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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,

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