Help with change of variables in multivariable calculus/analysis

In summary: I see what you're saying. So then ##T^{-1}## would be the function that takes the unit square to the parallelogram. Hmmm, I'll have to think about that. Thanks for the clarification, I think it's all finally clicking now! In summary, the problem involves finding an affine map T that sends a parallelogram H to a unit square and computing the Jacobian of T to be 5. Then, the integral over H is converted to an integral over the unit square by using the inverse of T, T-1, which transforms both the region and the integrand. This is because T maps the unit square to the parallelogram, and the goal is to go from H
  • #1
lus1450
40
1

Homework Statement


Let ##H## be the parallelogram in ##\mathbb{R}^2## whose vertices are ##(1,1), (3,2), (4,5), (2,4).## Find the affine map ##T## which sense ##(0,0)## to ##(1,1), (1,0)## to ##(3,2), (0,1)## to ##(2,4)##. Show that ##J_T=5## (the Jacobian). Use ##T## to convert the integral
$$
\alpha = \int_H e^{x-y}dxdy
$$
to an integral over ##I^2## (unit square) and thus compute ##\alpha##.


Homework Equations





The Attempt at a Solution



So I'm not sure why I'm having some trouble with this. I've done everything up to solving the integral. The affine map I found is ##T(x,y) = (1,1) + (2x+y,x+3y)##, and I've checked that it is correct. However, in my mind, it makes sense that in changing the integral to be over the square, I take ##T^{-1}(H) = I^2##, that is, obtaining:
$$
\alpha = \int_H e^{x-y}dxdy = \int_{I^2} f(T^{-1}(x,y))|J_{T^{-1}}|dxdy
$$

However, the integral is supposed to be ##f(T(x,y))|J_T|##, not of ##T^{-1}##. Can someone explain why this is so? I think I'm just forgetting something from lower division calculus, as it's been a while. It made sense back then, but for some reason this problem is screwing me up. Thanks in advance.
 
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  • #2
Zaculus said:

Homework Statement


Let ##H## be the parallelogram in ##\mathbb{R}^2## whose vertices are ##(1,1), (3,2), (4,5), (2,4).## Find the affine map ##T## which sense ##(0,0)## to ##(1,1), (1,0)## to ##(3,2), (0,1)## to ##(2,4)##. Show that ##J_T=5## (the Jacobian). Use ##T## to convert the integral
$$
\alpha = \int_H e^{x-y}dxdy
$$
to an integral over ##I^2## (unit square) and thus compute ##\alpha##.


Homework Equations





The Attempt at a Solution



So I'm not sure why I'm having some trouble with this. I've done everything up to solving the integral. The affine map I found is ##T(x,y) = (1,1) + (2x+y,x+3y)##, and I've checked that it is correct. However, in my mind, it makes sense that in changing the integral to be over the square, I take ##T^{-1}(H) = I^2##, that is, obtaining:
$$
\alpha = \int_H e^{x-y}dxdy = \int_{I^2} f(T^{-1}(x,y))|J_{T^{-1}}|dxdy
$$

However, the integral is supposed to be ##f(T(x,y))|J_T|##, not of ##T^{-1}##. Can someone explain why this is so? I think I'm just forgetting something from lower division calculus, as it's been a while. It made sense back then, but for some reason this problem is screwing me up. Thanks in advance.
I'm pretty rusty on this stuff, but since no one else has answered, I'll take a shot at it. Your original integral is over the parallelogram H. You want the new integral to be over the unit square, so you need to transform both the H region and the integrand to get a new integral with the same value as the old. IOW, you want to apply your transformation T to both H and the function in the integrand. You don't want to apply T to the region and T-1 to the function's inputs.

The idea is similar to integration by substitution where you start with one integral:
$$\int_a^b f(x)dx$$

and use a substitution u = u(x), du = u'(x)dx
to rewrite the integral and the limits of integration in a form that is more easily solvable...
$$ \int_{u(a)}^{u(b)} f(u(x))du$$
 
  • #3
Thanks Mark44. I mean, I get you want to apply it to BOTH region and integrand, but I'm not sure why applying T to H would give me the integral over the unit square, as T is from the unit square to the parallelogram.
In my lower division calculus, when we learned about this, our transformations were always from the "harder" region to the "easier" region. However, in this case, our T maps the "easier" region (unit square) to the "harder" region (parallelogram H), and for some reason it's just not clicking in my head. I should be getting this if my past class is any , but I'm not sure why it isn't now. Perhaps I'm just recalling the definition incorrectly...

