Question Number 98520 by bobhans last updated on 14/Jun/20
$$\mathrm{Integrate}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{xy}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{over}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{R}:\left\{−\mathrm{3}\leqslant\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{3},\:\mathrm{1}\leqslant\mathrm{xy}\leqslant\mathrm{4}\right\} \\ $$
Answered by john santu last updated on 14/Jun/20
![I=∫∫_R xy(x^2 +y^2 )dxdy put x^2 −y^2 =u ; xy=v limits : −3≤u≤3; 1≤v≤4 which bounds region R J = ((J(u,v))/(J(x,y))) = determinant ((((∂u/∂x) (∂u/∂y))),(((∂v/∂x) (∂v/∂y))))= determinant (((2x −2y)),((y x))) = 2x^2 +2y^2 J(x,y) = ((J(u,v))/(2(x^2 +y^2 ))) I = (1/2) ∫_(u=−3) ^(u=3) ∫_(v=1) ^(v=4) v dvdu = (1/2) [3−(−3))((v^2 /2)) ]_1 ^4 = 3 ( ((15)/2)) = ((45)/(2 )) ■](Q98522.png)
$$\mathrm{I}=\int\underset{\mathrm{R}} {\int}{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy}\: \\ $$$${put}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={u}\:;\:{xy}={v} \\ $$$${limits}\::\:−\mathrm{3}\leqslant{u}\leqslant\mathrm{3};\:\mathrm{1}\leqslant{v}\leqslant\mathrm{4}\:\mathrm{which}\: \\ $$$$\mathrm{bounds}\:\mathrm{region}\:\mathrm{R} \\ $$$$\mathrm{J}\:=\:\frac{\mathrm{J}\left(\mathrm{u},\mathrm{v}\right)}{\mathrm{J}\left(\mathrm{x},\mathrm{y}\right)}\:=\:\begin{vmatrix}{\frac{\partial\mathrm{u}}{\partial\mathrm{x}}\:\:\:\:\frac{\partial\mathrm{u}}{\partial\mathrm{y}}}\\{\frac{\partial\mathrm{v}}{\partial\mathrm{x}}\:\:\:\:\frac{\partial\mathrm{v}}{\partial\mathrm{y}}}\end{vmatrix}=\:\begin{vmatrix}{\mathrm{2x}\:\:\:−\mathrm{2y}}\\{\mathrm{y}\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}}\end{vmatrix} \\ $$$$=\:\mathrm{2x}^{\mathrm{2}} +\mathrm{2y}^{\mathrm{2}} \: \\ $$$$\mathrm{J}\left(\mathrm{x},\mathrm{y}\right)\:=\:\frac{\mathrm{J}\left(\mathrm{u},\mathrm{v}\right)}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{\mathrm{u}=−\mathrm{3}} {\overset{\mathrm{u}=\mathrm{3}} {\int}}\:\underset{\mathrm{v}=\mathrm{1}} {\overset{\mathrm{v}=\mathrm{4}} {\int}}\:\mathrm{v}\:\mathrm{dvdu}\: \\ $$$$\left.=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\mathrm{3}−\left(−\mathrm{3}\right)\right)\left(\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}\right)\:\right]_{\mathrm{1}} ^{\mathrm{4}} \\ $$$$=\:\mathrm{3}\:\left(\:\frac{\mathrm{15}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{45}}{\mathrm{2}\:}\:\blacksquare \\ $$