Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 12.13 : Spherical Coordinates
2. Convert the Cartesian coordinates for \(\left( { - 2, - 1, - 7} \right)\) into Spherical coordinates.
Show All Steps Hide All Steps
Start SolutionFrom the point we’re given we have,
\[x = - 2\hspace{0.5in}y = - 1\hspace{0.5in}z = - 7\] Show Step 2Let’s first determine \(\rho \).
\[\rho = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 7} \right)}^2}} = \sqrt {54} \] Show Step 3We can now determine \(\varphi \).
\[\cos \varphi = \frac{z}{\rho } = \frac{{ - 7}}{{\sqrt {54} }}\hspace{0.5in}\varphi = {\cos ^{ - 1}}\left( {\frac{{ - 7}}{{\sqrt {54} }}} \right) = 2.8324\] Show Step 4Let’s use the \(y\) conversion formula to determine \(\theta \).
\[\sin \theta = \frac{{ - 1}}{{\rho \sin \varphi }} = \frac{{ - 1}}{{\sqrt {54} \sin \left( {2.8324} \right)}} = - 0.4472\hspace{0.25in} \to \hspace{0.25in}{\theta _1} = {\sin ^{ - 1}}\left( { - 0.4472} \right) = - 0.4636\]This angle is in the fourth quadrant and if we sketch a quick unit circle we see that a second angle in the third quadrant is \({\theta _2} = \pi + 0.4636 = 3.6052\).
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 third quadrant. This in turn means that we need to use \({\theta _2}\) for our point.
The Spherical coordinates are then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\sqrt {54} ,3.6052,2.8324} \right)}}\]