Understanding the Riemann Integral and Its Application to Area Under Curves

In summary: The integral over C1, C2, and C3 of x^2+ y^2? That can be done in polar coordinates, of course. Or in cartesian coordinates, in the first two cases. But I don't see anything to "summarize". I agree with Dick, you have not made it at all clear what the question is. Please try again.In summary, the question is asking to find the integrals of x^2 + y^2 over three two-dimensional sets, C1, C2, and C3. C1 is a square with limits -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1, C2 is a
  • #1
bizoid
8
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.
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
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  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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?
 

1. What is the concept of "area under curve" in Calculus II?

The area under curve is a fundamental concept in Calculus II that represents the total area enclosed by a continuous curve and the x-axis within a given interval. It is also known as the definite integral and is denoted by ∫f(x)dx. It is used to calculate the total change in a quantity over a specific interval.

2. How do you find the area under a curve using Calculus II?

To find the area under a curve using Calculus II, you need to first identify the function and the interval over which you want to find the area. Then, you need to integrate the function with respect to x over that interval. The resulting value is the area under the curve.

3. What is the significance of calculating the area under a curve in Calculus II?

The calculation of the area under a curve in Calculus II has many practical applications in fields such as physics, engineering, economics, and more. It allows us to determine the total change in a quantity, which can help in making predictions, analyzing data, and solving real-world problems. It is also a crucial tool for finding the volumes of irregular shapes and calculating probabilities in statistics.

4. Can the area under a curve be negative in Calculus II?

Yes, the area under a curve can be negative in Calculus II. This can happen when the function is below the x-axis within the given interval. In this case, the negative area represents a decrease in the total quantity being measured. However, when calculating the total area, the negative and positive areas are considered separately, and the final result is always positive.

5. What are some common methods for finding the area under a curve in Calculus II?

There are several methods for finding the area under a curve in Calculus II, such as the Riemann sum, integration by substitution, integration by parts, and using definite integrals. These methods involve different techniques and formulas, and the choice of method depends on the complexity of the function and the interval. It is essential to practice and understand each method to become proficient in finding the area under a curve in Calculus II.

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