Understanding the Riemann Integral and Its Application to Area Under Curves

Click For Summary

Homework Help Overview

The discussion revolves around the Riemann integral and its application to calculating areas under curves in a two-dimensional context. The original poster presents a problem involving three sets defined in R² and seeks to evaluate the integral of the function f(x,y) = x² + y² over these sets.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use iterated integrals and line integrals to evaluate the integral for the sets, particularly focusing on C2, which is defined as the line x=y. They express confusion regarding the area under this line segment and its relation to the area of the square defined in C1.
  • Some participants question the appropriateness of using a line integral for a one-dimensional set and clarify that the integral over a line segment results in zero area.
  • Others suggest reviewing the definition of a Riemann integral to understand why the area under the curve along the line x=y is considered zero.

Discussion Status

The discussion is ongoing, with participants providing clarifications and exploring the implications of the definitions involved. There is recognition of the misunderstanding regarding the nature of the sets and the integrals being evaluated, particularly the distinction between two-dimensional and one-dimensional integrals.

Contextual Notes

There are indications of confusion regarding the definitions of the sets C1, C2, and C3, particularly with respect to their dimensionality and the implications for integration. The original poster expresses uncertainty about the correctness of their interpretations and calculations, and there is a focus on understanding the foundational concepts of integration in this context.

bizoid
Messages
8
Reaction score
0

Homework Statement

The question verbatim from the text is as follows:

For every two-dimensional set C contained in R2 for which the integral exists, let Q(C) = ∫∫c x2 + y2 dxdy.

If C1 = {(x,y): -1 ≤ x ≤ 1, -1 ≤ y ≤ 1},
C2 = {(x,y): -1 ≤ x=y ≤ 1},
C3 = {(x,y): x2 + y2 ≤ 1},

Find Q(C1), Q(C2), Q(C3).

My post describes what I did for C2.

I used itterated integrals for C1 with an answer of 4/3, and
I switch to polar coordinates for C1 with an answer of 2pi

Homework Equations



None

The Attempt at a Solution



OK. So the domain appears to be the line x=y. Thus I will try a line integral.

Parametrize the curve x = t, thus y= t. So f(x,y) = f(t) = 2t2.

The Line integral states that I take the function * position of the curve

and we end up with

[-1,1][tex]\int[/tex] 2t2 * Sqrt( (t dx/dt)^2 + (t dy/dt)^2)

= [-1,1][tex]\int[/tex] 2t2 * Sqrt(2)

= (2Sqrt(2)*t3)/3 from [-1,1]

= 2Sqrt(2) / 3 + 2Sqrt(2) / 3

= 4Sqrt(2) / 3

My questions:
(1) I assume this is incorrect, because if C = {(x,y): -1 <= x <= 1, -1<= y <= 1} we get 4/3. And it seems unintuitive in this case that the entire area in a square region is less than the area along one path. Since we have a parabolic bowl shape.

Is this intuition wrong. Why?

(2) I understand the line integral to be the sum of the areas along each point in the curve's path. And I intuitively see how this works, if I have a non-constant curve. But in this case I have a scalar increasing the area. I see that the length of the curve is clearly 2Sqrt(2), from basic geometry, but it doesn't "feel" right. However, it is synonymous to the case of a circle cos^2 + sin^2 which will give an arc length of 1, noting that the radius is 1.

What am I missing, is it because I am parametrizing the function that changes the perspective?

(3) What is the correct answer? What other ways can this be solved.

Thank you in advance for reviewing my over analysis of a basic calc II question and any input that you provide.

Also, are there any good links to the PHP codes for all math symbols? This way my next question can be better formed.
 
Last edited:
Physics news on Phys.org
To say what the answer is, you are going to have to state what the question is a lot more clearly than that. My first guess is that you are trying to integrate the function f(x,y)=x^2+y^2 underneath the curve x=y in the square -1<=x<=1, -1<=y<=1. Could that be it? That wouldn't involve a line integral at all. And wouldn't have much to do with an 'area' either.
 
Sorry for the confusion.

The question verbatim from the text is as follows:

For every two-dimensional set C contained in R2 for which the integral exists, let Q(C) = ∫∫c x2 + y2 dxdy.

