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Let $D=\{(x,y)\space|\space1\leq x^2+y^2 \leq 2 \text{ and }y\geq0\}$.

Evaluate $\int\int_D(1+xy) dA$.

So I stated that $D$ is a y-simple region because for all $(x,y)\in D$, $\sqrt{1-x^2} \leq y \leq \sqrt{2-x^2}$.

My book states that for a y-simple region $D$, where $\phi_1(x)\leq y \leq \phi_2(x)$, the integral

$\int\int_Df(x,y) dA=\int_a^b\int_{\phi_1(x)}^{\phi_2(x)}f(x,y)dydx$.

I tried doing this with $$\int_{-\sqrt{2}}^{\sqrt{2}}\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}(1+xy)dydx$$ but I am not getting the correct result.

Could someone explain what I am doing wrong?

Bobby Lee
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  • You need $\phi_1, \phi_2$ defined in a certain interval. What interval is that? – Git Gud Mar 30 '14 at 17:32
  • $[0,\sqrt(2)]$ on the y-axis? – Bobby Lee Mar 30 '14 at 17:34
  • I'll let you answer your own question: what is $\phi_1(x)$ for $1<x\leq \sqrt 2$. – Git Gud Mar 30 '14 at 17:35
  • I think $\sqrt{1-x^2}$? Why did you choose the interval $1<x \leq \sqrt{2}$? – Bobby Lee Mar 30 '14 at 17:42
  • What is $\phi_1(\sqrt 2),$ for instance? – Git Gud Mar 30 '14 at 17:44
  • Oh wait. Is it $y = 0$? – Bobby Lee Mar 30 '14 at 17:44
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    Yes, but there are still adjustments to be made. I gotta run. What you want is $$\phi _1\colon [-\sqrt 2, \sqrt 2]\to \mathbb R, x\mapsto \begin{cases} \sqrt{1-x^2}, &\text{if }x\in [-1,1]\ 0, &\text{if }x\in [-\sqrt 2, \sqrt 2]\setminus [-1, 1]\end{cases}$$ I'll help you later if you still need it. If you manage with this, it would be good if you posted an answer yourself, so this doesn't come up as unanswered. – Git Gud Mar 30 '14 at 17:48

1 Answers1

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The mistake I made was assuming that $\phi_1(x)=\sqrt{1-x^2}$ over the entire interval of $-\sqrt{2} \leq x \leq \sqrt{2}$.

Graphing the region helps, it will make you realize that $\phi_1(x)=0$ over the intervals of

$[-\sqrt{2},-1)$ and $(1,\sqrt{2}]$.

You end up with the evaluation of

$$\int_{-\sqrt{2}}^{-1}\int_0^{\sqrt{2-x^2}}f(x,y)dydx+\int_{-1}^{1}\int_{\sqrt{-x^2}}^{\sqrt{2-x^2}}f(x,y)dydx+\int_{1}^{\sqrt{2}}\int_0^{\sqrt{2-x^2}}f(x,y)dydx=\frac{\pi}{2}$$

Bobby Lee
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