# Intuitive proof of multivariable changing of variables formula (jacobian) without using mapping and/or measure theory?

iWhat is a intuitive proof of multivariable changing of variables formula (jacobian) without using mapping and/or measure theory?

I was thinking that textbooks make the proofs over complicate.

If possible, using linear algebra and calculus to solve it, that way would be simplest for me?

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If there was a simpler proof, don't you think the books would use it? –  Potato Dec 29 '12 at 20:47
@Potato - Couldn't the author also give the intuitions? –  Victor Dec 29 '12 at 20:49
What exactly do you want? A different proof, or an intuitive explanation of the standard proof (say, the one that is in Folland). –  Potato Dec 29 '12 at 20:50
A lengthy proof of the change of variables formula for Riemann integrals in $\mathbb R^n$ (that does not use measure theory) is given in Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach by Hubbard and Hubbard. A discussion of the intuition behind it is given on page 493. –  Potato Dec 29 '12 at 21:04
@Tim A proof for Lebesgue integrals can be found in any standard book on measure theory and integration, including Folland's book. –  Potato Dec 29 '12 at 21:14

## 3 Answers

A lengthy proof of the change of variables formula for Riemann integrals in $\mathbb R^n$ (that does not use measure theory) is given in Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach by Hubbard and Hubbard. A discussion of the intuition behind it is given on page 493.

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The multivariable change of variables formula is nicely intuitive, and it's not too hard to imagine how somebody might have derived the formula from scratch. However, it seems that proving the theorem rigorously is not as easy as one might hope.

Here's my attempt at explaining the intuition -- how you would derive or discover the formula.

The first thing to understand is that if $A$ is an $N \times N$ matrix with real entries and $S \subset \mathbb R^N$, then $m(AS) = |\det A| \, m(S)$. (Technically I should assume that $S$ is measurable.) This is intuitively clear from the SVD of $A$: $$A = U \Sigma V^T$$ where $U$ and $V$ are orthogonal and $\Sigma$ is diagonal with nonnegative diagonal entries. Multiplying by $V^T$ doesn't change the measure of $S$. Multiplying by $\Sigma$ scales along each axis, so the measure gets multiplied by $\det \Sigma = | \det A|$. Multiplying by $U$ doesn't change the measure.

Next suppose $\Omega$ and $\Theta$ are open subsets of $\mathbb R^N$ and suppose $g:\Omega \to \Theta$ is $1-1$ and onto. We should probably assume $g$ and $g^{-1}$ are $C^1$ just to be safe. (Since we're just seeking an intuitive derivation of the change of variables formula, we aren't obligated to worry too much about what assumptions we make on $g$.) Suppose also that $f:\Theta \to \mathbb R$ is, say, continuous (or whatever conditions we need for the theorem to actually be true).

Partition $\Theta$ into tiny subsets $\Theta_i$. For each $i$, let $u_i$ be a point in $\Theta_i$. Then $$\int_{\Theta} f(u) \, du \approx \sum_i f(u_i) m(\Theta_i).$$

Now let $\Omega_i = g^{-1}(\Theta_i)$ and $x_i = g^{-1}(u_i)$ for each $i$. The sets $\Omega_i$ are tiny and they partition $\Omega$. Then \begin{align} \sum_i f(u_i) m(\Theta_i) &= \sum_i f(g(x_i)) m(g(\Omega_i)) \\ &\approx \sum_i f(g(x_i)) m(g(x_i) + Jg(x_i) (\Omega_i - x_i)) \\ &= \sum_i f(g(x_i)) m(Jg(x_i) \Omega_i) \\ &\approx \sum_i f(g(x_i)) |\det Jg(x_i)| m(\Omega_i) \\ &\approx \int_{\Omega} f(g(x)) |\det Jg(x)| \, dx. \end{align}

We have discovered that $$\int_{g(\Omega)} f(u) \, du \approx \int_{\Omega} f(g(x)) |\det Jg(x)| \, dx.$$ By using even tinier subsets $\Theta_i$, the approximation would be even better -- so we see by a limiting argument that we actually have equality.

At a key step in the above argument, we used the approximation $$g(x) \approx g(x_i) + Jg(x_i)(x - x_i)$$ which is a good approximation when $x$ is close to $x_i$

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Let there be some vector function $f(x) = x'$, which can be interpreted as remapping points or changing coordinates. For example, $f(x) = \sqrt{x \cdot x} e_1 + \arctan \frac{x^2}{x^1} e_2$ remaps the cartesian coordinates $x^1, x^2$ to polar coordinates on the basis vectors $e_1, e_2$.

Now, let $c(\tau)$ be a path parameterized by the scalar parameter $\tau$. Let $f(c) = c'(\tau)$ be the image of this path under the transformation. The chain rule tells us that

$$\frac{dc'}{d\tau} = \Big(\frac{dc}{d\tau} \cdot \nabla \Big) f$$

Define $a \cdot \nabla f \equiv \underline f(a)$ as the Jacobian operator acting on a vector $a$, and the equation can be rewritten as

$$\frac{dc}{d\tau} = \underline f^{-1} \Big(\frac{dc'}{d\tau} \Big)$$

(Note that the primes have switched, so we use the inverse Jacobian.)

This is all we need to show that a line integral in the original coordinates is related to a line integral in the new coordinates by using the Jacobian. For some scalar field $\phi$, if $\phi(x) = \phi'(x')$, then

$$\int_c \phi \, d\ell = \int_{c'} \phi' \, \underline f^{-1}(d\ell')$$

because $d\ell'$ can be converted to $\frac{d\ell'}{d\tau} \, d\tau$.

Edit: didn't see the word intuitive. As far as intuitive explanations go, you can think of a coordinate transformation like so. Imagine the lines of a polar coordinate system being warped and stretched so that they become rectangular instead. This makes working with them easier, but because the shapes of coordinate lines, paths, and areas have changed (and because you don't want them to change the result, since changing coordinates should not change the result), the naive errors introduced must be corrected for with a factor of the Jacobian operator.

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