Editing Jacobian matrix and determinant (section)

== Examples ==

=== Example 1 ===

Consider a function {{math|'''f''' : '''R'''<sup>2</sup> → '''R'''<sup>2</sup>,}} with  {{math|(''x'', ''y'') ↦ (''f''<sub>1</sub>(''x'', ''y''), ''f''<sub>2</sub>(''x'', ''y'')),}} given by

<math display="block">\mathbf f\left(\begin{bmatrix} x\\y\end{bmatrix}\right) = \begin{bmatrix} f_1(x,y)\\f_2(x,y)\end{bmatrix} =
  \begin{bmatrix}  x^2 y \\5 x + \sin y 
  \end{bmatrix}.</math>

Then we have

<math display="block">f_1(x, y) = x^2 y</math>

and

<math display="block">f_2(x, y) = 5 x + \sin y.</math>

The Jacobian matrix of {{math|'''f'''}} is

<math display="block">\mathbf J_{\mathbf f}(x, y) = \begin{bmatrix}
  \dfrac{\partial f_1}{\partial x} & \dfrac{\partial f_1}{\partial y}\\[1em]
  \dfrac{\partial f_2}{\partial x} & \dfrac{\partial f_2}{\partial y} \end{bmatrix}
= \begin{bmatrix}
  2 x y & x^2    \\
  5     & \cos y \end{bmatrix}</math>

and the Jacobian determinant is

<math display="block">\det(\mathbf J_{\mathbf f}(x, y)) = 2 x y \cos y - 5 x^2.</math>

=== Example 2: polar-Cartesian transformation ===

The transformation from [[polar coordinate system|polar coordinates]] {{math|(''r'', ''φ'')}} to [[Cartesian coordinate system|Cartesian coordinates]] (''x'', ''y''), is given by the function {{math|'''F''': '''R'''<sup>+</sup> × [0, 2{{pi}}) → '''R'''<sup>2</sup>}} with components

<math display="block">\begin{align}
x &= r \cos \varphi ; \\
y &= r \sin \varphi .
\end{align}</math>

<math display="block">\mathbf J_{\mathbf F}(r, \varphi)
= \begin{bmatrix}
    \frac{\partial x}{\partial r} & \frac{\partial x}{\partial\varphi}\\[0.5ex]
    \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\varphi}
  \end{bmatrix}
= \begin{bmatrix}
    \cos\varphi & - r\sin \varphi \\
    \sin\varphi &   r\cos \varphi
  \end{bmatrix}</math>

The Jacobian determinant is equal to {{math|''r''}}. This can be used to transform integrals between the two coordinate systems:

<math display="block">\iint_{\mathbf F(A)} f(x, y) \,dx \,dy = \iint_A f(r \cos \varphi, r \sin \varphi) \, r \, dr \, d\varphi .</math>

=== Example 3: spherical-Cartesian transformation ===

The transformation from [[spherical coordinate system|spherical coordinates]] {{math|(''ρ'', ''φ'', ''θ'')}}<ref>Joel Hass, Christopher Heil, and Maurice Weir. ''Thomas' Calculus Early Transcendentals, 14e''. Pearson, 2018, p. 959.</ref> to [[Cartesian coordinate system|Cartesian coordinates]] (''x'', ''y'', ''z''), is given by the function {{math|'''F''': '''R'''<sup>+</sup> × [0, ''π'') × [0, 2''π'') → '''R'''<sup>3</sup>}} with components

<math display="block">\begin{align}
x &= \rho \sin \varphi \cos \theta ; \\
y &= \rho \sin \varphi \sin \theta ; \\
z &= \rho \cos \varphi .
\end{align}</math>

The Jacobian matrix for this coordinate change is

<math display="block">\mathbf J_{\mathbf F}(\rho, \varphi, \theta)
= \begin{bmatrix}
  \dfrac{\partial x}{\partial \rho} & \dfrac{\partial x}{\partial \varphi} & \dfrac{\partial x}{\partial \theta} \\[1em]
  \dfrac{\partial y}{\partial \rho} & \dfrac{\partial y}{\partial \varphi} & \dfrac{\partial y}{\partial \theta} \\[1em]
  \dfrac{\partial z}{\partial \rho} & \dfrac{\partial z}{\partial \varphi} & \dfrac{\partial z}{\partial \theta}
  \end{bmatrix}
= \begin{bmatrix}
      \sin \varphi \cos \theta & \rho \cos \varphi \cos \theta & -\rho \sin \varphi \sin \theta \\
      \sin \varphi \sin \theta & \rho \cos \varphi \sin \theta & \rho \sin \varphi \cos \theta \\
      \cos \varphi & - \rho \sin \varphi & 0
  \end{bmatrix}.</math>

