Change of Variables | Sahithyan's Notes

The Jacobian determinant of a transformation $T$ measures the local scaling of area. A change of variables moves an integral from the target region onto the source region.

Theorem

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 $x - y = 2$ .

Solution Reveal Hide

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.

Solution Reveal Hide

\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