Change of Variables in Double Integrals

Recall from the Evaluating Double Integrals in Polar Coordinates page that if the region $D = \{ (r, \theta) : h_1(\theta) ≤ r ≤ h_2(\theta), \alpha ≤ \theta ≤ \beta \}$ then:

(1)

\begin{align} \quad \iint_D f(x, y) \: dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r \cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

What we have really done is defined a transformation in using these substitution to make evaluating certain integrals simpler. We can apply this technique more generally by changing variables. Suppose that $x$ and $y$ are functions of $u$ and $v$, that is $x = x(u, v)$ and $y = y(u, v)$. Then every $(x, y) \in S \subset \mathbb{R}^2$ is mapped to $(u, v) \in D \subset \mathbb{R}^2$. More importantly, we will want a transformation that is one-to-one from $S$ and onto $D$ so that points in $S$ are mapped to distinct points in $D$ and every point in $D$ is mapped to.

Consider a one-to-one transformation defined by the equations $x = x(u, v)$ and $y = y(u, v)$. Then:

(2)

\begin{equation} f(x, y) = f(x(u, v), y(u,v)) \end{equation}

We now need to express the area element, $dA$, from the variables $x$ and $y$ to the variables $u$ and $v$. Suppose that the value of $u$ is fixed such that $u = c$. Then we get that $x = x(c, v)$ and $y = y(c, v)$ from the transformation equations, and these equations together define a parametric curve $\left\{\begin{matrix} x = x(c, v) \\ y = y(c, v) \end{matrix}\right.$ in $\mathbb{R}^3$ that we'll call a $u$-curve. Similarly, if the value of $v$ is fixed such that $v = c$ then $x = x(u, c)$ and $y = y(u, c)$ from the transformation equations, and these equations together define a parametric curve $\left\{\begin{matrix} x = x(u, c) \\ y = y(u, c) \end{matrix}\right.$ in $\mathbb{R}^3$ that we'll call a $v$-curve.

Now consider the area element that is bounded by the $u$-curves for $u$ and $u + du$ and the $v$-curves for $v + dv$:

Let $P$, $Q$, and $R$ be the points shown in the diagram above. As the area elements get smaller, the area of the parallelogram formed by the vectors $\vec{PQ}$ and $\vec{PR}$ approximates the area of the area elements and the error approaches zero since for small values of $du$ and $dv$, the $u$ and $v$ curves are approximately straight. Thus:

(3)

\begin{align} \quad dA = \mid \vec{PQ} \times \vec{PR} \mid \end{align}

We will not go too much into detail, but it can be shown that thus:

(4)

\begin{align} \quad dA = \biggr \rvert \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert \: du \: dv \end{align}

The notation $\biggr \rvert \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert$ represents the absolute value of the Jacobian Determinant. The following theorem summarizes the change of variables for double integrals with respect to the transformation equations $x = x(u, v)$ and $y = y(u, v)$.

Theorem 1: Let $D$ be a region on the $xy$-plane and let $S$ be a region on the $uv$-plane, and let the equations $x = x(u, v)$ and $y = y(u, v)$ define a one-to-one transformation. Suppose that the functions $x$, $y$, and their first partial derivatives with respect to the variables $u$ and $v$ are continuous on the domain $S$ and that $f(x, y)$ is integrable on $D$. Then: $\iint_D f(x, y) \: dx \: dy = \iint_S f(x(u, v), y(u, v)) \biggr \rvert \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert \: du \: dv$.