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So I have been trying to understand the intuitive proof of change of variables in double integrals.

Equations $$ x=x(u, v) \ \mathrm{and} \ y=y(u,v) $$ define a mapping from uv-Cartesian plane to points in xy-plane. The mapping is one-to-one.

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The differential area element $$ dA = dx dy$$ is bounded in xy-plane by $$ u, \ u+du,\ v, \ \mathrm{and} \  v+dv.$$ Now the question is, what is dA in uv-plane?

In most intuitive proofs it is said that since du and dv are small and the area element is bounded by smooth curves, the area element is approximately a parallelogram and therefore its are is $$ dA = \left|\vec{AC} \times \vec{AB} \right|=\dots= \left|\frac{\partial(x, y)}{\partial(u, v)} \right| du dv,$$ where $$ \vec{AB}=\frac{\partial x}{\partial u}du\hat{i}+\frac{\partial y}{\partial u}du\hat{j} \ \mathrm{and} \ \vec{AC}=\frac{\partial x}{\partial v}dv\hat{i}+\frac{\partial y}{\partial v}dv\hat{j}.$$

But if we approximate the area element as a parallelogram, shouldn't it be $$ \left \Vert  \vec{AB} \right \Vert = du \ \mathrm{and} \ \left \Vert  \vec{AC} \right \Vert = dv$$ and therefore we would have a different result for dA?

Edit: So my question really is, are $$ du=\left \Vert \frac{\partial x}{\partial u}du\hat{i}+\frac{\partial y}{\partial u}du\hat{j} \right \Vert \ \mathrm{and} \ dv=\left \Vert \frac{\partial x}{\partial v}dv\hat{i}+\frac{\partial y}{\partial v}dv\hat{j} \right \Vert ?$$

mathslover
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1 Answers1

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How so? The reason we use the cross product is because the area element spanned by two vectors is given by the cross product of the two. If you do the way you are doing, then you need to take one side as base and find the 'height' of the parellogram, which will end up in the same formula. Here is a more explicit calculation of the proof you mentioned, consider a coordinate system $(x,y)$ and another old coordinate system $(u,v)$ , let's say we have a map from $(u,v)$ system to $(x,y)$ system then we can write:

$$ dx= < \frac{\partial x}{\partial u} du , \frac{\partial x}{\partial v} dv>$$

The idea here is the amount of increase in $u$ direction as we increase little bit in x in original system, is just slope of $x$ with $u$ times $u$ and similarly for in v direction

$$ dy = < \frac{\partial y}{\partial u} du , \frac{\partial y}{\partial v} dv>$$

So, it's like we have an area spanned by $dx$ and $dy$ in our original system, then the expression equivalent area produced in the other system can be found by cross product the two vectors, that is:

$$ |dA| = dx \times dy = < \frac{\partial x}{\partial u} du , \frac{\partial x}{\partial v} dv> \times < \frac{\partial y}{\partial u} du , \frac{\partial y}{\partial v} dv>$$

tryst with freedom
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  • thank you! I tried to calculate dA both ways but ended up with different results. I added a new question to the post. So is du the length of vector <dx/du*du, dy/du*du> ? I guess that would solve my problem. – mathslover Mar 01 '21 at 07:34
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    No no , in the post I've written (x,y) as a function of the coordniates (u,v) and I have found the length dx and dy vectors caused by moving a little bit in the u,v coordinates. – tryst with freedom Mar 01 '21 at 07:48
  • I could also have done du = < du/dx * dx, du/dy * dy> and that works because jacobian for forward and backward transform arei nverses of each other – tryst with freedom Mar 01 '21 at 07:49