Calculus III - Cylindrical Coordinates

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Section 12.12 : Cylindrical Coordinates

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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,

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