Change of Variables - Finding New Limits

• erok81
In summary, the student attempted to solve a homework problem but did not understand how to find new limits. He was able to use straightforward arithmetic to find new limits.
erok81

Homework Statement

Integrate the following over the set E.

$\int_E \frac{2x+y}{x+3y} dA$

Bounded by the lines:
y = −x/3+1
y = −x/3+2/3
y = −2x
y = −2x + 1

None.

The Attempt at a Solution

I can up to the same point everytime, but always get stuck on finding the new limits. Here is where I get to.

Let u=2x+y and v=x+3y

I take the Jacobian to find the distortion factor (or whatever you want to call it) and invert it. Please excuse my poor attempt at a matrix.

$\left| 2 \ 1 \right| \\ \left|1 \ 3 \right|$

This is 5, which inverted it 1/5. This gives me the following:

$\frac{1}{5} \int_E \frac{u}{v} du \ dv$

Now the part I don't get. My new limits for integration. On the solution page for this problem, there is another Jacobian presented that doens't seem to match anything, except slightly matching the new limits.

So back to the question at hand - how does one figure out those new limits?

And as a secondary question - how do you choose u and v? In this case it is easy. But in other cases can you just choose whatever you want (within reason of course) and run it through the same steps to get your new limits?

I've seen some examples where an easy shape is drawn onto x-y co-ords and then mapped onto the u v plane.

For example take a unit square located at the origin. (0,0), (0,1), (1,0), (1,1). Is a correct method then to put those values into u and v and get new points?

Solve for ##x## and ##y## in terms of ##u## and ##v## and then plug them into the expressions for the boundaries.

1 person
Okay, this should be right.

I have x=(3u-v)/5 and y=(2v-u)/5

But I think I am doing something wrong as this is just getting messier and messier the further I go.

Using the first boundary y=-(x/3)+1 and my two equations there I end up with v=(5-2u)/3.

Which I am pretty sure is wrong because in the answer the limits are single digit numbers...

Last edited:
erok81 said:
Okay, this should be right.

I have x=(3u-v)/5 and y=(2v-u)/5

But I think I am doing something wrong as this is just getting messier and messier the further I go.

You have u = 2x + y and v = x + 3y.

The given boundaries are these:
y = −x/3+1
y = −x/3+2/3
y = −2x
y = −2x + 1

There are straightforward arithmetic operations which will turn the first two into "v = x + 3y = ..." and the second two into "u = 2x + y = ...".

1 person
Ooooh yes. I see now. That is very straightforward. :)

Now I have:
2 ≤ v ≤ 3
0 ≤ u ≤ 1

I am guessing you choose which for v and which for u based on plotting the graph and seeing which corresponds to which axis? I tried swapping them and it wasn't working so well (I got results with x instead of values).

Now for some practice problems.

1. What is the purpose of change of variables in finding new limits?

The purpose of change of variables is to simplify the calculation of a limit by replacing the original variable with a new variable that makes the limit easier to evaluate.

2. How do you choose the new variable in change of variables?

The new variable should be chosen in a way that cancels out the complicated or problematic parts of the original expression in the limit. This can be done by using algebraic manipulations or by making substitutions based on known trigonometric identities.

3. Can change of variables be used for all types of limit problems?

No, change of variables is typically used for limits involving rational functions, trigonometric functions, or exponential functions. It may not be effective for limits involving complicated functions or indeterminate forms like 0/0 or ∞/∞.

4. Are there any limitations to using change of variables?

Yes, there are some limitations to using change of variables. It may not always be possible to find a suitable new variable that simplifies the limit. Additionally, the new variable may introduce new points of discontinuity, so it is important to check the validity of the new limit.

5. How can change of variables be applied in real-world problems?

Change of variables can be applied in real-world problems involving rates of change, optimization, or physical systems. For example, in physics, it can be used to transform the coordinates of a system to make solving equations of motion easier.

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