# Integration regions

1. Feb 1, 2015

### Nicolaus

1. The problem statement, all variables and given/known data
What type of region(s) do the following classify as?

2. Relevant equations

3. The attempt at a solution
I would classify D1 as both types; my reasoning is that by the definition of a convex polygon (i.e. all x,y in D1, the lie segment connecting x and y is entirely in D1), this therefore qualifies as both types.
For D2: Let c be a point in interval between x-value endpoints of region, cross section of D2@c, projected onto y-axis is closed interval and depends on c continuously. (This is true for an arbitrary c in between y-value endpoints), so would this be both types as well?

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2. Feb 1, 2015

### LCKurtz

So tell us what a region's "type" means. We aren't reading your book.

3. Feb 1, 2015

### Nicolaus

Type I: y varies as a function of x; x is bounded by 2 constants -> y(x) < y < y(x)2
Type II: x varies as a function of y; y is bounded by 2 constants -> x(y) < x < x(y)2

4. Feb 1, 2015

Bump

5. Feb 1, 2015

### LCKurtz

I would agree that both areas are both type 1 and type 2. In the right hand figure (D2) the lower boundary is not a function because of the vertical segment. But you can just leave the vertical segent out making a function of $x$ that has a finite jump. Still, area is continuous as a function of $c$ if I understand what you are saying.

6. Feb 2, 2015

### Nicolaus

Can you expand on the part about negating the vertical segment by making a function of x that has a finite jump? Does this still count as being both type I and II? If I integrate D2 (starting from the left), until I hit the line segment, then add another double integral that starts from that vertical line segment and continues to the right-most bound, will this still be considered both Type I and II (the fact that I partitioned the region into 2 segments of type I (or II) and added them)?

7. Feb 2, 2015

### LCKurtz

The fact that you have to break the regions into two parts to do the integration doesn't prevent it from being a type 1 or 2 region. You still have an upper and lower function. They are just two piece formulas.