How do you convert \[{x^2} + {y^2} = 4y\] to polar form?

Hint: The given question belongs to the polar geometry of the equation. We can represent any Cartesian form of equation into polar form. The polar form is another method of representing the coordinate in space. The main difference between polar form and Cartesian form is that, the polar form has an angle to represent with $x - y$ plane.in polar coordinate we do not have to move in a straight line like in a Cartesian form of $x - y$ coordinate. We know that, we represent the Cartesian coordinates $\left( {x,y} \right)$ to polar coordinate $\left( {r,\theta } \right)$. We will use basic trigonometric ratios identities to convert the given equation into polar form. The Cartesian coordinates are also called rectangular coordinates.

Complete step by step solution:
Step: 1 the given Cartesian equation is,
\[{x^2} + {y^2} = 4y\]
Let us assume that,
$
  x = r\cos \theta \ldots \ldots \ldots \ldots 1 \\
  y = r\sin \theta \ldots \ldots \ldots \ldots 2 \\
 $
Now we have to find the value of the angle with the $x - y$ axis.
To find the angle, divide the second equation by the first equation.
$ \Rightarrow \dfrac{y}{x} = \dfrac{{r\sin \theta }}{{r\cos \theta }}$
Cancel the same term from denominator and numerator of the equation.
$ \Rightarrow \dfrac{y}{x} = \dfrac{{\sin \theta }}{{\cos \theta }}$
We know that, in polar coordinate, $\tan \theta = \dfrac{y}{x}$ . So substitute the value of $\theta $ in the given equation.
$ \Rightarrow \theta = \arctan \left( {\dfrac{y}{x}} \right)$
Step: 2 to find the given Cartesian form of the equation into polar form, substitute the value of $x$ and $y$ in the given Cartesian equation.
So the given Cartesian form of equation is,
\[{x^2} + {y^2} = 4y\]
Hence substitute the value of $x = r\cos \theta $ and $y = r\sin \theta $ in the given equation.
${r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 4r\sin \theta $
Simplify the trigonometric equation by using the formula of ${\sin ^2}\theta + {\cos ^2}\theta = 1$ .
$
   \Rightarrow {r^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 4r\sin \theta \\
   \Rightarrow {r^2} \times 1 = 4r\sin \theta \\
 $
Cancel the term $r$ which is present on both sides of the equation.
$r = 4\sin \theta $

Final Answer:
Therefore the polar form of the given equation is $r = 4\sin \theta $.

Note:
Use the basic trigonometric identities ratio formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to solve the equation. Students are advised to remember the coordinate in the polar form, that is $\left( {r,\theta } \right)$. They should understand the importance of polar coordinate. Polar coordinates are widely used nowadays in most of the industry.