The aim of this question is to find the surface corresponding to the **Spherical Coordinates** $p=sin\theta sin\phi$ by utilizing the **Cartesian Coordinate system** and **Equation of Sphere**.

First, we will explain the concept of **Sphere**, its **Equation**, and its **Coordinates in the Cartesian Coordinate System**.

A** Sphere** is defined as a $3D$ geometrical structure have a constant radius $\rho$ across all three dimensions and its center point is fixed. Therefore, the **equation of sphere** is being derived by considering the position coordinates of sphere centers with their constant radius $\rho$

\[{(x-a)}^2+{(y-b)}^2+{(z-c)}^2= \rho^2\]

This is the **Equation of Sphere** where

$Center = A(a, b, c)$

$Radius = \rho$

For a **Standard Sphere** in standard form, we know that the center has coordinates as $O(0,0,0)$ with $P(x,y,z)$ being any point on the sphere.

\[A(a, b, c) = O(0, 0, 0)\]

By substituting the coordinates of the center in the above equation we get:

\[{(x-0)}^2+{(y-0)}^2+{(z-0)}^2= \rho^2\]

\[x^2+y^2+z^2= \rho^2\]

In **Cartesian Coordinate System**, we **convert** the equation given in **spherical coordinates** to **rectangular coordinates** to identify its surface.

In physics, $\theta$ is defined as the **Polar Angle** (from the positive z-axis) and $\phi$ is defined as the **Azimuthal Angle**. By utilizing the concept of **spherical coordinates**, we know that a sphere having a radius is defined by **3 coordinates**

\[x=\rho\ sin\theta\ cos\phi\]

\[y=\rho\ sin\theta\ sin\phi\]

\[z=\rho\ cos\theta\]

## Expert Answer

Given As:

\[p= sin\theta\ sin\phi\]

By multiplying both sides with $\rho$, we get

\[\rho^2= \rho\ sin\theta\ sin\phi\]

As we know as per the **Cartesian Coordinate System**

\[y= \rho\ sin\theta\ sin\phi\]

Hence,

\[\rho^2=y\]

By substituting the value of $\rho^2$ in the **Equation of Sphere**, we get:

\[x^2+y^2+z^2 = y\]

\[x^2+y^2-y+z^2 = 0\]

Adding $\dfrac{1}{4}$ on both sides:

\[x^2+{(y}^2-y+\dfrac{1}{4})+z^2 = \dfrac{1}{4}\]

As we know that:

\[y^2-y+\dfrac{1}{4} = {(y-\dfrac{1}{2})}^2\]

By substituting the value in above equation

\[{(x-0)}^2+{(y-\dfrac{1}{2})}^2+{(z-0)}^2 = {(\dfrac{1}{2})}^2\]

By comparing it with the **equation of sphere**

\[{(x-a)}^2+{(y-b)}^2+{(z-c)}^2 = \rho^2\]

We get the coordinates for the **center of the sphere** and **radius** $\rho$ as follows:

\[Center\ A(a, b, c)=A(0, \dfrac{1}{2}, 0)\]

\[Radius\ \rho= \dfrac{1}{2}\]

## Numerical Result

The surface that corresponds to $p=sin\theta sin\phi$ is a **Sphere** with $Center\ A(a, b, c)=A(0, \dfrac{1}{2}, 0)$ and $Radius\ \rho=\dfrac{1}{2}$.

Figure 1

## Example

Identify the surface whose equation is given as $r = 2sin\theta$

We know that:

**Cylindrical Coordinates** $(r,\theta,z)$ with **Center** $A(a, b)$ are represented by equation:

\[{(x-a)}^2+{(y-b)}^2 = r^2\]

\[\tan{\theta = \dfrac{y}{x}}\]

\[z=z\]

Where:

\[x= rcos\theta\]

\[y= rsin\theta\]

Given that:

\[r= 2sin\theta\]

\[r^2=4\sin^2\theta\]

\[r^2=2sin\theta\times2sin\theta=2sin\theta\times \ r=2rsin\theta\]

Substituting the value of $y=rsin\theta$, we get

\[r^2=2y\]

Putting the value in the equation of **Cylindrical Coordinates**, we get

\[x^2+y^2=2y\]

\[x^2+y^2-2y=0\]

Adding $1$ on both sides

\[x^2+(y^2-2y+1)=1\]

\[x^2+(y^2-2y+1)=1\]

As we know that:

\[y^2-2y+1={(y-1)}^2\]

By substituting the value in the above equation

\[{(x-0)}^2+{(y-1)}^2=1\]

We get the coordinates for the **center of the circle** and **radius** $r$ as follows:

\[Center\ A(a,b)=A(0,1)\]

\[Radius\ r=1\]

Hence, the surface that corresponds to $r=2sin\theta$ is a circle with $Center\ A(a,b)=A(0,1)$ and $Radius\ r=1$.

Figure 2

*Image/Mathematical drawings are created in Geogebra.*