Evaluating the following integral: $ \iiint_R x \, dA$ using a change of variable. I'm calculating the following integral: 
$$ \iiint_R x \, dA$$
Where $S$ is the region in the first quadrant delimited by the portion of the circle $x^2+y^2=4$ and the lines $y=1$ and $y=2$. This specific problem suggests to use the variable change: $u=x^2+y^2$, $v=x^2-y^2$. However, I don't know how to express the area $S$ using this new variables. I would appreciate any help. 
 A: HINT: $y=\sqrt{\dfrac{u-v}2}$ for $y>0$.
A: This region is perfectly defined for the circle of radius $2$ because $y=2$ lies tangent to the circle. Perhaps it's more intuitive to define the region as bounded below by $y = 1$ and above by $x^2 + y^2 = 4$. Since the region is inside the circle, this means $0 \le x^2 + y^2 \le 4$, or $0 \le u \le 4$. The other condition is:
$$ 1 \le y \le 2 \Rightarrow 1 \le y^2 \le 4 \Rightarrow 1\le \frac{u-v}{2} \le 4 \Rightarrow 2 \le u - v \le 8 $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#66f}{\large\iint_{\rm S}x\,\dd y\,\dd x}&
=\int_{0}^{\root{3}}\int_{1}^{\root{4 - x^{2}}}x\,\dd y\,\dd x
=\int_{0}^{\root{3}}x\pars{\root{4 - x^{2}} - 1}\,\dd x
\\[5mm]&=\half\int_{0}^{3}\pars{\root{4 - x} - 1}\,\dd x
=\half\bracks{-\,{2 \over 3}\,\pars{4 - x}^{3/2} - x}_{0}^{3}
\\[5mm]&=\half\pars{-\,{2 \over 3} - 3 + {2 \over 3}\,4^{3/2}}
=\color{#66f}{\large{5 \over 6}} \approx {\tt 0.8333}
\end{align}
