# Deriving multivariate change of variables using vector calculus

I'm trying to derive the formula (?) for multivariate change of variables using vector calculus but am getting stuck on the final steps. I am hoping that someone can help me finish the derivation from where I'm stuck rather than provide an alternative derivation.

Show: $\iint f(x,y)dxdy = \iint f(u,v)\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$ where x=x(u,v) and y=y(u,v)

suppose we have $\iint f(x,y)dxdy$ and we want to integrate using (u,v) coordinates.

Let $\mathbf{r}(u,v)=\begin{bmatrix}x(u,v)\\y(u,v)\end{bmatrix}$, then the integral can be written $\iint f(\mathbf{r}(u,v))dxdy$

if we change u by some small amount du, we can approximate the change in $\mathbf{r}(u,v)$ by $\mathbf{r}_udu = \begin{bmatrix}x_u\\y_u\end{bmatrix}du$.

Likewise we can approximate the change in $\mathbf{r}(u,v)$ as $\mathbf{r}_vdv = \begin{bmatrix}x_v\\y_v\end{bmatrix}dv$ for some small change in v.

The area spanned by these two vectors can be found from the absolute value of the determinant, giving the area differential:

$dA = \left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$

So we can approximate the volume under f of this chunk as:

$f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$

From the above, I can see the jacobian and how it relates to change of variables. I like the way I arrived there. The problem is that I don't really understand why I would equate this with the corresponding volume in the (x,y) plane. The area differentials for the two coordinate systems being equal does not seem immediately obvious to me:

$f(x,y)dxdy \overset{?}{=} f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$

Then to get the full area under f we would just need to integrate those volumes over their boundaries

$\iint f(x,y)dxdy \overset{?}{=} \iint f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$

• Here's a relevant thread. math.stackexchange.com/questions/267267/… – littleO Aug 10 '16 at 6:21
• The change of variables formula is a hard theorem, and your derivation is nowhere even close to a proof. I also can't tell what you're really asking. But maybe this is what you want: say $dx=x_udu+x_vdv$, $dy=y_udu+y_vdv$. Then $$dxdy=(x_udu+x_vdv)(y_udu+y_vdv)=x_uy_udu^2+x_uy_vdudv+x_vy_udvdu+x_vy_vdv^2$$ But in the algebra of differential forms, $du^2=dv^2=0$ and $dvdu=-dudv$. Therefore $dxdy=(x_uy_v-y_ux_v)dudv$. – symplectomorphic Aug 10 '16 at 6:26
• Yes proof is not the right word. I meant more along the lines of an intuitive derivation or train of thought for how to integrate the original integral if I change coordinate systems – ApplePi Aug 10 '16 at 6:33
• @ApplePi: then what exactly is your question? You say "The area differentials for the two coordinate systems being equal does not seem immediately obvious to me." But you just "showed" that $dA=\det J\,dudv$. What else could $dA$ be but $dxdy$? – symplectomorphic Aug 10 '16 at 6:36
• dA=det J dudv in the uv plane. The whole problem with changing coordinate systems is that there's a stretching of space.I found an expression for dA in the uv plane for a small change in u and v, but I don't see why that area is necessarily equal to dxdy in the xy plane as the change of coordinate systems has stretched the space – ApplePi Aug 10 '16 at 6:54