Jacobian Determinant

Let ξ
1
= r and ξ2
= θ . Therefore,
x1 = x= rcosθ = ξ1cosξ2
x2 = y = rsinθ = ξ1sinξ2
Then the vector obtained by making a differential change dξ 1
= dr has components
( dr cos θ , dr sinθ ), where I have placed dr in front of each term for clarity. Similarly,
the vector obtained by a differential change in the polar angle, which is dξ2
= dθ
, has
components ( −rsin θ dθ, rcosθ
d θ ). From your sketch, verify that indeed these are the
correct components.
The extension to three dimensions is straightforward. We make differential displacements
in the three coordinates ξ 1
, ξ 2
and ξ 3
. We write the components of the infinitesimal
vectors so obtained as
⎛ ∂x
∂x
∂x
⎜
⎝∂ξ ∂ξ ∂ξ
1 2
3
dξ1, dξ1,
dξ
1
1 1 1
⎞
⎟
⎠
⎛ ∂x1 ∂x2
∂x
⎞
3
⎜ dξ2, dξ2,
dξ2⎟
⎝∂ξ2 ∂ξ2 ∂ξ2
⎠
⎛ ∂x1 ∂x2
∂x
⎞
3
⎜ dξ3, dξ3,
dξ3⎟
⎝∂ξ3 ∂ξ3 ∂ξ3
⎠
The volume of the parallelepiped formed with these three vectors as its sides is given by
the magnitude of the triple scalar product, which is the absolute value of the determinant
formed by the components of the vectors. From this, by taking out the common factor
dξ1dξ2dξ 3
, we find that the volume dV is given by
( , , )
dV = dx dx dx = J ξ ξ ξ dξ dξ dξ
=
1 2 3 1 2 3 1 2 3
∂x1 ∂x1 ∂x1
∂ξ ∂ξ ∂ξ
1 2
∂x2 ∂x2 ∂x2
∂ξ ∂ξ ∂ξ
1 2
∂x3 ∂x3 ∂x3
∂ξ ∂ξ ∂ξ
1 2
3
3
3
dξ 1 dξ 2 dξ 3
where the inner set of vertical lines stands for the determinant, and the outer set is needed
to yield the absolute value.