May I know what is the Implicit equation to define the upper bound of a circle of radius 1?
Is it $ y^2 - \sqrt{1-x^2} = 0$? and for the lower bound, $ y^2 + \sqrt{1-x^2} = 0$?
What is the implicit equation of the upper bound of an ellipse of radius 0.3 then?
The implicit equation of a circle of radius $1$ centered at the origin is $x^2+y^2=1$, so you can find: $y^2=1-x^2$ and the upper esmicircle ($y\ge 0$) is given by: $y=\sqrt{1-x^2}$ ( that you can write as $ y-\sqrt{1-x^2}=0$ if you want). The other semicircle ($y\le 0$) is $y=-\sqrt{1-x^2}$.
For an ellipse with center in the origin and symmetry axes parallel to the coordinate axis you can start from the equation $$ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 $$ where $a$ and $b$ are the semiaxis, and, in the same way, you find $$ y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}=\pm \dfrac{b}{a}\sqrt{a^2-x^2} $$ that with the $+$ sign represent the upper ($y\ge 0$) semiellipse.
Finally, note has no sense to talk of an ellipse of radius $0.3$