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[tex]

\int\limits_{}^{} {\int\limits_R^{} {f(x,y)dA = \int\limits_{}^{} {\int\limits_{}^{} {f(x(u,v),y(u,v))\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv} } } }

[/tex]

I've got two examples to illustrate my problem.

[tex]

\int\limits_{}^{} {\int\limits_R^{} {x^2 dA} }

[/tex]

Where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36. I am given the transfromations x = 2u and y = 3v. The determination of the image of this transformation from the xy plane to the uv plane is easy because the original ellipse is one continuous curve, specified by one equation.

However I am having much more trouble with the following integral.

[tex]

\int\limits_{}^{} {\int\limits_R^{} {\left( {3x + 4y} \right)dA} }

[/tex]

Where R is the region bounded by the lines y = x, y = x - 2, y = -2x and y = 3 - 2x. I am given the transformations x = (1/3)(u+v) and y = (1/3)(v-2u).

The solution just computes the Jacobian(which is easy enough) but then writes 3x + 4y = (u+v) + (4/3)(v-2u) = (1/3)(7v-5u). Then S is the region bounded by the lines u = 0, (1/3)(v-2) = (1/3)(u+v) - 2 or u = 2, (1/3)(v-2u) = -(2/3)(u+v) or v = 0, and (1/3)(v-2u) = 3 - (2/3)(u+v) or v = 3. Thus [tex]\int\limits_{}^{} {\int\limits_R^{} {\left( {3x + 4y} \right)dA} } = \int\limits_0^3 {\int\limits_0^2 {\left( {\frac{1}{3}} \right)} \left( {7v - 5u} \right)} \left( {\frac{1}{3}dudv} \right) = \frac{{11}}{3}[/tex].

I understand the first part with 3x + 4y, its just to rewrite the integrand in terms of u and v. However, afetr that I don't really follow the solution. In each case I think they've just used the original y and x equations and subtituted in the u and v transformations for x and y. So I guess I can kind of follow that part. However, I am at a loss as to how, after performing those algebraic manipulations, the solution suddenly writes out the required integral.

I was under the impression that some kind of sketch and appropriate 'transformations' from x and y to u and v values were needed. For example say I had x = 2u and y = 4u. Suppose also that I am given 0 <= x <= 2. Then the 'range' of values of u would be 0 <= u <= 1. The solution seems to have skipped this particular step. Could this be because the appropriate range of u and v values can be determined after finding the equations for the uv plane?