Question

In Exercises \(41-46,\) convert the point from cylindrical coordinates to spherical coordinates. $$ (2,2 \pi / 3,-2) $$

Short answer

The spherical coordinates of the point are \((2\sqrt{2}, \frac{2 \pi}{3}, -\frac{\pi}{4})\).

Step-by-step solution

- Recall the Conversions Formulas

The switch from cylindrical coordinates \((r, \phi, z)\) to spherical coordinates \((\rho, \phi, \theta)\) is given by the formulas: \[ \rho = \sqrt{r^2 + z^2} \] \[ \phi = \phi \] \[ \theta = \arctan{\left(\frac{r}{z}\right)} \] Here \(r, \phi, z\) are the cylindrical coordinates and \(\rho, \phi, \theta\) are the spherical coordinates.

- Apply the Conversion Formulas

Substitute the values \((r = 2, \phi = \frac{2 \pi}{3}, z = -2)\) into the conversion formulas and calculate the values for \(\rho, \phi, \theta\). \[ \rho = \sqrt{(2)^2 + (-2)^2} = 2\sqrt{2} \] \[ \phi = \frac{2 \pi}{3} \] \[ \theta = \arctan{\left(\frac{2}{-2}\right)} = -\frac{\pi}{4} \] Note that the range of \(\theta\) is \([-\pi / 2, \pi / 2]\).

- Write the Final Answer in Spherical Coordinates

The spherical coordinates are therefore \((\rho, \phi, \theta) = (2\sqrt{2}, \frac{2 \pi}{3}, -\frac{\pi}{4})\).