Answer
$(4, \displaystyle \frac{\pi}{4})$
Work Step by Step
To change from rectangular to polar coordinates, use
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$r^{2}=(\sqrt{8})^{2}+(\sqrt{8})^{2}=8+8=16$
$r=\pm 4$
$\displaystyle \tan^{-1}(\frac{\sqrt{8}}{\sqrt{8}})=\tan^{-1}(1)=\frac{\pi}{4}$, which is in quadrant I.
The point P$(\sqrt{8},\sqrt{8})$ is also in quadrant I, so we take $\theta= \displaystyle \frac{\pi}{4}$.
The terminal end of $\theta$ passes through P, so for this angle, r is positive.
Polar coordinates: $(4, \displaystyle \frac{\pi}{4})$