Oh a quick edit, on your comment about applying ##T## to the region and ##T^{-1}## to the function inputs, I applied ##T^{-1}## to both the region AND the inputs
 
Last edited:
  • #4
I misread (or actually, didn't read closely enough) your first post, thinking that T transformed H to the unit square.

Did applying T-1 to the region and the inputs work out for you? I would think that's what you need to do.
 
  • #5
Well the thing is, nothing went wrong per say, but there was a dispute in office hours about whether it was applying ##T## or ##T^{-1}## to the region/integrand, and the majority was on the side of applying ##T##, but it just didn't make sense to me as to why. So, applying ##T^{-1}## is correct then?
 
  • #6
You want the integral to be evaluated over the unit square. T sends the unit square to the parallelogram - you want to go the other way, from H to the unit square.

BTW, it's per se, Latin for "in itself".
 
  • #7
Zaculus said:
Well the thing is, nothing went wrong per say, but there was a dispute in office hours about whether it was applying ##T## or ##T^{-1}## to the region/integrand, and the majority was on the side of applying ##T##, but it just didn't make sense to me as to why. So, applying ##T^{-1}## is correct then?

Not sure which you mean by T, so I'll use R instead: that is, we can change variables from (x,y) to (u,v), where
[tex] \pmatrix{x\\y} = R \pmatrix{u\\v} [/tex]
Then
[tex] \alpha = \int_H f(x,y) \, dx \, dy = \int_{[0,1]^2} f(x(u,v),y(u,v))
\left| \frac{ \partial(x,y)}{\partial(u,v)}\right| \, du \, dv [/tex]
 
  • #8
Ray Vickson said:
Not sure which you mean by T, so I'll use R instead: that is, we can change variables from (x,y) to (u,v), where
[tex] \pmatrix{x\\y} = R \pmatrix{u\\v} [/tex]
Then
[tex] \alpha = \int_H f(x,y) \, dx \, dy = \int_{[0,1]^2} f(x(u,v),y(u,v))
\left| \frac{ \partial(x,y)}{\partial(u,v)}\right| \, du \, dv [/tex]

Alright, so then your R goes from H to the unit square?
 
  • #9
Zaculus said:
Alright, so then your R goes from H to the unit square?

The way I wrote it, R takes (u,v) into (x,y), so takes the unit square in (u,v)-space into the quadrilateral in (x,y)-space. Never mind using canned formulas; think about the concepts and meanings first.
 
  • #10
I think Zalculus is getting confused by his own notation. The parallelogram is given in ##(x,y)## space and is being transformed into the unit square in the ##(u,v)## space. The transformation that is requested in the original problem is to find the transformation that takes the unit square in the ##(u,v)## space to the parallelogram in the ##(x,y)## space. So instead of writing$$
T(x,y) = (1,1) + (2x+y,x+3y)$$he should have written$$
(x,y) = T(u,v) = (1,1) + (2u+v,u+3v)$$
 

Related to Help with change of variables in multivariable calculus/analysis

1. What is a change of variables in multivariable calculus/analysis?

A change of variables in multivariable calculus/analysis is a method used to transform a given function or equation from one set of variables to another set of variables. This is often done to simplify the problem or make it easier to solve.

2. Why do we need to use a change of variables in multivariable calculus/analysis?

A change of variables can be helpful in simplifying a problem, making it easier to solve, or gaining new insights into the problem. It can also be used to transform a problem into a more familiar form or to solve a problem that is difficult to solve using the original set of variables.

3. What are the steps for performing a change of variables in multivariable calculus/analysis?

The steps for performing a change of variables in multivariable calculus/analysis are as follows:

  1. Identify the original set of variables in the given problem.
  2. Choose a new set of variables that will make the problem easier to solve or understand.
  3. Write down the transformation equations that relate the original variables to the new variables.
  4. Substitute the new variables into the original problem and simplify as needed.
  5. Solve the problem using the new variables.
  6. Translate the solution back into the original variables, if needed.

4. What are some common examples of using a change of variables in multivariable calculus/analysis?

Some common examples of using a change of variables in multivariable calculus/analysis include:

  • Changing from Cartesian coordinates to polar coordinates
  • Transforming a double integral into polar coordinates
  • Using a linear transformation to simplify a problem
  • Changing to a more convenient coordinate system, such as cylindrical or spherical coordinates

5. Are there any limitations or restrictions when using a change of variables in multivariable calculus/analysis?

Yes, there are some limitations and restrictions when using a change of variables in multivariable calculus/analysis. These may include:

  • The transformation equations must be one-to-one and differentiable.
  • The domain of the new variables must be consistent with the original problem.
  • The transformation should not introduce singularities or discontinuities in the problem.
  • The transformation should not change the type of the problem (e.g. from a scalar function to a vector function).

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