# Calc II Area Under Curve

1. Sep 11, 2009

### bizoid

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

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

2. Relevant equations

None

3. 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]$$\int$$ 2t2 * Sqrt( (t dx/dt)^2 + (t dy/dt)^2)

= [-1,1]$$\int$$ 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: Sep 12, 2009
2. Sep 11, 2009

### Dick

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. Sep 11, 2009

### bizoid

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. Sep 11, 2009

### Dick

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: Sep 11, 2009
5. Sep 11, 2009

### bizoid

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. Sep 11, 2009

### Dick

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. Sep 12, 2009

### HallsofIvy

Staff Emeritus
Are you sure its not "$-1\le x\le 1, -1\le y\le 1$"? Saying "$-1\le x\le 1, -1\le x\le 1$", that is just repeating the same thing, doesn't make sense to me. If it really is just $-1\le x\le 1$, that is an infinite strip and the integral does not exist.

Again, are you sure it is not $\{(x, y): -1\le x\le 1, -1\le y\le 1\}$? 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.

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!

8. Sep 12, 2009

### bizoid

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.