What is a Suitable Transformation for a Double Integral on a Parallelogram?

Jacobian as the appropriate factor in your integration. This will change the limits into ones that are easy to use.
  • #1
vintwc
24
0

Homework Statement


Let Ω ⊂ R^2 be the parallelogram with vertices at (1,0), (3,-1), (4,0) and (2,1). Evaluate ∫∫_Ω e^x dxdy.

Hint: It may be helpful to transform the integral by a suitable (affine) linear change of variables.

Homework Equations


The Attempt at a Solution


Ok here is what I have done:
From the sketch of the parallelogram, I have found the limits to be x=y+1 to (4-2y) and y=-1 to 1. With this, I am able to determine the solution of the integral which is 3/2*e^2 -1 -1/2*e^6.

Could anyone please verify this for me? Also, if my solution turns out to be right, how would I approach the hint to find the solution of this double integral? Thanks.
 
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  • #2
vintwc said:

Homework Statement


Let Ω ⊂ R^2 be the parallelogram with vertices at (1,0), (3,-1), (4,0) and (2,1). Evaluate ∫∫_Ω e^x dxdy.

Hint: It may be helpful to transform the integral by a suitable (affine) linear change of variables.


Homework Equations





The Attempt at a Solution


Ok here is what I have done:
From the sketch of the parallelogram, I have found the limits to be x=y+1 to (4-2y) and y=-1 to 1. With this, I am able to determine the solution of the integral which is 3/2*e^2 -1 -1/2*e^6.

Could anyone please verify this for me? Also, if my solution turns out to be right, how would I approach the hint to find the solution of this double integral? Thanks.

The sides of your parallogram are not parallel to the coordinate axes and you would have to break up the integrals into two parts; you can't go from y = -1 to 1 in one swoop. The sides of your parallelogram are of the form x - y = constants and x + 2y = constants. So if you want to transform your problem into one with constant limits you might want to consider the transformation:

u = x - y
v = x + 2y
 

Related to What is a Suitable Transformation for a Double Integral on a Parallelogram?

1. What is a double integral parallelogram?

A double integral parallelogram is a mathematical concept used in multivariable calculus to calculate the volume under a surface. It is a two-dimensional shape formed by two intersecting lines, and is used to represent the limits of integration for a double integral.

2. How is a double integral parallelogram different from a regular parallelogram?

A double integral parallelogram is a mathematical concept used in calculus, while a regular parallelogram is a two-dimensional shape commonly used in geometry. The main difference is that a double integral parallelogram represents the limits of integration for a double integral, while a regular parallelogram is used to calculate area and perimeter.

3. What is the purpose of using a double integral parallelogram?

The purpose of using a double integral parallelogram is to determine the volume under a surface in three-dimensional space. It is a key concept in multivariable calculus and is used to solve various real-world problems in physics, engineering, and economics.

4. How is a double integral parallelogram calculated?

To calculate a double integral parallelogram, the limits of integration for the two variables must be determined. This is usually done by setting up a coordinate system and finding the intersection points of the two lines that form the parallelogram. The integral is then evaluated using these limits of integration.

5. Can a double integral parallelogram have different shapes?

Yes, a double integral parallelogram can have different shapes depending on the function being integrated and the limits of integration. It can be a rectangle, a trapezoid, or any other shape formed by two intersecting lines. The important thing is that it represents the limits of integration for the double integral.

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