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Question Number 61785 by aliesam last updated on 08/Jun/19
∫_0 ^(√(3−x^2 )) ((xy(4−x^2 −y^2 )(√(4−x^2 −y^2 ))−xy)/3) dy
$$\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}−{x}^{\mathrm{2}} }} {\int}}\frac{{xy}\left(\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }−{xy}}{\mathrm{3}}\:{dy} \\ $$
Answered by MJS last updated on 09/Jun/19
∫((xy(4−x^2 −y^2 )(√(4−x^2 −y^2 ))−xy)/3)dy=  =(x/3)∫y(4−x^2 −y^2 )^(3/2) dy−(x/3)∫ydy=  =−(x/(15))(4−x^2 −y^2 )^(5/2) −(x/6)y^2   ⇒  ∫_0 ^(√(3−x^2 )) ((xy(4−x^2 −y^2 (√(4−x^2 −y^2 ))−xy)/3)dy=  =(x/(10))(5x^2 −17+2(4−x^2 )^(5/2) )
$$\int\frac{{xy}\left(\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }−{xy}}{\mathrm{3}}{dy}= \\ $$$$=\frac{{x}}{\mathrm{3}}\int{y}\left(\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} {dy}−\frac{{x}}{\mathrm{3}}\int{ydy}= \\ $$$$=−\frac{{x}}{\mathrm{15}}\left(\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{5}/\mathrm{2}} −\frac{{x}}{\mathrm{6}}{y}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{3}−{x}^{\mathrm{2}} }} {\int}}\frac{{xy}\left(\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }−{xy}\right.}{\mathrm{3}}{dy}= \\ $$$$=\frac{{x}}{\mathrm{10}}\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{17}+\mathrm{2}\left(\mathrm{4}−{x}^{\mathrm{2}} \right)^{\mathrm{5}/\mathrm{2}} \right) \\ $$
Commented by aliesam last updated on 08/Jun/19
thanks sir
$${thanks}\:{sir} \\ $$
Commented by Prithwish sen last updated on 09/Jun/19
I think it will   − (x/(15))(4−x^2 −y^2 )^(5/2)
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\:\:−\:\frac{\mathrm{x}}{\mathrm{15}}\left(\mathrm{4}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$
Commented by MJS last updated on 09/Jun/19
you′re right, thanks for checking  I corrected it
$$\mathrm{you}'\mathrm{re}\:\mathrm{right},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{checking} \\ $$$$\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$

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