The [[determinant]] is {{math|''ρ''<sup>2</sup> sin ''φ''}}. Since {{math|''dV'' {{=}} ''dx'' ''dy'' ''dz''}} is the volume for a rectangular differential volume element (because the volume of a rectangular prism is the product of its sides), we can interpret {{math|''dV'' {{=}} ''ρ''<sup>2</sup> sin ''φ'' ''dρ'' ''dφ'' ''dθ''}} as the volume of the spherical [[differential volume element]]. Unlike rectangular differential volume element's volume, this differential volume element's volume is not a constant, and varies with coordinates ({{math|''ρ''}} and {{math|''φ''}}). It can be used to transform integrals between the two coordinate systems:

<math display="block">\iiint_{\mathbf F(U)} f(x, y, z) \,dx \,dy \,dz = \iiint_U f(\rho \sin \varphi \cos \theta, \rho \sin \varphi\sin \theta, \rho \cos \varphi) \, \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta .</math>

=== Example 4 ===

The Jacobian matrix of the function {{math|'''F''' : '''R'''<sup>3</sup> → '''R'''<sup>4</sup>}} with components

<math display="block">\begin{align}
y_1 &= x_1 \\
y_2 &= 5 x_3 \\
y_3 &= 4 x_2^2 - 2 x_3 \\
y_4 &= x_3 \sin x_1
\end{align}</math>

is

<math display="block">\mathbf J_{\mathbf F}(x_1, x_2, x_3) = \begin{bmatrix}
  \dfrac{\partial y_1}{\partial x_1} & \dfrac{\partial y_1}{\partial x_2} & \dfrac{\partial y_1}{\partial x_3} \\[1em]
  \dfrac{\partial y_2}{\partial x_1} & \dfrac{\partial y_2}{\partial x_2} & \dfrac{\partial y_2}{\partial x_3} \\[1em]
  \dfrac{\partial y_3}{\partial x_1} & \dfrac{\partial y_3}{\partial x_2} & \dfrac{\partial y_3}{\partial x_3} \\[1em]
  \dfrac{\partial y_4}{\partial x_1} & \dfrac{\partial y_4}{\partial x_2} & \dfrac{\partial y_4}{\partial x_3} \end{bmatrix}
= \begin{bmatrix}
  1 & 0 & 0 \\
  0 & 0 & 5 \\
  0 & 8 x_2 & -2 \\
  x_3\cos x_1 & 0 & \sin x_1 \end{bmatrix}.</math>

This example shows that the Jacobian matrix need not be a square matrix.

=== Example 5 ===

The Jacobian determinant of the function {{math|'''F''' : '''R'''<sup>3</sup> → '''R'''<sup>3</sup>}} with components

<math display="block">\begin{align}
  y_1 &= 5x_2 \\
  y_2 &= 4x_1^2 - 2 \sin (x_2 x_3) \\
  y_3 &= x_2 x_3
\end{align}</math>

is

<math display="block">\begin{vmatrix}
  0 & 5 & 0 \\
  8 x_1 & -2 x_3 \cos(x_2 x_3) & -2 x_2 \cos (x_2 x_3) \\
  0 & x_3 & x_2
\end{vmatrix} = -8 x_1 \begin{vmatrix}
  5 & 0 \\
  x_3 & x_2
\end{vmatrix} = -40 x_1 x_2.</math>

From this we see that {{math|'''F'''}} reverses orientation near those points where {{math|''x''<sub>1</sub>}} and {{math|''x''<sub>2</sub>}} have the same sign; the function is [[locally]] invertible everywhere except near points where {{math|''x''<sub>1</sub> {{=}} 0}} or {{math|''x''<sub>2</sub> {{=}} 0}}. Intuitively, if one starts with a tiny object around the point {{math|(1, 2, 3)}} and apply {{math|'''F'''}} to that object, one will get a resulting object with approximately {{math|40 × 1 × 2 {{=}} 80}} times the volume of the original one, with orientation reversed.