<s>[INST] Change of variables in Double integral - explanation of solved example. I'm exercising double integrals and here's an solved example that I'm not understand something.

Calculate $\iint_D xy(x^2 + y^2) dx\,dy$ where $D$ in the first quadrant bounded by:
$1 \le xy \le 2$ and $5 \le x^2 - y^2 \le 9.$ 

Here's the solution:

Let $u = xy$ and $v = x^2 - y^2$.
$1 \le u \le 2$ and $5 \le v \le 9$.
Calculate the Jacobian. $$
        \begin{vmatrix}
        u'_x & u'_y \\
        v'_x & v'_y \\
        \end{vmatrix} = 
        \begin{vmatrix}
        y & x \\
        2x & -2y \\
        \end{vmatrix}  =  -2y^2 - 2x^2 = -2(x^2 + y^2).$$
$J = \frac{1}{-2(x^2 + y^2)}$ so need to plug $dx\, dy = -\frac{1}{2(x^2+y^2)}du\,dv$.
$$\iint_D xy(x^2 + y^2)dx\,dy = -\int^9_5\int^2_1 u(x^2 + y^2) \frac{1}{2(x^2+y^2)} = \color{red}{-\frac{1}{2}\int^9_5\int^2_1udu\,dv} = -\frac{1}{2}\int^9_5\bigg(\frac{y^2}{2}\bigg|^2_1\bigg)dv = -\frac{1}{2}\int^9_5(2-\frac{1}{2})dv = -\frac{3}{4}\bigg(v\bigg|^9_5\bigg) = -3.$$

My question:
In the part I marked in $\text{$\color{red}{red}$}$, how has $(x^2 + y^2)$ just gone away?
Thanks in advance!!
 [/INST] It was cancelled. There is an $x^2+y^2$ in the numerator and in the denominator.
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