Integral calculus -- regions

[Nicolaus]

Homework Statement

Find all values of a such that this set is a type II region (i.e. the bounds of x can be represented as functions of y, while the bounds of y are constant valued)
-1<y<0, Y<x<-y (union) 0<y<1, -y+a<x<y+a

Homework Equations

The Attempt at a Solution

I arrived at a being any value in the bound (-inf, 0}, since if a is a value greater than 0, then the y bounds (0<y<1) start to exist outside of the region bounded by -y+a and y+a. In other words, a must be negative or 0.

 

[haruspex]

Check a = - 1/2.

 

[Nicolaus]

I've tried this, and it appears to work. Maybe I'm missing something. What's your reasoning for this?

 

[Nicolaus]

Bump

 

[haruspex]

the y bounds (0<y<1) start to exist outside of the region bounded by -y+a and y+a

I don't understand that statement.
Call the regions A = {-1<=y<=0 & y<=x<=-y}, B = {0<=y<=1 & -y+a<=x<=y+a}. (Allowing equality won't affect integration.)
What do A and B look like, in words?
Where do they meet?
What happens if you flip the signs of a and x consistently throughout?
However, I'm not sure how to interpret the question. What sort of function of x is allowed? I originally was thinking only of 'nice' functions, so drew attention to a = -1/2. If anything is allowed, check a = +1/2.

 

[Nicolaus]

So Region A looks like a triangle with its top vertex at the origin (the entire triangle is below the horizontal x-axis and its y coordinates range from -1 to 0); this region is bounded by x=y to the left and x=-y to the right. I am asking what values of a are allowed such that the union of regions A and B is a type II region (i.e. the x bounds can be represented as functions of y, while the y bounds are constant values).
If a is >0, the region does not fall in the 0<=y<=1 bound because the linear functions shift up away from the origin, but it works for any negative a values, I think?

 

[haruspex]

If a is >0, the region does not fall in the 0<=y<=1 bound because the linear functions shift up away from the origin, but it works for any negative a values, I think?

Check what solutions there are for y=0 in region B.
Anyway, even if there appeared a gap in the range for y, there would still be a way to treat it as a contiguous range for integration purposes. If the lower bound is expressed as y=f(x) and the upper bound as y=g(x) then through the gap region specify f=g. If that's not allowed then this categorisation strikes me as somewhat arbitrary and pointless.