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Section 12.13 : Spherical Coordinates

7. Identify the surface generated by the given equation : \(\displaystyle \varphi = \frac{{4\pi }}{5}\)

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Okay, as we discussed this type of equation in the notes for this section. We know that all points on the surface generated must be of the form \(\left( {\rho ,\theta ,\frac{{4\pi }}{5}} \right)\).

So, we can rotate around the \(z\)-axis as much as want them to (i.e. \(\theta \) can be anything) and we can move as far as we want from the origin (i.e. \(\rho \) can be anything). All we need to do is make sure that the point will always make an angle of \(\frac{{4\pi }}{5}\) with the positive ­z-axis.

In other words, we have a cone. It will open downwards and the “wall” of the cone will form an angle of \(\frac{{4\pi }}{5}\) with the positive \(z\)-axis and it will form an angle of \(\frac{\pi }{5}\) with the negative \(z\)-axis.