Question Number 78284 by msup trace by abdo last updated on 15/Jan/20
$${calculate}\:\int\int_{{D}} {xy}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{dxdy} \\ $$$${D}=\left\{\left({x},{y}\right)/\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right\} \\ $$
Commented by abdomathmax last updated on 17/Jan/20
![I =∫∫_D xy(√(x^2 +2y^2 ))dxdy ⇒ I =∫_0 ^1 (∫_0 ^(√(1−x^2 )) y(√(x^2 +2y^2 ))dy)xdx but ∫_0 ^(√(1−x^2 )) y(√(x^2 +2y^2 ))dy =[(1/6)(x^2 +2y^2 )^(3/2) ]_0 ^(√(1−x^2 )) =(1/6){ (x^2 +(x^2 +2−2x^2 )^(3/2) −x^3 } =(1/6){ (x^2 −x^3 +(2−x^2 )^(3/2) } ⇒ 6I =∫_0 ^1 x(x^2 −x^3 +(2−x^2 )^(3/2) )dx =∫_0 ^1 (x^3 −x^4 )dx+∫_0 ^1 x(2−x^2 )^(3/2) dx =[(x^4 /4)−(x^5 /5)]_0 ^1 +∫_0 ^1 x(2−x^2 )^(3/2) dx =(1/(20)) +∫_0 ^1 x(2−x^2 )^(3/2) dx we have ∫_0 ^1 x(2−x^2 )^(3/2) dx =_(x=(√2)sint) ∫_0 ^(π/4) (√2)sint(2−2sin^2 t)^(3/2) (√2)costdt =2×2^(3/2) ∫_0 ^(π/4) sint cos^4 t dt =2^(5/2) [−(1/5)cos^5 t]_0 ^(π/4) =−((4(√2))/5)( ((1/( (√2))))^5 −1) =−((4(√2))/5)( (1/(4(√2)))−1) =−(1/5)+((4(√2))/5) ⇒ 6I =(1/(20))−(1/5) +((4(√2))/5) =((−3)/(20)) +((4(√2))/5) ⇒ I =−(1/(40)) +((2(√2))/(15))](https://www.tinkutara.com/question/Q78430.png)
$${I}\:=\int\int_{{D}} {xy}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{dxdy}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{dy}\right){xdx}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{dy}\:=\left[\frac{\mathrm{1}}{\mathrm{6}}\left({x}^{\mathrm{2}} \:+\mathrm{2}{y}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left\{\:\left({x}^{\mathrm{2}} +\left({x}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −{x}^{\mathrm{3}} \right\}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left\{\:\left({x}^{\mathrm{2}} −{x}^{\mathrm{3}} +\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}\:\Rightarrow\right. \\ $$$$\mathrm{6}{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left({x}^{\mathrm{2}} −{x}^{\mathrm{3}} +\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −{x}^{\mathrm{4}} \right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dx} \\ $$$$=\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{20}}\:+\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dx}\:\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dx}\:=_{{x}=\sqrt{\mathrm{2}}{sint}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{2}}{sint}\left(\mathrm{2}−\mathrm{2}{sin}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \sqrt{\mathrm{2}}{costdt} \\ $$$$=\mathrm{2}×\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {sint}\:{cos}^{\mathrm{4}} {t}\:{dt} \\ $$$$=\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{2}}} \:\:\left[−\frac{\mathrm{1}}{\mathrm{5}}{cos}^{\mathrm{5}} {t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=−\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{5}}\left(\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{5}} −\mathrm{1}\right) \\ $$$$=−\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{5}}\left(\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}−\mathrm{1}\right)\:=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{5}}\:\Rightarrow \\ $$$$\mathrm{6}{I}\:=\frac{\mathrm{1}}{\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{5}}\:+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{5}}\:=\frac{−\mathrm{3}}{\mathrm{20}}\:+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{5}}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{40}}\:+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{15}} \\ $$