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Question Number 27595 by abdo imad last updated on 10/Jan/18

find  ∫∫_D   xy(√( x^2 +y^2 ))  dxdy   with  D={ (x,y)∈R^2 / x^2  +2y^2  ≤1  ,x≥0 ,y ≥0}

$${find}\:\:\int\int_{{D}} \:\:{xy}\sqrt{\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\:{dxdy}\:\:\:{with} \\ $$$${D}=\left\{\:\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:{x}^{\mathrm{2}} \:+\mathrm{2}{y}^{\mathrm{2}} \:\leqslant\mathrm{1}\:\:,{x}\geqslant\mathrm{0}\:,{y}\:\geqslant\mathrm{0}\right\} \\ $$

Commented by abdo imad last updated on 15/Jan/18

we use the polar coordinate let use the ch.x=rcosθ  and y= (1/(√2)) sinθ   due to the diffeomorphisme we must have   0≤θ≤ (π/2) and  0≤r≤1  I=∫∫_D xy(√(x^2 +2y^2 )) dxdy= ∫∫_w  Φof/j_Φ  /dr dθ  (r,θ)−(f_1 (r,θ) ,f_2 (r,θ))=(x y ) =(rcosθ ,(r/(√2)) sinθ)  M_j =   (_((1/(√2))sinθ) ^(cosθ)     _((r/(√2))cosθ)^(−rsinθ)  )  I=∫∫_(0≤r≤1 and 0≤θ≤(π/2))    rcosθ.(r/(√2)) sinθ r .(r/(√2)) drdθ  I= (1/2)∫_0^  ^1  r^4   ∫_0 ^(π/(2 ))  cosθ sinθ dθ = (1/4) [(1/5) r^5 ]_0 ^1  ∫_0 ^(π/2)  sin(2θ)dθ  = (1/(40)) [−cos(2θ)]_0 ^(π/2) =  (1/(20)) .

$${we}\:{use}\:{the}\:{polar}\:{coordinate}\:{let}\:{use}\:{the}\:{ch}.{x}={rcos}\theta \\ $$$${and}\:{y}=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{sin}\theta\:\:\:{due}\:{to}\:{the}\:{diffeomorphisme}\:{we}\:{must}\:{have} \\ $$$$\:\mathrm{0}\leqslant\theta\leqslant\:\frac{\pi}{\mathrm{2}}\:{and}\:\:\mathrm{0}\leqslant{r}\leqslant\mathrm{1} \\ $$$${I}=\int\int_{{D}} {xy}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }\:{dxdy}=\:\int\int_{{w}} \:\Phi{of}/{j}_{\Phi} \:/{dr}\:{d}\theta \\ $$$$\left({r},\theta\right)−\left({f}_{\mathrm{1}} \left({r},\theta\right)\:,{f}_{\mathrm{2}} \left({r},\theta\right)\right)=\left({x}\:{y}\:\right)\:=\left({rcos}\theta\:,\frac{{r}}{\sqrt{\mathrm{2}}}\:{sin}\theta\right) \\ $$$${M}_{{j}} =\:\:\:\left(_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{sin}\theta} ^{{cos}\theta} \:\:\:\:_{\frac{{r}}{\sqrt{\mathrm{2}}}{cos}\theta} ^{−{rsin}\theta} \:\right) \\ $$$${I}=\int\int_{\mathrm{0}\leqslant{r}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:{rcos}\theta.\frac{{r}}{\sqrt{\mathrm{2}}}\:{sin}\theta\:{r}\:.\frac{{r}}{\sqrt{\mathrm{2}}}\:{drd}\theta \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}^{} } ^{\mathrm{1}} \:{r}^{\mathrm{4}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \:{cos}\theta\:{sin}\theta\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\frac{\mathrm{1}}{\mathrm{5}}\:{r}^{\mathrm{5}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left(\mathrm{2}\theta\right){d}\theta \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{40}}\:\left[−{cos}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\:\frac{\mathrm{1}}{\mathrm{20}}\:. \\ $$

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