How to graph/visualize complicated inequalities I'm having trouble visualizing areas defined by for example,
$$
x^2 + y^2 \leq 2y
$$
Or
$$
(x^2 +y^2)^2 \leq 2a^2(x^2 - y^2)
$$
What is the thought process in picturing these regions?
 A: Here are some methods:


*

*If possible, express y explicitly in terms of x (or vice versa) on the line of equality. For instance in the first example write:
$$
x = \sqrt{y(2-y)}
$$
Plot this function (not forgetting the negative square root). It's a circle.

*Picture each side as a two dimensional surface and imagine where they intersect (and where one is higher than the other). Not always easy, but in the first example $x^2+y^2$ is a rotationally symmetric bowl and $2y$ is a sloping plane. They intersect in an ellipse.

*To help the visualisation process in method 2, set x to a series of constants and plot each side of the inequality as a function of y. Do the same for a series of y values. These plots are sections through the 3-dimensional landscape.

*Alternatively set each side to the same constant and plot the resulting contours.

*Shift terms from one side of the inequality to the other, or change the variables by shifting the origin or rotating, in an attempt to simplify the picture.
A: $x^2+y^2 \leq 2y \Rightarrow x^2+y^2 - 2y \leq 0 \Rightarrow x^2 + (y-1)^2 -1 \leq 0 \Rightarrow $
$\Rightarrow x^2 + (y-1)^2 \leq 1$
We have the interior and boundary of a circle with center in (0,1)
A: For $(x^2 +y^2)^2 \leq 2a^2(x^2 - y^2)$ I would immediately go to polar coordinates, and rewrite as
$$r^4\le 2a^2(r^2\cos^2\theta-r^2\sin^2\theta)=2a^2r^2\cos(2\theta).$$
Remembering that $r=0$ is the origin, we can almost rewrite the inequality as 
$r^2\le 2a^2\cos^2(2\theta)$. The polar curve is easy to draw, by just imagining $\theta$ growing, and following what happens to $2a^2\cos(2\theta)$. We are inside or on the curve.  
A: For the second, try polar coordinates: $x^2 + y^2 = r^2$ while $x^2 - y^2 = r^2 (\cos^2(\theta) - \sin^2(\theta)) = r^2 \cos(2\theta)$.  So the equation becomes
$$ (r/a)^2  \le 2 \cos(2\theta)$$
Note that the right side is positive for $-\pi/4 < \theta < \pi/4$ and for
$3\pi/4 < \theta < 5\pi/4$, negative for $\pi/4 < \theta < 3\pi/4$ and $5 \pi/4 < \theta < 7\pi/4$.  The region exists in the sectors where the right side is positive.  It looks like this:

The boundary curve is called a lemniscate.
A: Type in RegionPlot[x^2+y^2<=2y,{x,-1,1},{y,0,2}] link or RegionPlot[(x^2+y^2)^2<=2*2^2(x^2-y^2),{x,-3,3},{y,-3,3}] (using $a=2$) link to WolframAlpha or Wolfram Cloud.
