Why do we compute the partial derivatives in terms on the new variables and not with respect to the old variables? Can someone say this intuitively, so that I have the best chance of remembering why?
For instance, when going from x,y coordinates to, say, u,v coordinates, I am tempted to compute the "Jacobian" as the determinant of the partial derivatives $u_x, u_y, v_x, v_y$, which would be wrong, I think - and that the correct Jacobian is gotten from computing the partials $x_u, x_v, y_u, y_v$.
Thanks,
Edit: is this a good way of thinking about it (from Wikipedia): I am computing the factor by which x,y shrinks or expands u and v?
Think of how it would go for a single variable
$\int f(x) dx = \int f(u) \cdot \frac{dx}{du} du$ ( you can see the $du$'s cancelling )
rather than $\int f(u) \cdot \frac{du}{dx} du$
The Jacobean is the factor by which the transformation from $(u,v)$ to $(x,y)$ expands or contracts areas, so that's what you need to multiply the $(u,v)$ area to recover the $(x,y)$ area, which is what the original question is asking for.
I remember when I first learned about Jacobeans I had a eureka realization about the nature of determinants, for a linear transformation the Wronskian matrix is constant and its determinant is the factor by which areas are multiplied so that a 2x2 matrix that maps the plane onto a line has determinant zero, this is why a matrix must have determinant zero in order to map any point other than $(0,0)$ onto $(0,0)$ ( because a matrix with a non-zero determinant must be one-to-one and zero always maps to zero under any linear transformation )
Edit: ( response to comment )
the short answer is that you are just conveniently taking advantage of the fact that
$$\frac{du}{dx} \frac{dx}{du}=1 $$
consider the example $f(x)=x^2$, Let $u=x^2$ so that $du = 2 x dx =2 \sqrt{u} dx$
Our Jacobean is now $\frac{dx}{du}=\frac{1}{2 \sqrt{u}}$
Notice that we could has got the same Jacobean by expressing $x$ in terms of $u$ from the outset
$$x=u^\frac 12 \Rightarrow dx = \frac{1}{2 \sqrt{u} } du$$
So
$$ \int x^2 dx = \int u \left( \frac{dx}{du} \right ) du = \frac 12 \int u^{\frac 12} du = \frac 13 u^\frac32 =\frac 13 x^3 $$