Solving Change of Variables for Triangular Region

In summary, the conversation discusses showing the equality between two double integrals, one evaluated over a triangular region and the other over a function involving u as the variable. The triangular region has boundaries x=0, y=0, and x+y=1, and the problem requires finding a suitable u that satisfies these boundaries. The problem is incomplete as the limits of integration in the second integral are not specified.
  • #1
bodensee9
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I am wondering if someone could help me with the following? I am supposed to show that ∫∫f(x+y)dA evaluated from the triangular region with the vertices (0,0), (1,0) and (0,1) is equal to ∫∫uf(u)du.

This triangular region has the equations, x = 0, x = 1, and y = -x + 1. If I set x+y = u, then I know that u must satisfy the equations u = 1, u = 0. But, I’m not sure what to do after that? If someone could give me some hints, that be great. Thanks!
 
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  • #2
bodensee9 said:
I am wondering if someone could help me with the following? I am supposed to show that ∫∫f(x+y)dA evaluated from the triangular region with the vertices (0,0), (1,0) and (0,1) is equal to ∫∫uf(u)du.

This triangular region has the equations, x = 0, x = 1, and y = -x + 1. If I set x+y = u, then I know that u must satisfy the equations u = 1, u = 0. But, I’m not sure what to do after that? If someone could give me some hints, that be great. Thanks!
No, the triangular region has boundaries x= 0, y= 0, and x+ y= 1. The upper line, x+ y= 1 corresponds to u= 1 but neither x=0 nor y= 0 corresponds to u= 0. In any case, your problem is incomplete. ∫∫f(x+y)dA is a double integral. In ∫∫uf(u)du du is not an area differential. Did you mean show that it is equal to ∫∫uf(u)dudx? If so then the limits of integration are irrelevant. You need only show that dA= dydx= ududx. for some choice of u.
 

FAQ: Solving Change of Variables for Triangular Region

What is the purpose of solving change of variables for triangular region?

The purpose of solving change of variables for triangular region is to transform a given integral in a triangular region into an equivalent integral in a simpler region, such as a rectangle or a circle. This can make the integration process easier and more manageable.

What are the steps involved in solving change of variables for triangular region?

The steps involved in solving change of variables for triangular region are:
1. Identify the triangular region and determine the limits of integration.
2. Choose a suitable change of variables, which typically involves transforming the triangular region into a rectangle or a circle.
3. Apply the appropriate change of variables formula to transform the integral.
4. Solve the resulting integral in the new region.
5. Convert the final answer back to the original variables, if necessary.

Can any change of variables be used for solving triangular regions?

No, not all changes of variables can be used for solving triangular regions. The change of variables must be one-to-one and have a Jacobian determinant that is non-zero in the triangular region. In addition, the transformed region should be simpler to integrate in, such as a rectangle or a circle.

How can solving change of variables for triangular region be applied in real-life situations?

Solving change of variables for triangular region can be applied in various real-life situations, such as calculating the area of a triangular plot of land, finding the center of mass of a triangular-shaped object, or determining the volume of a triangular prism. It can also be used in more complex scenarios, such as in physics and engineering problems.

Are there any limitations or challenges when solving change of variables for triangular region?

One potential limitation is that not all integrals in triangular regions can be transformed into simpler regions using a change of variables. Some integrals may require specialized techniques or may not have closed-form solutions. Additionally, choosing the correct change of variables and setting up the integral can be challenging, especially for more complex triangular regions.

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