If C1 = {(x,y): -1 ≤ x ≤ 1, -1 ≤ x ≤ 1},
C2 = {(x,y): -1 ≤ x=y ≤ 1},
C3 = {(x,y): x2 + y2 ≤ 1},

Find C1, C2, C3.

My post describes what I did for C2. Maybe I just missed the boat all together?
 
That's a lot clearer. Thanks. C2 is actually the easiest of all. You are doing a two dimensional integral. Write it as an iterated integral. One of your integrals is, for example, integral from x to x of (x^2+y^2)*dy. That's zero since the two limits are the same. You are doing a two dimensional integral over a one dimensional set. You can't just change it into a line integral, that's a different problem. Do you see what I'm saying? The other two should be straightforward.
 
Last edited:
Sorry for being dense. It probably so easy that I missing it.

>> You are doing an integrated integral.

So my integrands are from [-1,1] ∫ [x,y]∫ x2 + y2 dxdy

>> One of your integrals is, for example, integral from x to x of (x^2+y^2)*dy.

x=y, so I have [-1,1] ∫ [x,x]∫ x2 + y2 dxdy

= [-1,1] ∫ y2 dy + [x,x]∫ x2 dx

>> That's zero since the two limits are the same.

= [-1,1] ∫ y2 + 0

= y3 / 3 from [-1,1]

= 2/3>> You are doing a two dimensional integral over a one dimensional set.

So what you are getting at is that...I am integrating from x to y, but there is no area since x=y. Oops I get it.

Is there an easier way to do this with parameterization?

>> You can't just change it into a line integral, that's a different problem.

When do I want to use the line integral then? Shouldn't a line integral work for any curve?

Thank you for the help. I really appreciate it.
 
No, a line integral is a different problem. If one of your iterated integrals is zero then the integral over the region is zero. The other iterated integral is just integrating 0. The answer is 0.
 
bizoid said:
Sorry for the confusion.

The question verbatim from the text is as follows:

For every two-dimensional set C contained in R2 for which the integral exists, let Q(C) = ∫∫c x2 + y2 dxdy.

If C1 = {(x,y): -1 ≤ x ≤ 1, -1 ≤ x ≤ 1},
Are you sure its not "[itex]-1\le x\le 1, -1\le y\le 1[/itex]"? Saying "[itex]-1\le x\le 1, -1\le x\le 1[/itex]", that is just repeating the same thing, doesn't make sense to me. If it really is just [itex]-1\le x\le 1[/itex], that is an infinite strip and the integral does not exist.

C2 = {(x,y): -1 ≤ x=y ≤ 1}[/itex]
Again, are you sure it is not [itex]\{(x, y): -1\le x\le 1, -1\le y\le 1\}[/itex]? that would be a square while what you have is a line segment and the integral of "dxdy" over a line segment is, as Dick said, 0.

C3 = {(x,y): x2 + y2 ≤ 1},

Find C1, C2, C3.
Finally, does the problem say "Find C1, C2, C3" or "Find Q(C1), Q(C2), Q(C3)", the integral of the function over those sets? If it is only to find or graph those sets, no integration is required!

My post describes what I did for C2. Maybe I just missed the boat all together?
 
I'm sorry the question does say find Q(C1), Q(C2), Q(C3).

It is from Introduction to Mathematical Statistics by Hogg, Mckean and Craig.(1) Sorry your C1 is correct {(x,y): -1 <= x <= 1, -1 <= y <= 1}. I missed typed.

(2) C2 is written correctly.

Sorry for being stupid. In regards to C2, how can the area under the curve be zero. Granted its not a 3D volume. But there should still be area under the curve f(x,y) = x^2 + y^2 along this line segment x=y.

What am I missing? I will pull out the old Stewart Calc book on Sunday and have a look through at what I am missing.

Thank you again for the comments. I appreciate your feedback.
 
You might review the definition of a Riemann integral, if you're not willing to trust the results of the iterated integral. The curve x=y has zero area. You can make a set of rectangles enclosing it of vanishingly small area. x^2+y^2 is bounded by 2. How can that sum to anything positive in the limit?
 

Similar threads

Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K