Change of variables in integrals

1. Jun 21, 2005

Benny

Hi I'm having trouble with using change of variables.

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

I've got two examples to illustrate my problem.

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

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.

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

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 $$\int\limits_{}^{} {\int\limits_R^{} {\left( {3x + 4y} \right)dA} } = \int\limits_0^3 {\int\limits_0^2 {\left( {\frac{1}{3}} \right)} \left( {7v - 5u} \right)} \left( {\frac{1}{3}dudv} \right) = \frac{{11}}{3}$$.

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?

2. Jun 21, 2005

HallsofIvy

Staff Emeritus
Do you see the reason for that particular choice of transform? If you solve the equations for u and v you get u= x-y and v= v= 2x+y. Then u= 0 on y= x and u= 2 on y= x-2. v= 0 on y= -2x and v= 3 on y= 3- 2x. That's where you get the limits of integration (you might find it simpler to understand if the limits were written $$\int_{u=0}^{u=2}\int_{v=0}^{v=3}$$) and the whole point of the transformation: it maps the given parallelogram in the xy-system to a rectangle in the uv-system. In fact, if you weren't given the transformation you do the reverse of that: write the equations of the lines as x-y= 0, x-y= 2, y+2x= 0, and y+2x= 3 so you can see easily that you want u= x-y and v= y+ 2x.

Of course, with x= (1/3)(u+ v) and y= (1/3)(v- 2u), 3x+ 4y= u+ v+ (4/3)v- (8/3)u=
(7/3)v- (5/3)u as shown.

Finally, the Jacobian is the determinant of the matrix formed by taking the partial derivatives: xu= 1/3, xv= 1/3, yu= -2/3, and yv= 1/3. The Jacobian is 1/9+ 3/9= 1/3 so dxdy= (1/3)dudv.

3. Jun 21, 2005

Benny

Thanks for the help HallsofIvy. So with these types of problems there are various substitutions that can be made but the simplest or most convenient ones are chosen?

4. Jun 21, 2005

arildno

Correct; choosing the simplest/smartest substitution is a skill to be developed.
There are many valid ways of substituting, most of them are unhelpful.

5. Jun 21, 2005

Benny

Ok thanks. Well its definitely something that I'll need to put some work into if I'm to even get to a decent level of competency with this topic. I think once I get good grasp of the idea of the transformations I'll be able to tackle this topic effectively.

6. Jun 21, 2005

OlderDan

I drew a diagram that illustrates the transformation in your second problem. The area of integration in x-y is a parallelogram illustrated in the upper left in black, plotted in x-y coordinates. The square grid represents x-y area elements having dimensions 0.1 by 0.1. The corners of the parallologram are labeled in u-v coordinates, and there are little parallelograms with x-y dimensions mapped to 0.1 by 0.1 u-v coordinates. The area is divided into 20 x 30 = 600 little parallelograms. Although they do not line up well with the boundaries, each square x-y grid cell cleary encloses 3 u-v area cells. When you integrate over u-v, you are covering an area that is 2 by 3 in u-v for an area of 6, but the area in terms of x-y is only 1/3 of that, or 2. The factor of 1/3 that you get from evaluating the Jacobian compensates for integrating over a u-v area that is three times larger than the original x-y area.

The blue grid on the right has the same height and base as the original parallelogram, so it occupies exactly the same area on the page. The now rectangular u-v cells correspond to 200 x-y cells, and divide the 2 by 3 u-v region into 600 0.1 by 0.1 u-v cells. The red figure on the bottom shows this same u-v grid expanded to make the 0.1 by 0.1 u-v cells squares occupy the same page space as the 0.1 by 0.1 x-y cells in the original figure. The area of the red rectangle is three times the area of the original parallelogram.

Imagine an approximate numerical integration using these grids. There would be 200 values of 3x + 4y multiplied by the .01 area grid area to evaluate the integral in terms of x-y, but there would be 600 values of 3x(u,v) + 4y(u.v) multiplied by the .01 grid area to evaluate the integral in terms of u-v. To the extent that the value of 3x + 4y in each x-y cell matches the averge of the three values used for the three nearby u-v cells, the u-v integral would be 3 times more than the x-y integral were it not for the 1/3 scale factor that compensates for the additional cells.

Of course you don't want to go through a geometric comparison for every integral you transform. Not all geometries lend themselves to simple diagrams, but in all cases the Jacobian factor compensates for the scaling of the different areas covered in the different representations of the function wrt different variables.

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7. Jun 22, 2005

Benny

Thanks for the in depth response and diagram OlderDan. I'll have a further look into it at a later time.