# Describe in words the surface whose equation is given:

$\phi = \dfrac{\pi}{4}$

– The upper half of the right circular cone whose vertex lies at the origin and axis at the positive z axis.

– The plane perpendicular to the xz plane crossing z = x, where $x \geq 0$.

– The plane perpendicular to the xz plane crossing y= x, where $x \geq 0$.

– The bottom of the right circular cone whose vertex lies at the origin and axis at the positive z axis.

– The plane perpendicular to the $yz$ plane crossing z = y, where $y \geq 0$.

This problem aims to describe the surface of a circular cone whose equation is given. To better understand the problem, you should be familiar with cartesian coordinate systems, spherical coordinates, and cylindrical coordinate systems.

Spherical coordinates are the 3 coordinates that determine the location of a point in a 3 dimensional trajectory. These 3 coordinates are the length of its internal radius vector r, the angle $\theta$ between the vertical plane having this vector and the x axis, and the angle $\phi$ between this vector and the horizontal x-y plane.

We can relate cylindrical coordinates with spherical coordinates such that if a point contains cylindrical coordinates $\left( r, \theta, z \right)$, $\left( r, \theta, z \right)$, then these equations describe the association between cylindrical and spherical coordinates. $r = \rho \sin\phi$ These type of equations are used to convert from $\phi = \theta$, spherical coordinates to cylindrical $z = \rho \sin\phi$ coordinates.

Spherical Coordinates are given as:

$x = Rcos\theta sin\phi = \dfrac {Rcos\theta}{\sqrt{2}}$

$y = Rsin\theta sin\phi = \dfrac {Rsin\theta} {\sqrt{2}}$

$z = Rcos\phi = \dfrac {R} {\sqrt{2}}$

$x^2 + y^2 = \dfrac {R^2} {2} = z^2$

$z^2 = x^2 + y^2$

$z = \sqrt{x^2 + y^2}$

Now,

$z = +\sqrt{x^2 + y^2}$ is the upper bond and $z = -\sqrt{x^2 + y^2}$ is the lower bond.

We have only had the upper part of the cone that is $z = +\sqrt{x^2 + y^2}$.

if $\phi$ is representing the lower part of the cone, then the correct option comes out to be $1$.

## Numerical Result

The correct option is the option no. $1$ that is:

• The upper half of the right circular cone with vertex at the origin and axis at the positive $z$ axis.

## Example

An equation for a surface is given, elaborate it in verbal context: $\phi = \dfrac{\pi}{3}$.

Spherical Coordinates are  $\phi = \dfrac{\pi}{3}$:

$cos\phi = cos \left( \dfrac{\pi}{3}\right) = \dfrac{1}{2} \hspace{3ex} … (1)$

$x = \rho sin\phi cos\theta$

$cos^2 \phi = \dfrac{1}{4} \hspace{3ex} … (2)$

$y = \rho sin\phi sin\theta$

$\rho^2cos^2\theta = \dfrac{1}{4} \rho^2 \hspace{3ex} … (3)$

$z^2 = \dfrac{1}{4}(x^2 + y^2 + z^2) \hspace{3ex} … (4)$

$x^2 + y^2 + z^2 = \rho^2$

$4z^2 = x^2 + y^2 + z^2$

$3z^2 = x^2 + y^2$

so $3z^2 = x^2 + y^2$ is a double cone.
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