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Evaluate the following integral by changing to polar coordinates: $\iint x dxdy$, where $0\leq x \leq y$ and $0 \leq y \leq 1$. Above integral can be evaluated directly without changing the variables. I am getting the answer $\frac{1}{6}$. But I have to evaluate it by changing in polar coordinates. So let $x=r\cos\theta$ and $y=r\sin\theta$. Then $dxdy=rdrd\theta$. But what will be the limits on $r$ and $\theta$ ?

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  • $\begingroup$ You want the region between the lines y=x and y=0 and your radius is going between 0 and 1. Try drawing it out and see what u get $\endgroup$
    – Henry Lee
    Jan 12, 2021 at 3:22
  • $\begingroup$ @HenryLee I think the limit of $\theta $ will be $\frac{\pi}{4}$ to $\frac{\pi}{2}$. I am not sure whether I am correct or not and I have no idea for $r$. $\endgroup$ Jan 12, 2021 at 3:26
  • $\begingroup$ @HenryLee I drew the region and the limits of r will be $0\leq r \leq \frac {1}{\sin\theta}$ $\endgroup$ Jan 12, 2021 at 3:46

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Hint: Draw the region is key to understand the integral. If you do the graph you will find $$\frac{\pi}{4}\leq \theta\leq \frac{\pi}{2}$$ and $$0\leq y\leq 1$$ $$0\leq r \text{sin}(x)\leq 1$$ $$0\leq r\leq\frac{1}{\text{sin}(x)}$$

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  • $\begingroup$ How $0\leq r \ leq 1$? $\endgroup$ Jan 12, 2021 at 3:31
  • $\begingroup$ @Mathfun, remember $r=\sqrt{(x^2+y^2)}$ and $0\leq x\leq y\leq 1$ $\endgroup$
    – 00GB
    Jan 12, 2021 at 3:34
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    $\begingroup$ I drew the region and the limits of r will be $0\leq r \leq \frac {1}{\sin\theta}$ $\endgroup$ Jan 12, 2021 at 3:46
  • $\begingroup$ @Mathfun, you are right. I made a mistake. you can even get without drawing the graph. I fixed $\endgroup$
    – 00GB
    Jan 12, 2021 at 3:54

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