Jacobian in three variables to change variables

Formula for the 3x3 Jacobian matrix in three variables

In the past we’ve converted multivariable functions defined in terms of cartesian coordinates $x$ and $y$ into functions defined in terms of polar coordinates $r$ and $\theta$.

Similarly, given a region defined in $uvw$-space, we can use a Jacobian transformation to redefine it in $xyz$-space, or vice versa.

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Given three equations $x=f(u,v,w)$, $y=g(u,v,w)$, and $z=h(u,v,w)$, the Jacobian is

How to calculate the Jacobian for a function in three variables

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Changing a function in x, y, and z into a function in u, v, and w

Example

Find the Jacobian of the transformation.

$x=uw^2$

$y=v^3-3w$

$z=\frac{2uv}{w}$

Our functions tell us that we have a $3\times3$ set-up, so we’ll use the formula

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=\frac{\partial{x}}{\partial{u}}\left(\frac{\partial{y}}{\partial{v}}\cdot\frac{\partial{z}}{\partial{w}}-\frac{\partial{y}}{\partial{w}}\cdot\frac{\partial{z}}{\partial{v}}\right)-\frac{\partial{x}}{\partial{v}}\left(\frac{\partial{y}}{\partial{u}}\cdot\frac{\partial{z}}{\partial{w}}-\frac{\partial{y}}{\partial{w}}\cdot\frac{\partial{z}}{\partial{u}}\right)$

$+\frac{\partial{x}}{\partial{w}}\left(\frac{\partial{y}}{\partial{u}}\cdot\frac{\partial{z}}{\partial{v}}-\frac{\partial{y}}{\partial{v}}\cdot\frac{\partial{z}}{\partial{u}}\right)$

We need to start by finding the partial derivatives of $x$, $y$ and $z$ with respect to $u$, $v$ and $w$.

$\frac{\partial{x}}{\partial{u}}=w^2$

$\frac{\partial{x}}{\partial{v}}=0$

$\frac{\partial{x}}{\partial{w}}=2uw$

and

$\frac{\partial{y}}{\partial{u}}=0$

$\frac{\partial{y}}{\partial{v}}=3v^2$

$\frac{\partial{y}}{\partial{w}}=-3$

and

$\frac{\partial{z}}{\partial{u}}=\frac{2v}{w}$

$\frac{\partial{z}}{\partial{v}}=\frac{2u}{w}$

$\frac{\partial{z}}{\partial{w}}=-\frac{2uv}{w^2}$

given a region defined in uvw-space, we can use a Jacobian transformation to redefine it in xyz-space, or vice versa.

We’ll plug the partial derivatives into our formula and get

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=w^2\left[3v^2\left(-\frac{2uv}{w^2}\right)-(-3)\left(\frac{2u}{w}\right)\right]-0\left[0\left(-\frac{2uv}{w^2}\right)-(-3)\left(\frac{2v}{w}\right)\right]$

$+2uw\left[0\left(\frac{2u}{w}\right)-3v^2\left(\frac{2v}{w}\right)\right]$

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=w^2\left(-\frac{6uv^3}{w^2}+\frac{6u}{w}\right)+2uw\left(-\frac{6v^3}{w}\right)$

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=-\frac{6uv^3w^2}{w^2}+\frac{6uw^2}{w}-\frac{12uv^3w}{w}$

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=-6uv^3+6uw-12uv^3$

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=-18uv^3+6uw$

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=6uw-18uv^3$

$\frac{\partial{(x,y,z)}}{\partial{(u,v,w)}}=6u\left(w-3v^3\right)$

This is the Jacobian of the transformation.

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