**\[ \phi = \dfrac{\pi}{4} \]**

**Choose the correct answer:**

**– 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.

## Expert Answer

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.**