Choosing and finding limits of integration

Click For Summary

Homework Help Overview

The discussion revolves around determining the limits of integration for a double integral over a specified triangular region defined by the points (-1,1), (-1,-2), and (3,-2). Participants are exploring how to set up the integral correctly based on the boundaries of the region.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the boundaries for the integral, with some suggesting the limits for x and y based on the vertices of the triangle. There is also a debate about whether to integrate with respect to x or y first and how to determine which approach might be easier.

Discussion Status

Some participants have provided guidance on writing the integral in terms of the correct variables and limits. There is an ongoing exploration of the implications of choosing different orders of integration, and participants are encouraged to visualize the region to aid in understanding.

Contextual Notes

There is mention of potential confusion regarding the slopes and boundaries, as well as the need for clarity in describing the region and the integration process. The discussion reflects a collaborative effort to clarify these points without reaching a definitive conclusion.

inner08
Messages
48
Reaction score
0

Homework Statement


For the given region R, find intR f(x) dA. The region has the following points:
(-1,1), (-1,-2) and (3,-2)


Homework Equations





The Attempt at a Solution



I'm having problems finding the boundaries for the integral. I know that we have:
-1<=x<=3 and -2<=y<=1.

I can also find the slope that it has and I found: y=-3/4x + 1/4.

From there, I don't know what to do. I know I'm supposed to choose if I want to integrate first, does it matter? or how can I know which one is easier?

I thought I could have something like: int(-1,3) int(1, -3/4x + 1/4) f(x) dA.

Hope someone can help me with this!
 
Physics news on Phys.org
It looks like you already chosen dy as the inner variable and dx as the outer. You should probably write f(x,y) for the function and dydx for dA. Now for your problem. Why do you think the lower limit for the dy integration is 1?
 
I just thought..the slope starts at y=1 and then it goes downwards with the -3/4x + 1/4 slope till it reaches the point (3,-2).
 
The lower limit in y is the lower y value connecting the points (-1,-2) and (3,-2). Did you draw a picture of the region? Always recommended.
 
So it would simply be -2. Yes I do have the region drawn.

So then I would have...

int(-1,3) int(-2, -3/4x + 1/4) f(x) dA.?
 
Yes. Now write f(x,y) for f(x) (it's not just a function of x) and dA=dydx (to indicate the order of integration).
 
I don't mean to sound dumb, but I'm not quite sure I know what you mean. Sorry. Maybe its just late :S
 
inner08 said:

Homework Statement


For the given region R, find intR f(x) dA. The region has the following points:
(-1,1), (-1,-2) and (3,-2)


Homework Equations





The Attempt at a Solution



I'm having problems finding the boundaries for the integral. I know that we have:
-1<=x<=3 and -2<=y<=1.

I can also find the slope that it has and I found: y=-3/4x + 1/4.
The slope of what? You haven't mentioned a line!

From there, I don't know what to do. I know I'm supposed to choose if I want to integrate first, does it matter?
Please, please, please, write what you really mean! You can't possibly mean you want to "choose if I want to integrate first"! What else can you do? You need to choose whether you want to integrate with respect to x or y first.
or how can I know which one is easier?
Try both ways and see! Here it doesn't really matter.

I thought I could have something like: int(-1,3) int(1, -3/4x + 1/4) f(x) dA.
Again, what you are writing makes no sense because you haven't said which variable you are using first! Write either int(x=-1, 3) int(y= 1, (-3/4)x+ 1/4) f(x)dA, or int(-1,3)int 1, -(3/4)x+ 1/4) f(x) dydx or, even better, both: int(x=-1, 3) int(y= 1, (-3/4)x+ 1/4) f(x)dydx.

Hope someone can help me with this!
First, explain what in the world you are talking about! You start by saying "The region has the following points: (-1,1), (-1,-2) and (3,-2)" Do you mean they are the vertices of a triangular region? If so, then you have a right triangle as your region. There are two different ways you can do this.

If you want to do the "dy" integral first, then you choose x-values as constants to cover the entire triangle: x will go from -1 to 3. Now, draw a vertical line through the region, representing the small rectangle you would use, for each x, in setting up a "Riemann sum" for the problem. Clearly, for all x, the lower boundary is on the line y= -1. Also, for all x, the upper boundary is on the line from (-1,1) to (3, -2) which is, as you say, y= (-3/4)x+ 1/4:
\int_{x=-1}^1 \int_{y= -1}^{(-3/4)x+ 1/4} f(x)dydx

If you want to do the "dx" integral first, you do just the opposite: first choose y-values as constants to cover the entire triangle: y will go from -2 to 1. Now draw a horizontal line representing the small rectangle you would use, for each y, in setting up the Riemann sum for the problem. Clearly, for all y, the left boundary is the line x= -1. Also, for all y, the right boundary is on the line from (-1,1) to (3, -2) which you can get by solving the equation above for x: x= (-4/3)y+ 1/3. The integral is:
\int_{y= -2}^1 \int_{x= -1}^{(-4/3)y+ 1/3} f(x)dxdy
 

Similar threads

Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
24
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K