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


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Question Number 130478 by liberty last updated on 26/Jan/21
$Integrate the function f(x,y)=xy(x^2 +y^2 ) over the domain R={−3≤x^2 −y^2 ≤3, 1≤y≤4 }$
$I n t e g r a t e t h e f u n c t i o n f (x, y) = x y (x^{2} + y^{2})$ $o v e r t h e d o m a i n R = {- 3 ⩽ x^{2} - y^{2} ⩽ 3, 1 ⩽ y ⩽ 4}$
	Answered by EDWIN88 last updated on 26/Jan/21
	$I=∫∫_R xy(x^2 +y^2 )dxdy let u=x^2 −y^2 ; v=xy → { ((−3≤u≤3)),((1≤v≤4)) :} J = ((J(u,v))/(J(x,y)))= determinant (((2x −2y)),((y x)))= 2(x^2 +y^2 ) J(x,y)=(1/(2(x^2 +y^2 )))J(u,v) I=(1/2)∫_(−3) ^( 3) ∫_1 ^( 4) v dvdu = (1/2)[ 3−(−3)]((v^2 /2))_1 ^4 I= ((45)/2)$
	$I = \int \int_{R} xy (x^{2} + y^{2}) dxdy$ $let u = x^{2} - y^{2}; v = xy \to {\begin{cases} - 3 ⩽ u ⩽ 3 \\ 1 ⩽ v ⩽ 4 \end{cases}$ $J = \frac{J (u, v)}{J (x, y)} = \| \begin{matrix} 2 x - 2 y \\ y x \end{matrix} \| = 2 (x^{2} + y^{2})$ $J (x, y) = \frac{1}{2 (x^{2} + y^{2})} J (u, v)$ $I = \frac{1}{2} \int_{- 3}^{3} \int_{1}^{4} v dvdu = \frac{1}{2} [3 - (- 3)] {(\frac{v^{2}}{2})}_{1}^{4}$ $I = \frac{45}{2}$
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