Double Integration: Can I finish?

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Homework Help Overview

The problem involves evaluating a double integral by making an appropriate change of variables. The integral is defined over a specific rectangular region determined by linear equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to change variables to simplify the integral, selecting u and v based on the sums and differences of x and y. They express concern about the order of integration and whether they can proceed with their current setup.

Discussion Status

Some participants provide guidance on recognizing patterns in the integral and suggest using the Fundamental Theorems of Calculus to continue. The original poster indicates progress in solving the problem, but there is no explicit consensus on the best approach moving forward.

Contextual Notes

The original poster notes a realization about the order of integration, which may affect their approach. There is also a mention of a completed solution, but details on the correctness or validity of that solution are not discussed.

Char. Limit
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Homework Statement


Evaluate the integral by making an appropriate change of variable: \int \int_R (x + y) e^{x^2 - y^2} dA, where R is the rectangle enclosed by the lines x-y=0, x-y=2, x+y=0, and x+y=3


Homework Equations


If x=x(u,v) and y=y(u,v), then \int \int f(x,y) dx dy = \int \int g(u,v) |J(u,v)| du dv, where J(u,v) is the Jacobian determinant of x and y wrt u and v.


The Attempt at a Solution



So, I begin by selecting my u and v as u=x+y and v=x-y, which gives my bounds as a nice rectangle: 0<v<2 and 0<u<3. With that, I get my new function as:

\int_0^2 \int_0^3 u e^{u v} |J(u,v)| du dv

Finding the Jacobian, I get it to be -1/2, and thus |J(u,v)| = 1/2. Okay, so far so good. I take my first integration:

\frac{1}{2} \int_0^2 \left(\frac{u}{v} e^{u v} - \frac{1}{v^2} e^{u v}\right)|_0^3 dv = \frac{1}{2} \int_0^2 \frac{3 e^{3v}}{v} - \frac{e^{3v}}{v^2} + \frac{1}{v^2} dv

Now, at this point I realized I should have changed the order of integration. However, I'm wondering if I can still find this integral. The first two terms look suspiciously like the derivative of e^(3v)/v, after all... so my question is, is it still easy to continue from here, or should I just start all over and change the order of integration?
 
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Yes, the first two terms are the derivative of \displaystyle\frac{e^{3v}}{v}. Rewrite them as such and simply use the Fundamental Theorems of Calculus. It is find the way it is.
 
Excellent! I finished the second half of the problem and got (e^6 - 7)/4 as my answer. Danke!
 
From your sig:
Char. Limit said:
I banned Micromass.
Yeah, right...:smile:
 
Mark44 said:
From your sig:

Yeah, right...:smile:

But it's true! Albeit indirectly and not from PF.
 

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