Calculating Double Integrals over Two-Dimensional Sets

[RJLiberator]

Homework Statement

For every two-dimensional set C contained in R^2 for which the integral exists, let $Q(C) = \int \int_C (x^2+y^2) dxdy$. If $C_1 = [{(x,y) : -1 ≤ x = y ≤ 1}],$ find Q(c).

Homework Equations

The Attempt at a Solution

This was a tougher one for me (the other 2 on this part were easy).
So we have the condition that x must = y.

Now, would the double integral become:

$Q(C) = \int (2x^2) dx$ ?

Where the bounds are from -1 to 1, thus resulting in an answer of 4/3.

Or am I going about this wrong?

 

[Charles Link]

I think you only evaluated the first term of the integral and the answer should be twice that. Perhaps someone else will concur with me. $\\$ I think the condition $x=y$ really just meant they both have the same interval requirements for the boundaries.

 

[RJLiberator]

Hm I do not think that they both have the same interval for boundaries.
This is part of a 3 part question and it is part 2. Part 1 actually has the same set up with -1 ≤ x ≤1, -1 ≤ y ≤ 1.

That would mean part 2 has to have x = y otherwise they are the same question.

For part 1 I did end up getting 8/3 which is (as you said) 2 * the answer.

 

[LCKurtz]

I think you only evaluated the first term of the integral and the answer should be twice that. Perhaps someone else will concur with me. $\\$ I think the condition $x=y$ really just meant they both have the same interval requirements for the boundaries.

Wouldn't that be written $-1\le x,y\le 1$? The definition of the region $C_1$ in this question needs to be clarified. I would gather from what is written that the region is just the points on the straight line $y=x$ for $-1\le x \le 1$.

 

[LCKurtz]

@RJLiberator: So what do you think the answer is?

 

[Charles Link]

Wouldn't that be written $-1\le x,y\le 1$? The definition of the region $C_1$ in this question needs to be clarified. I would gather from what is written that the region is just the points on the straight line $y=x$ for $-1\le x \le 1$.

It is a double integral, so that should mean the area of integration would be finite, instead of having zero area. The statement of the problem isn't completely clear.

 

[RJLiberator]

Since x must equal y and it's from -1 to 1 we have a line as you described.
We'd do the integral from -1 to 1 of 2x^2 and get 4/3, correct?

Although we need an area here. Under this logic, would it be 0?

 

[Stephen Tashi]

It is a double integral, so that should mean the area of integration would be finite, instead of having zero area. The statement of the problem isn't completely clear.

The point of the exercise may be to show what happens when we evaluate the double integral of a function over a line segment. If we take the problem at face value, that's what it's asking us to do.

 

[LCKurtz]

@RJLiberator: If $\epsilon > 0$ is small, think about a very narrow strip of area $A$ about the line $y=x$, so narrow that its area is $\epsilon$. How big could $\iint_A x^2+y^2~dydx$ be? How would that compare with your integral?

 

[Charles Link]

The point of the exercise may be to show what happens when we evaluate the double integral of a function over a line segment. If we take the problem at face value, that's what it's asking us to do.

If that is indeed what it is asking, the answer is zero, and no evaluation is necessary. If on the other hand, it is the square region, with $-1<x<1$ and $-1<y<1$, then I believe the OP was off by a factor of 2.

 

[Stephen Tashi]

Since x must equal y and it's from -1 to 1 we have a line as you described.

LCKurtz gave a good suggestion.

You can also approach the problem formally. Think about the general procedure used to evaluate double integrals. When it's convenient, we can do them as $\int_{y0}^{y1} ( \int_{x0=g(y)}^{x1= h(y)} f(x,y) dx) dy$. You are given $f(x,y) = x^2 + y^2$. You need to determine the functions $h(y)$ and $g(y)$ that give the upper and lower limits for $x$ when $y$ has a given value.

 

[RJLiberator]

LCKurtz gave a good suggestion.

You can also approach the problem formally. Think about the general procedure used to evaluate double integrals. When it's convenient, we can do them as $\int_{y0}^{y1} ( \int_{x0=g(y)}^{x1= h(y)} f(x,y) dx) dy$. You are given $f(x,y) = x^2 + y^2$. You need to determine the functions $h(y)$ and $g(y)$ that give the upper and lower limits for $x$ when $y$ has a given value.

Sorry guys, I am having a troubling time understanding this problem still. The problem is due (and i'll have the solution) in about an hour. So, I'm just going to try once more here then if I get the solution I'll post the proper answer here.

So, we know the domain is from -1 to 1.
The range is a function 2x^2 as x=y and we have the function x^2+y^2 = x^2+x^2 = 2x^2.

The actual graph is just the parabola 2x^2 from -1 to 1 but it's only the line and not the area underneath it that we are concerned about since y must equal x.

 

[Stephen Tashi]

Sorry guys, I am having a troubling time understanding this problem still.

If you do the problem the formal way, you will have an expression with an integral of the form $\int_{x=y}^{x=y} f(x,y) dx$ inside it. This integral is equal to zero - just as any integral with equal lower and upper limits is zero (e.g. $\int_{17}^{17} x^2 dx = 0$)

 

[RJLiberator]

AH, so I was on to it when the range is a function 2x^2, but inputting the domain into this function results in x(-1) = 2 and x(1) = 2 thus the upper and lower limits = 2 and the integral is equal to zero.
Nice!

 

[Stephen Tashi]

AH, so I was on to it when the range is a function 2x^2, but inputting the domain into this function results in x(-1) = 2 and x(1) = 2 thus the upper and lower limits = 2 and the integral is equal to zero.
Nice!

No, the upper and lower limits of integration do not depend on the function f(x,y). They only depend on the area over which you are integrating. The problem asks you integrate over an "area" that is a line segment. For a given value of y, you are integrating from x = y to x = y. Visualizing the shape of the function x^2 + y^2 as a function of a location on the line has nothing to do with the fact that the "area" of a line is zero.

 

[Stephen Tashi]

Just think about the "normal" way you do a double integral problem. It asks you to integrate some function f(x,y) over some area. What makes such a problem tricky? It's usually the shape of the area that makes it tricky. You have to figure out equations that describe the area in order to get the upper and lower limits of integration. The equations describing the area have nothing to with the function f(x,y).

How would you integrate $f(x,y)$ over the area defined by $\{(x,y): -1 \le x \le 1, \ x-0.1 \le y \le x+0.1\}$?

This can be done as $\int_{-1}^{1} ( \int_{y=x-0.1}^{y=x+0.1} f(x,y) dy) dx$

The limits $y = x -0.1, y = x + 0.1$ don't depend on $f(x,y)$.

 

[LCKurtz]

Or, as I implied in post #9, you could notice the answer is so small it must be $0$.