## Calculus 8th Edition

a) $(2, \dfrac{3\pi}{4}, 1)$ b) $(2\sqrt 2, \dfrac{\pi}{4}, 2)$
a) In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$ Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$ $r=\sqrt{x^2+y^2} \implies r= 2$ Here, $\theta=\tan^{-1} (\dfrac{\sqrt 2}{-\sqrt 2})$ and $\theta=\arctan (-1)=\dfrac{3\pi}{4}$ Therefore, $(2, \dfrac{3\pi}{4}, 1)$ b) In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$ $r=\sqrt{x^2+y^2} \implies r=\sqrt{(2)^2+(2)^2}=8=2\sqrt 2$ Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$ $\theta=\arctan (\dfrac{2}{2}) \implies \theta=\dfrac{\pi}{4}$ Hence, the rectangular coordinates are: $(2\sqrt 2, \dfrac{\pi}{4}, 2)$