Jacobian | Sahithyan's Notes

For a transformation $T$ from the $uv$ -plane to the $xy$ -plane defined by $x = x(u,v)$ , $y = y(u,v)$ , the Jacobian of $T$ is

J(u,v) = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}

For a transformation $T$ from $uvw$ -space to $xyz$ -space defined by $x = x(u,v,w)$ , $y = y(u,v,w)$ , $z = z(u,v,w)$ , the Jacobian of $T$ is

J(u,v,w) = \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\[6pt] \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix}

Examples:

$\dfrac{\partial(x,y)}{\partial(r,\theta)} = r$ , for polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$
$\dfrac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)} = \rho^2\sin\phi$ , for spherical coordinates

Change of Variables

Let $T$ be a transformation from the $uv$ -plane to the $xy$ -plane, mapping a region $S$ onto a region $R$ . If:

$T$ is one-to-one on $S$
$T$ has continuous first partial derivatives
$\dfrac{\partial(x,y)}{\partial(u,v)}$ is non-zero and does not change sign on $S$

then

\iint_R f(x,y)\,\text{d}A = \iint_S f(x(u,v), y(u,v)) \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \text{d}A

Worked Examples

Parallelogram

Evaluate $\iint_R (x^2 - y^2)\,\text{d}A$ over the parallelogram $R$ bounded by the lines $x + y = 1$ , $x + y = 3$ , $x - y = 0$ and $x - y = 2$ .

The boundary lines are constant values of $x + y$ and $x - y$ , and the integrand factors as $(x+y)(x-y)$ . Let $u = x + y$ and $v = x - y$ . Then $R$ becomes the rectangle $S: 1 \le u \le 3,\ 0 \le v \le 2$ and the integrand becomes $uv$ .

Solve for $x$ and $y$ .

x = \frac{u+v}{2}, \quad y = \frac{u-v}{2}

Compute the Jacobian.

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\[6pt] \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{2}

The transformation is one-to-one, has continuous first partial derivatives, and its Jacobian is a non-zero constant, so the change of variables applies with $\left| -\frac{1}{2} \right| = \frac{1}{2}$ .

\iint_R (x^2 - y^2)\,\text{d}A = \int_1^3 \int_0^2 uv \cdot \frac{1}{2}\,\text{d}v\,\text{d}u

The integral separates.

\frac{1}{2} \int_1^3 u\,\text{d}u \cdot \int_0^2 v\,\text{d}v = \frac{1}{2} \cdot 4 \cdot 2 = 4