The problem asks to convert Cartesian coordinates to cylindrical coordinates for two points: (a) $(2, 2, 3)$ and (b) $(4\sqrt{3}, -4, 6)$.

1. Problem Description

The problem asks to convert Cartesian coordinates to cylindrical coordinates for two points: (a) $(2, 2, 3)$ and (b) $(4\sqrt{3}, -4, 6)$.

2. Solution Steps

To convert Cartesian coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$, we use the following formulas:

$r = \sqrt{x^2 + y^2}$

$\theta = \arctan(\frac{y}{x})$

$z = z$

(a) For the point $(2, 2, 3)$:

$x = 2, y = 2, z = 3$

$r = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

$\theta = \arctan(\frac{2}{2}) = \arctan(1) = \frac{\pi}{4}$

$z = 3$

So the cylindrical coordinates are $(2\sqrt{2}, \frac{\pi}{4}, 3)$.

(b) For the point $(4\sqrt{3}, -4, 6)$:

$x = 4\sqrt{3}, y = -4, z = 6$

$r = \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{16 \cdot 3 + 16} = \sqrt{48 + 16} = \sqrt{64} = 8$

$\theta = \arctan(\frac{-4}{4\sqrt{3}}) = \arctan(\frac{-1}{\sqrt{3}}) = \arctan(-\frac{\sqrt{3}}{3}) = -\frac{\pi}{6}$

Since $x > 0$ and $y < 0$, the angle is in the fourth quadrant. Thus, $\theta = -\frac{\pi}{6}$. This can also be written as $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

$z = 6$

So the cylindrical coordinates are $(8, -\frac{\pi}{6}, 6)$ or $(8, \frac{11\pi}{6}, 6)$.

3. Final Answer

(a) The cylindrical coordinates for $(2, 2, 3)$ are $(2\sqrt{2}, \frac{\pi}{4}, 3)$.

(b) The cylindrical coordinates for $(4\sqrt{3}, -4, 6)$ are $(8, -\frac{\pi}{6}, 6)$ or $(8, \frac{11\pi}{6}, 6)$.