Question

In Exercises 63 and 64 , sketch the solid that has the given description in spherical coordinates. $$ 0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi / 6,0 \leq \rho \leq a \sec \phi $$

Short answer

The solid is ice-cream cone shaped, facing downwards from the origin along the z-axis.

Step-by-step solution

Setting \(\theta\) values

Here, \(\theta\) ranges from 0 to 2\(\pi\). This covers all values in the xy-plane, indicating it forms a complete circle around the z-axis.

Setting \(\phi\) values

Our solid covers a range of \(\phi\) values from 0 to \(\pi / 6\). Since \(\phi\) is the angle from the z-axis, this means the solid exists from the positive z-axis down to a pi/6 angle below the postive z-axis.

\(\rho\) is dependent upon \(\phi\)

This indicates that as \(\phi\) changes, so does \(\rho\). As \( \phi = 0 \) , \(\rho = a (sec 0) = a\). Thus, the radius increases as \( \phi \) increases. Meaning, the greater the angle from the positive z-axis, the greater the radius of the solid from the origin.

Sketch the solid

Considering these determined ranges, the solid is formed above the xy-plane within the region described by the \(\phi\) and \(\theta\) angles with a radius that increases as \(\phi\) increases. It's basically the sector of a sphere from the north pole down to the parallel at latitude angle \( \pi / 6 \) with radius increasing from a to infinity as \(\phi\) increases. This results in an ice-cream cone shape with the tip at the origin, pointing upwards along the z-axis, and the round surface facing downwards (opening down).