have}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\mathrm{0}<{r}<\sqrt{\mathrm{2}} \\ $$$${but}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\geqslant\mathrm{0}\:\Rightarrow\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\Rightarrow \\ $$$$\int\int_{{D}} \sqrt{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }{dxdy}\:=\int\int_{{D}} \sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy} \\ $$$$=\int\int_{\mathrm{0}<{r}<\sqrt{\mathrm{2}\:\:}{and}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}} \:\:\:\sqrt{{r}^{\mathrm{2}} \left({cos}^{\mathrm{2}} \theta\:−{sin}^{\mathrm{2}} \theta\right.}{r}\:\:\:{r}\:{drd}\theta \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {r}^{\mathrm{3}} {dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\sqrt{{cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta}{dr}\:{d}\theta \\ $$$$=\left[\frac{{r}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} .\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\left(\mathrm{2}\theta\right)}{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cos}\left(\mathrm{2}\theta\right)}{d}\theta \\ $$$$….{be}\:{continued}… \\ $$