Changing the variable in multiple integrals

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Homework Help Overview

The problem involves evaluating a double integral of the function \((x-y)^2 \sin^2(x+y)\) over a specified square region defined by its vertices. The subject area pertains to multiple integrals and variable transformations in calculus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply a transformation to simplify the integral and calculates the Jacobian. They express uncertainty about the new limits of integration and the correctness of their approach. Some participants question the accuracy of the Jacobian factor used in the transformation.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's method. There is a recognition of a potential error in the Jacobian, but no consensus has been reached regarding the overall correctness of the approach.

Contextual Notes

Participants are navigating the complexities of variable transformations and the implications for the limits of integration, with some expressing doubts about their calculations and assumptions.

kidsmoker
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Homework Statement



Evaluate

[tex]\int\int(x-y)^2sin^2(x+y)dxdy[/tex]

taken over a square with successive vertices (pi,0), (2pi,pi), (pi,2pi), (0,pi).


Homework Equations



[tex]I = \int\int_{K} f(x,y)dxdy = \int\int_{K'} g(u,v)*J*dudv[/tex]

where J is the Jacobian.

The Attempt at a Solution



Okay so I've just been learning this for the first time, so I may be doing it completely wrong!

I used the transformations u=x-y, v=x+y which give the Jacobian as 2.

Now i wasn't sure how to get the new limits for the integrals. What I did was apply the transformation above to give new vertices:

(pi,0) -> (pi,pi)
(0,pi) -> (-pi,pi)
(pi,2pi) -> (-pi,3pi)
(2pi,pi) -> (pi,3pi)

This gives a simple rectangle, so then i just wrote

[tex]I = 2*\int^{3\pi}_{\pi}\int^{\pi}_{-\pi}u^2sin^2(v)dudv = \frac{4\pi^{4}}{3}.[/tex]

I wish this was right, but I've a feeling it's not :-(

Any help greatly appreciated!
 
Last edited:
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The rectangle looks ok. But haven't you got the jacobian factor upside down?
 
Ah yeah, should be 1/2. Other than that though does my method look correct?

Thanks.
 
kidsmoker said:
Ah yeah, should be 1/2. Other than that though does my method look correct?

Thanks.

Looks ok to me.
 

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