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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

f(x,y) D

jcai
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XogoX
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  • Can you graph $D$? – Git Gud Sep 06 '14 at 11:56
  • Yes sure, I forgot to upload it – XogoX Sep 06 '14 at 11:59
  • Look at the bottom graph. Your $\phi_1$ is the function determined by the bottom line and dually for $\phi_2$. – Git Gud Sep 06 '14 at 12:09
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    But how can I find the bounds and the functions without any drawing? – XogoX Sep 06 '14 at 12:13
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    I'd like to know that too... I wasn't taught that and I've never seen it anywhere. I developed a method when I was a student, but it doesn't always work. The best I can do to make it rigorous is to suggest that you come up $\phi _1$ and $\phi_2$ somehow (let it be by drawings, divine inspiration or other methods) and then prove that $D={(x,y)\in \mathbb R^2\colon \phi_1(x)\leq y\leq \phi_2(x)}$. Proving this equality makes it rigorous. As a practical matter, I suggest that you stick to graphing the sets. Though lacking in rigour, it's easier, faster and whoever marks your exam will accept it – Git Gud Sep 06 '14 at 12:54

1 Answers1

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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.

Jeybe
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