Integrating Change Variables: Triangular Region

[zhuyilun]

Homework Statement

let f be continuous on [0,1] and let R be triangular region with vertices (0,0),(1,0) and (0,1) show that\int\intf(x+y) dA= \intu*f(u)du from 0 to 1

Homework Equations

The Attempt at a Solution

i don't know how to start this question. can anyone give me a hint to help me start this question?

 

[zhuyilun]

i think i didnt type the question correctly, its supposed to be a double integral of f(x+y) over the region R
and that value is equal to a single integral of u*f(u) du from u=0 to u=1

 

[Mark44]

From your two posts, I think this is what you're trying to prove:
\int \int_R f(x + y) dA~=~ \int_{u = 0}^1 u f(u) du
where f is continuous on [0, 1] and R is the triangular region defined by the points (0, 0), (1, 0), and (0, 1).

Here are some thoughts that might be of help to you. Assuming for the moment that f is positive on its domain, the double integral represents the volume of the region above the triangular region. Normally, to calculate a volume like this, we might use either vertical strips (vertical in relation to the x-axis) of width \Delta x and length 1 - x (since y = 1 - x), and height f(x). This would be our incremental volume, or \Delta V. To get the total volume we would integrate on y, as y runs from 0 to 1.

Alternatively, we might use horizontal strips (in relation to the x-axis) that are \Delta y by 1 - y by f(y) to get \Delta V. To get the total volume, we would integrate on x, as x runs from 0 to 1.

In this problem, I think that the idea is that they want you to use diagonal strips. Instead of working directly with x or y, I think you want to work with u = x + y. All of the line segments parallel to the hypotenuse of your triangle are such that at every point on the line segment, u is constant. For example, on the segment between (1, 0) and (0, 1), u = 1, where u = x + y. On the segment that joins (1/2, 0) and (0, 1/2), u = 1/2. And so on.

Using this line of reasoning, the incremental volume element has dimensions (length of diagonal from x-axis to y-axis) X (width of strip = \Delta u) X (height = f(u) = f(x + y)). It's not too big a jump to recognize that you can get the total volume by integrating the incremental volume element from u = 0 to u = 1. If you can convince yourself that the length of each of these diagonal strips is u, then you're pretty well on your way with this problem.

One other thing: I assumed for convenience that f > 0 so I could think about the double integral as a volume. You won't be able to make this assumption, since we don't know this about f.

 

[zhuyilun]

From your two posts, I think this is what you're trying to prove:
\int \int_R f(x + y) dA~=~ \int_{u = 0}^1 u f(u) du
where f is continuous on [0, 1] and R is the triangular region defined by the points (0, 0), (1, 0), and (0, 1).

Here are some thoughts that might be of help to you. Assuming for the moment that f is positive on its domain, the double integral represents the volume of the region above the triangular region. Normally, to calculate a volume like this, we might use either vertical strips (vertical in relation to the x-axis) of width \Delta x and length 1 - x (since y = 1 - x), and height f(x). This would be our incremental volume, or \Delta V. To get the total volume we would integrate on y, as y runs from 0 to 1.

Alternatively, we might use horizontal strips (in relation to the x-axis) that are \Delta y by 1 - y by f(y) to get \Delta V. To get the total volume, we would integrate on x, as x runs from 0 to 1.

In this problem, I think that the idea is that they want you to use diagonal strips. Instead of working directly with x or y, I think you want to work with u = x + y. All of the line segments parallel to the hypotenuse of your triangle are such that at every point on the line segment, u is constant. For example, on the segment between (1, 0) and (0, 1), u = 1, where u = x + y. On the segment that joins (1/2, 0) and (0, 1/2), u = 1/2. And so on.

Using this line of reasoning, the incremental volume element has dimensions (length of diagonal from x-axis to y-axis) X (width of strip = \Delta u) X (height = f(u) = f(x + y)). It's not too big a jump to recognize that you can get the total volume by integrating the incremental volume element from u = 0 to u = 1. If you can convince yourself that the length of each of these diagonal strips is u, then you're pretty well on your way with this problem.

One other thing: I assumed for convenience that f > 0 so I could think about the double integral as a volume. You won't be able to make this assumption, since we don't know this about f.

hi, i think i kinda get it. but i don't get where the extra u( infront of f(u)) come from

 

[Mark44]

I believe that the extra u comes from the diagonal length of the base of the approximating incremental volume elements. At a given value of u, say u = x + y = k, with 0 <= k <= 1, the diagonal line segment extends from (0, k) to (k, 0), and has a length of sqrt(2k^2) = k*sqrt(2) = (x + y) sqrt(2) = u sqrt(2).

Can you provide the exact wording of the problem, particularly what you need to prove?