Section 12.12 : Cylindrical Coordinates
1. Convert the Cartesian coordinates for \(\left( {4, - 5,2} \right)\) into Cylindrical coordinates.
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Start SolutionFrom the point we’re given we have,
\[x = 4\hspace{0.5in}y = - 5\hspace{0.5in}z = 2\]So, we already have the \(z\) coordinate for the Cylindrical coordinates.
Show Step 2Remember as well that for \(r\) and \(\theta \) we’re going to do the same conversion work as we did in converting a Cartesian point into Polar coordinates.
So, getting \(r\) is easy.
\[r = \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 5} \right)}^2}} = \sqrt {41} \] Show Step 3Finally, we need to get \(\theta \).
\[{\theta _1} = {\tan ^{ - 1}}\left( {\frac{{ - 5}}{4}} \right) = - 0.8961\hspace{0.5in}{\theta _2} = - 0.8961 + \pi = 2.2455\]If we look at the three dimensional coordinate system from above we can see that \({\theta _1}\) is in the fourth quadrant and \({\theta _2}\) is in the second quadrant. Likewise, from our \(x\) and \(y\) coordinates the point is in the fourth quadrant (as we look at the point from above).
This in turn means that we need to use \({\theta _1}\) for our point.
The Cylindrical coordinates are then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\sqrt {41} , - 0.8961,2} \right)}}\]