# Drawing a Region of Integration

1. Nov 30, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data

http://img803.imageshack.us/img803/6397/skjermbilde20111130kl50.png [Broken]

3. The attempt at a solutionI seriously can't even figure out how this region is drawn. I think in the second integral I need y to range from -x to 3. Then the answer could conceivably be a). But, as is, I can't do this.

Last edited by a moderator: May 5, 2017
2. Nov 30, 2011

### I like Serena

Hi TranscendArcu!

Perhaps you can start with a drawing in which you mark the end points of the boundaries?
And if you have that, a couple of points in between?

If you switch the integrals around, the outer integral(s) will be with respect to y.
y can not be dependent on x then, since x does not exist anymore when you integrate y.
The outer integral should typically cover the largest possible range, while the inner integral sets a condition on what to integrate.

3. Nov 30, 2011

### TranscendArcu

Okay. Here's what I have: in the first quadrant, a triangle with vertices (0,0), (3,3), (0,3). This is the region I get when I map out the region described by the first integral. In the second integral I have x≤y≤3, which is exactly the same constraint on y I had before. If I let -3≤x≤0, in the first quadrant I have a square with vertices (0,0), (0,3), (-3,3), (-3,0). In the third quadrant I have another triangle with vertices (-3,-3), (0,0), (-3,0).

!!!

Is it c?

4. Nov 30, 2011

### I like Serena

Hmm, before saying yeah or nay, let's explore your analysis.

If -3≤x≤0, then you're not in the first quadrant...
Furthermore, if x=-3, can you recheck the bounds?

So you think it's c?
What is your analysis then of the region that c covers?

5. Nov 30, 2011

### TranscendArcu

I think I mistyped something: If I let -3≤x≤0, in the second quadrant I have a square with vertices (0,0), (0,3), (-3,3), (-3,0).

But I still think it's c, since my picture did actually have the square in Q2.

6. Nov 30, 2011

### I like Serena

All right!

So your problem gives you 2 triangles and a square.
And c gives you a square in Q2.

Are these the same figures?

7. Nov 30, 2011

### TranscendArcu

I think so. If I graph the bounds in c I seem to get the same region.

8. Nov 30, 2011

### I like Serena

And what about a, b, and d?

9. Nov 30, 2011

### TranscendArcu

a is a triangle in the first and second quadrants. b I found more tricky, but I think it's the region in (a) and also its reflection over the x-axis. Who knows what d is? As you said in #2, it doesn't make sense to have variables in the outermost integral's bounds.

10. Nov 30, 2011

### I like Serena

Sounds like you've got things down!

So?

11. Nov 30, 2011

### I like Serena

You still seem unsure...?
I was kind of hoping for a statement like: "I'm sure it is c!".

12. Nov 30, 2011

### TranscendArcu

Pick c! I was pretty sure it was c after I really mapped out the region being described. Just by inspection, c seemed to be the appropriate answer once graphed correctly. Still, it took me a minute or two to realize what region b described.

13. Nov 30, 2011

Good!