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Question Number 130478 by liberty last updated on 26/Jan/21

 Integrate the function f(x,y)=xy(x^2 +y^2 )  over the domain R={−3≤x^2 −y^2 ≤3, 1≤y≤4 }

$$\:{Integrate}\:{the}\:{function}\:{f}\left({x},{y}\right)={xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$${over}\:{the}\:{domain}\:{R}=\left\{−\mathrm{3}\leqslant{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \leqslant\mathrm{3},\:\mathrm{1}\leqslant{y}\leqslant\mathrm{4}\:\right\} \\ $$

Answered by EDWIN88 last updated on 26/Jan/21

I=∫∫_R  xy(x^2 +y^2 )dxdy  let u=x^2 −y^2  ; v=xy → { ((−3≤u≤3)),((1≤v≤4)) :}  J = ((J(u,v))/(J(x,y)))= determinant (((2x     −2y)),((y            x)))= 2(x^2 +y^2 )  J(x,y)=(1/(2(x^2 +y^2 )))J(u,v)  I=(1/2)∫_(−3) ^( 3) ∫_1 ^( 4)  v dvdu = (1/2)[ 3−(−3)]((v^2 /2))_1 ^4   I= ((45)/2)

$$\mathrm{I}=\int\underset{\mathrm{R}} {\int}\:\mathrm{xy}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$$$\mathrm{let}\:\mathrm{u}=\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:;\:\mathrm{v}=\mathrm{xy}\:\rightarrow\begin{cases}{−\mathrm{3}\leqslant\mathrm{u}\leqslant\mathrm{3}}\\{\mathrm{1}\leqslant\mathrm{v}\leqslant\mathrm{4}}\end{cases} \\ $$$$\mathrm{J}\:=\:\frac{\mathrm{J}\left(\mathrm{u},\mathrm{v}\right)}{\mathrm{J}\left(\mathrm{x},\mathrm{y}\right)}=\begin{vmatrix}{\mathrm{2x}\:\:\:\:\:−\mathrm{2y}}\\{\mathrm{y}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}}\end{vmatrix}=\:\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{J}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)}\mathrm{J}\left(\mathrm{u},\mathrm{v}\right) \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{3}} ^{\:\mathrm{3}} \int_{\mathrm{1}} ^{\:\mathrm{4}} \:\mathrm{v}\:\mathrm{dvdu}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:\mathrm{3}−\left(−\mathrm{3}\right)\right]\left(\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}\right)_{\mathrm{1}} ^{\mathrm{4}} \\ $$$$\mathrm{I}=\:\frac{\mathrm{45}}{\mathrm{2}} \\ $$$$ \\ $$

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