Having difficulty deciding limits of integration for a joint to marginal pdf A joint pdf, $f_{X,Y}(x,y)=5$, is given with the following intervals:
$-1<x<1$
$x^2<y<x^2+{1\over{10}}$
I am trying to find marginal pdf of $f_Y(y)$ but I am stuck. Trying for hours.Help would be appreciated.

 A: The joint density can be expressed as $5$ times indicator functions:

 $$f_{X,Y}(x,y) = 5\times\mathbb{I}_{-1<x<1}\times\mathbb{I}_{x^2<y<x^2+1/10}$$ 

The marginal density of $Y$, $f_Y(y)$, is therefore the integral of the joint $f_{X,Y}(x,y)$ in $x$ over the conditional integration domain:

 $$\{x; -1<x<1,\ y-1/10<x^2<y\}$$

which can also be expressed as

 $$\{x; \max(-1,-\sqrt{y})<x<-\sqrt{\max(0,y-1/10)}\}\cup\{x; \min(1,\sqrt{y})>x>\sqrt{\max(0,y-1/10)}\}$$

with the marginal support of $Y$ obtained by the extreme values of its conditional domain:

 $$y>\min_{-1<x<1}x^2=0\quad\text{ and }\quad y<\max_{-1<x<1}x^2+1/10=11/10$$

meaning that$$f_Y(y)=\int_{\{x;\ f_{X,Y}(x,y)>0\}}f_{X,Y}(x,y)\,\text{d}x$$is given by [using the fact that $f_{X,Y}(x,y)=f_{X,Y}(-x,y)$ is symmetric]

 $$f_Y(y)=2\int_{\max(0,y-1/10)^{1/2}}^{\min(1,\sqrt{y})} 5\,\text{d}x$$

A: Each particular value of $y$ corresponds to a horizontal line.
If $0.1 < y < 1,$ then as you go from left to right, that line passes through the red curve, then the blue curve, then the blue curve again, then the red curve again, so you get
$$
f_Y(y) = \int_{-\sqrt y}^{-\sqrt{y-0.1}} 5 \, dx + \int_{\sqrt{y-0.1}}^{\sqrt y} 5 \, dx \quad \text{if } 0.1<y<1.
$$
If $0\le y \le 0.1$ then the horizontal line passes through the red curve twice, so you get
$$
f_Y(y) = \int_{-\sqrt y}^{+\sqrt y} 5\, dx \quad \text{if } 0\le y\le 0.1.
$$
If $1\le y\le 1.1$ then the horizontal line passes through the blue curve twice, so you get
$$
f_Y(y) = \int_{-1}^{-\sqrt{y-0.1}} 5\,dx + \int_{\sqrt{y - 0.1}}^1 5\, dx \quad \text{if } 1\le y\le1.1.
$$
