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Section 1-12 : Cylindrical Coordinates

2. Convert the Cartesian coordinates for \(\left( { - 4, - 1,8} \right)\) into Cylindrical coordinates.

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From the point we’re given we have,

\[x = - 4\hspace{0.5in}y = - 1\hspace{0.5in}z = 8\]

So, we already have the \(z\) coordinate for the Cylindrical coordinates.

Show Step 2

Remember 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.

\[r = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt {17} \] Show Step 3

Finally, we need to get \(\theta \).

\[{\theta _1} = {\tan ^{ - 1}}\left( {\frac{{ - 1}}{{ - 4}}} \right) = 0.2450\hspace{0.5in}{\theta _2} = 0.2450 + \pi = 3.3866\]

If we look at the three dimensional coordinate system from above we can see that \({\theta _1}\) is in the first quadrant and \({\theta _2}\) is in the third quadrant. Likewise, from our \(x\) and \(y\) coordinates the point is in the third quadrant (as we look at the point from above).

This in turn means that we need to use \({\theta _2}\) for our point.

The Cylindrical coordinates are then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\sqrt {17} ,3.3866,8} \right)}}\]