Question

In Exercises \(41-46,\) convert the point from cylindrical coordinates to spherical coordinates. $$ (-4, \pi / 3,4) $$

Short answer

The spherical coordinates for the point \(-4, \pi / 3, 4\) are \(4\sqrt{2}, \arccos(\sqrt{2}/2), \pi / 3\).

Step-by-step solution

Understand Coordinate System Transformations

Firstly, the understanding of how the cylindrical and spherical coordinate systems correspond to each other is paramount. Cylindrical coordinates (r, \theta, z) can be related to spherical coordinates (\rho, \phi, \theta) using the following formulas: \n\[ \rho = \sqrt{r^{2} + z^{2}} \] \n\[ \phi = \arccos(\frac{z}{\rho})\] \n\[ \theta = \theta \]

Conversion of coordinates

Now, replacing the values given in the exercise, \(-4, \pi / 3, 4\), into these formulas we get: \n\[ \rho = \sqrt{{(-4)}^{2} + 4^{2}} \] \n\[ \phi = \arccos(\frac{4}{\rho}) \] \n\[ \theta = \pi / 3 \]

Simplify the coordinates

Simplify these values to get the spherical coordinates for the given point. The square root of \(16 + 16\) is \( \sqrt{32} \), simplifying to \( 4\sqrt{2} \) which is the value of \( \rho \). Plugging the calculated value of \( \rho \) into the second formula gives us \( \phi = \arccos(4/{4\sqrt{2}}) \), which simplifies to \( \phi = \arccos(\sqrt{2}/2)\). Therefore, the final spherical coordinates for the point \(-4, \pi / 3, 4\) are \(4\sqrt{2}, \arccos(\sqrt{2}/2), \pi / 3 \).