Drawing a Region of Integration

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    Drawing Integration
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Homework Help Overview

The discussion revolves around determining the correct region of integration for a given problem involving multiple integrals. Participants are analyzing the geometric representation of the regions defined by the integrals and their respective bounds.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to visualize the regions defined by the integrals, discussing the vertices of the shapes formed in different quadrants. There are questions about the correctness of the bounds and the implications of switching the order of integration.

Discussion Status

There is an ongoing exploration of the regions represented by different options (a, b, c, d), with some participants expressing confidence in their interpretations while others are questioning assumptions and seeking clarification on the relationships between the regions.

Contextual Notes

Participants are navigating potential misunderstandings regarding the placement of variables in the bounds of the integrals and the implications of those placements on the regions being analyzed.

TranscendArcu
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Homework Statement



http://img803.imageshack.us/img803/6397/skjermbilde20111130kl50.png

The Attempt at a Solution

I 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.
 
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Hi TranscendArcu! :smile: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.
 
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?
 
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?
 
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.
 
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?
 
I think so. If I graph the bounds in c I seem to get the same region.
 
And what about a, b, and d?
 
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
Sounds like you've got things down!

So?
 
  • #11
You still seem unsure...?
I was kind of hoping for a statement like: "I'm sure it is c!".
 
  • #12
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
Good! :smile:
 

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