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Section 12.13 : Spherical Coordinates

1. Convert the Cartesian coordinates for \(\left( {3, - 4,1} \right)\) into Spherical coordinates.

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From the point we’re given we have,

\[x = 3\hspace{0.5in}y = - 4\hspace{0.5in}z = 1\] Show Step 2

Let’s first determine \(\rho \).

\[\rho = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {26} \] Show Step 3

We can now determine \(\varphi \).

\[\cos \varphi = \frac{z}{\rho } = \frac{1}{{\sqrt {26} }}\hspace{0.5in}\varphi = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {26} }}} \right) = 1.3734\] Show Step 4

Let’s use the \(x\) conversion formula to determine \(\theta \).

\[\cos \theta = \frac{x}{{\rho \sin \varphi }} = \frac{3}{{\sqrt {26} \sin \left( {1.3734} \right)}} = 0.6\hspace{0.25in} \to \hspace{0.25in}{\theta _1} = {\cos ^{ - 1}}\left( {0.6} \right) = 0.9273\]

This angle is in the first quadrant and if we sketch a quick unit circle we see that a second angle in the fourth quadrant is \({\theta _2} = 2\pi - 0.9273 = 5.3559\).

If we look at the three dimensional coordinate system from above we can see that from our \(x\) and \(y\) coordinates the point is in the fourth quadrant. This in turn means that we need to use \({\theta _2}\) for our point.

The Spherical coordinates are then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\sqrt {26} ,5.3559,1.3734} \right)}}\]