Why do you need Jacobian determinant to change variables in Vector Integral?

Question

Vector $\mathbf{y} = (y_1, y_2, ..., y_N)$ and vector $\mathbf{x} = (x_1, x_2, ..., x_N)$ are related according to equation:

$$p(\mathbf{y}) = \frac{1}{|\mathbf{A}|}q(\mathbf{x})$$

And linear equation:

$$\mathbf{y}=\mathbf{A}\mathbf{x}$$

Now I need to change variables for an integral below from $\mathbf{y}$ to $\mathbf{x}$.

$$g(\mathbf{y}) = \int p(\mathbf{y})~ d\mathbf{y}$$

Textbook says its like this:

$$g(\mathbf{y}) = \frac{1}{|A|} \int q(\mathbf{x})~ \bigg| \frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg|d\mathbf{x}$$

$$\frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \frac{\partial }{\partial x} Ax = A$$|

$$g(\mathbf{y}) = \frac{1}{|A|} \int q(\mathbf{x})~ d\mathbf{x}$$

I'm a little bit confused by the use of the Jacobian $\bigg| \frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg|$ to change variables in this case... mainly because most of the examples on the internet assume you have two integrals when using Jacobian to change variables...

Questions as follows:

(1) How does the Jacobian works in the case of a single integral where the variables x and y are vectors.

(2) I can see that $dy$ in numerator cancels with $\partial{y}$ in denominator... I'm not understanding how the determinate gets removed and why you can cancel a partial differential with a regular differential...

(3) why can't I just use the variable change method they teach in "calculus I" instead of using a Jacobin when applied to a vector integral?

Answer 1

Vector Integral Change of Variable Rules

The Jacobian determinant $\bigg|\frac{\partial y}{\partial x} \bigg|$ is needed to change variables of integration that are vectors.

Given:

$$\int_A f(\mathbf{y})~d\mathbf{y}$$

where:

$$\mathbf{y} = g(\mathbf{x})$$

We can change variables of integration from y to x by substitute the Jacobian determinate into the integral as follows::

$$d\mathbf{y} = \bigg|\frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg| d\mathbf{x}$$

Then Integrate as following:

$$\int_A f(\mathbf{y})~d\mathbf{y} = \int_{g^{-1}(A)} f(g(x))~\bigg|\frac{\partial y}{\partial x} \bigg| dx$$

Answer 2

$\newcommand{\R}{\mathbb{R}}$ Let me just summarize the comments and add a bit.

So lets say you have a function $p : \R^N \to \R$ and you would like to calculate the integral of a given (measurable set) $E \subset\R^N$. Then you would write $$ \int_E p(x) dx \qquad \text{which is a real number!} $$ Now, since we are in $\R^N$ this integrale is a $N$-D integral and so $dx$ is really $dx = dx_1dx_2\cdots dx_N$.

Sometimes this can be very difficult to calculate, especially when the set $E$ is a bit unconvenient as it is for example if you want to integrate over a Ball or similar things. Then it makes sense to make a transformation of variables, e.g. polar coordinates. So in general you would have an injective (one-to-one) differentiable function $A$, so that $x = A(y)$. Then $A : \R^N \to A(\R^N)$ is bijective and you can calculate the Jacobian of $A$ which I will denote by $DA$ (here you need a few more extra assumption on this transformaton $A$ to make the following formula work): This is a $N\times N$ matrix in every point $y=(y_1,\dots,y_N)$ and so you can calculate the determinant. Then you get the transformation rule: $$ \int_E p(x) dx = \int_{A^{-1}(E)} p(A(y)) |\text{det}(DA)(y)|dy $$ Now, the expression $|\text{det}(DA)(y)|$ (btw. the $| \cdot |$ mean the modulus here) might be a little bit to long and also there is nice analog from 1-D calculus, so some authors just like to introduce the notation: $$ \left|\frac{\partial y}{\partial x}\right| := |\text{det}(DA)(y)| $$ (I do not like it too much, but that is of course just taste).

Now lets do you special case: We have $$ x = Ay \qquad \text{where} \quad A \quad \text{is a } N\times N \text{matrix}, $$ with non vanishing determinant (so it is bijective!) also we know something about $p(x)$ namely that $$ p(x) = p(Ay) = \frac{1}{|\text{det}(A)|} q(y). $$ What is the first thing we have to do? We need to calculate $DA$. Now, since $A$ is just a constant linear map, the best linear approximation is $A$ itself so we get $DA(p) =A$ for any point $p\in \R^N$.

What is the $E$? Well, it is not mentioned in your setup, so lets just take $E = \R^N$, so $A^{-1}(\R^N) = \R^N$. Nice!

So we get: $$ \int_{\R^N} p(x) dx = \int_{\R^N} p(A(y)) |\text{det}(A)| dy = \int_{\R^N} q(y) dy $$