Domain and double integral Let $$D = \{(x,y)\in R^2 : 0<x<y<2x,x^2+y^2>4,xy<4\}$$ and $f : D \rightarrow R$ the continus and bounded function defined by $f(x,y)=xy$
I'm stucked to find some bounds for $\iint_D f(x,y) \,dx\,dy$
In the book I read they say :
Let define $\phi1,\phi2 : [2/\sqrt{5},2]\rightarrow R$ :
$$
\phi1(x)=\left\{
 \begin{array}{ll}
  \sqrt{4-x^2}  & \mbox{if } x \in [2/\sqrt{5},\sqrt{2}] \\
  x & \mbox{if } x \in [\sqrt{2},2]
 \end{array}
\right.$$
$$
\phi2(x)=\left\{
 \begin{array}{ll}
  2x  & \mbox{if } x \in [2/\sqrt{5},\sqrt{2}] \\
  4/x & \mbox{if } x \in [\sqrt{2},2]
 \end{array}
\right.$$
and so
$$
D=\{(x,y)\in R^2:2/\sqrt{5}<x<2,\phi1(x)<y<\phi2(x)\}
$$
And then it's easy to integrate.
My question : How did they found this ? I really don't understand



You need to find the corresponding intervals of $x$ such that your variable $y(x)$ changed from one curve to another. Let's check the intersection of some curves
When $y=2x; xy=4$. Then $y=\frac{4}{x}$, and by replacing, $\frac{4}{x}=2x \Rightarrow 4=2x^2 \Rightarrow x=\sqrt2$. Now, $y=x; xy=4$. Then $y=\frac{4}{x} \Rightarrow \frac{4}{x}=x \Rightarrow x=2$. Thus, in the interval $\sqrt2<x<2$ we've $y=\frac{4}{x}$.
Now, if you use the same argument to intercept $2x=y$ with $x^2+y^2=4$ you'll get $x=\frac{2}{\sqrt5}$ and so on. I'll let the rest for you.
