Chapter 6 - Review Exercises - Page 798: 32

$(3 \sqrt 2, \dfrac{7 \pi}{4})$

Here, $r=\sqrt {x^2+y^2}=\sqrt{(3)^2+(-3)^2}=3 \sqrt 2$ $\tan \theta =\dfrac{y}{x}$ and $\theta=arctan[\dfrac{3}{-3}]=arctan {-1}$ Thus, $\theta =\dfrac{7 \pi}{4}$ Hence, $(3 \sqrt 2, \dfrac{7 \pi}{4})$