 # Let D be the region between the circles of radius 6 and radius 8 centered at the Anonym 2021-09-25 Answered
Let D be the region between the circles of radius 6 and radius 8 centered at the origin that lies in the third quadrant.
Express D in polar coordinates.
1) $$\displaystyle{D}={\left\lbrace{\left({r},\theta\right)}{\mid}{6}\leq{r}\leq{8},{0}\leq\theta\leq{\frac{{\pi}}{{{2}}}}\right\rbrace}$$
2) $$\displaystyle{D}={\left\lbrace\begin{array}{cc} {r}&\theta\end{array}\right)}{\mid}{6}\leq{r}\leq{8},\pi\leq\theta\leq{\frac{{{3}\pi}}{{{2}}}}\rbrace$$
3) $$\displaystyle{D}={\left\lbrace{\left({r},\theta\right)}{\mid}{0}\leq{r}\leq{8},\pi\leq\theta\leq{\frac{{{3}\pi}}{{{2}}}}\right\rbrace}$$
4) $$\displaystyle{D}={\left\lbrace{\left({r},\theta\right)}{\mid}{6}\leq{r}\leq{8},\pi\leq\theta\leq{2}\pi\right\rbrace}$$

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Given:let D be the region between the circles of radius 6 and 8 centered at the origin that lies in the third quadrant express D in polar coordinates.
Here the region D is between the circles with radius r=6,r=8 centered at the origin
that is $$\displaystyle{6}\leq{r}\leq{8}$$ and that lies in quadrant 3
So,
$$\displaystyle\pi\leq\theta\leq{\frac{{{3}\pi}}{{{2}}}}$$
Therefore,
$$\displaystyle{D}={\left\lbrace{\left({r},\therefore\right)}{\mid}{6}\leq{r}\leq{8},\pi\leq\therefore\leq{3}\pi{2}\right\rbrace}$$
Therefore the 2 nd option is the correct one