Question

convert the point from rectangular coordinates to cylindrical coordinates. (6, 2\sqrt{3}, - 5) (r, \theta, z) = ()

Explanation:

Step1: Calculate $r$

The formula for $r$ in cylindrical - coordinates when converting from rectangular coordinates $(x,y,z)$ is $r = \sqrt{x^{2}+y^{2}}$. Here $x = 6$ and $y = 2\sqrt{3}$, so $r=\sqrt{6^{2}+(2\sqrt{3})^{2}}=\sqrt{36 + 12}=\sqrt{48}=4\sqrt{3}$.

Step2: Calculate $\theta$

The formula for $\theta$ is $\theta=\arctan(\frac{y}{x})$. Substituting $x = 6$ and $y = 2\sqrt{3}$, we get $\theta=\arctan(\frac{2\sqrt{3}}{6})=\arctan(\frac{\sqrt{3}}{3})$. Since $x>0$ and $y>0$, $\theta=\frac{\pi}{6}$.

Step3: Identify $z$

In rectangular coordinates $(x,y,z)$ and cylindrical coordinates $(r,\theta,z)$, the $z$ - value remains the same. Here $z=-5$.

Answer:

$(4\sqrt{3},\frac{\pi}{6}, - 5)$