The problem requires converting Cartesian coordinates to cylindrical coordinates. (a) Convert $(2, 2, 3)$ from Cartesian to cylindrical coordinates. (b) Convert $(4\sqrt{3}, -4, 6)$ from Cartesian to cylindrical coordinates.

1. Problem Description

The problem requires converting Cartesian coordinates to cylindrical coordinates.

(a) Convert $(2, 2, 3)$ from Cartesian to cylindrical coordinates.

(b) Convert $(4\sqrt{3}, -4, 6)$ from Cartesian to cylindrical coordinates.

2. Solution Steps

The conversion from Cartesian coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$ is given by:

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

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

$z = z$

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

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

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

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

$z = 3$

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

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

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

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

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

Since $x > 0$ and $y < 0$, the angle is in the fourth quadrant. Therefore, the angle is indeed $-\frac{\pi}{6}$.

$z = 6$

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

3. Final Answer

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

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