Change of variables in multiple integrals

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Sometimes, we need to integrate a function on a bounded closed region other than a rectangle or a region sandwiched by two curves. Even when the bounded closed region is simple, it may still be difficult to integrate a function on it. We may calculate the integral in such cases by changing the integration variables.



Consider two \(\mathbb{R}^2\) planes, the \((u,v)\) plane and \((x,y)\) plane. Suppose we have a map \(\Phi\)

\[\Phi(u,v) = (x(u,v), y(u,v))\]

defined on a neighbor of the bounded closed region \(E\) in the \((u,v)\) plane. Suppose \(\Phi\) maps \(E\) to the bounded closed region \(D\) in the \((x,y)\) plane (Figure 1):

\[\Phi(E) = D.\]

Figure 1. Change of variables by the mapping \(\Phi(u,v) = (x(u,v), y(u,v))\).

We further assume the following:

  1. \(\Phi(u,v)\) is injective.
  2. The functions \(x(u,v)\) and \(y(u,v)\) are of class \(C^1\), and at an arbitrary point \((u_0,v_0)\in E\), we have \[\frac{\partial x}{\partial u}(u_0,v_0)\frac{\partial y}{\partial v}(u_0,v_0)-\frac{\partial x}{\partial v}(u_0,v_0)\frac{\partial y}{\partial u}(u_0,v_0) \neq 0.\] That is, the determinant of the Jacobian of \(\Phi\) is non-zero in \(E\).

Let \(f(x,y)\) be a continuous function on \(D\). By composing with \(\Phi\), we have the function
\[(f\circ \Phi)(u,v) = f(x(u,v), y(u,v))\]
on \(E\).

Theorem (Change-of-variables formula for multiple integrals)

The following holds:

\[\iint_Df(x,y)dxdy = \iint_Ef(x(u,v),y(u,v))|J(u,v)|dudv\tag{Eq:chvar}\]

where

\[J(u,v) = \frac{\partial x}{\partial u}(u,v)\frac{\partial y}{\partial v}(u,v) -\frac{\partial x}{\partial v}(u,v)\frac{\partial y}{\partial u}(u,v).\]

Proof. Omitted. (Too technical and too long). ■

Instead of the proof, we give a hand-waving explanation of how the above change-of-variables formula works. Consider the situation as in Figure1 where the region \(E\) is rectangular, and \(D\) is of some general (i.e., not easily describable) shape. We would like to integrate a function \(f(x,y)\) on \(D\).

First, let us partition the rectangle \(E\) into smaller rectangles (Figure 2, left). Accordingly, the region \(D\) is also partitioned into corresponding smaller subregions (Figure 2, right).

Figure 2. Partitioning \(E\) and \(D\).

Figure 3. A close-up view of the highlighted regions in Figure 2.

Each small rectangular subregion \(E_{ij} = [a_{i}, a_{i+1}]\times [c_{j}, c_{j+1}] \subset E\) \((i = 0, 1, \cdots, n-1; j = 0, 1, \cdots, m-1)\) is mapped to a subregion \(D_{ij}\subset D\) (Figure 3). Each \(D_{ij}\) is a region surrounded by the pieces of four curves (Figure 3, right). Now, let \(p_{ij} = (a_i, c_j)\) be the bottom left vertex of the small rectangle \(E_{ij}\) and define \(q_{ij} = \Phi(p_{ij}) = (x(a_i,c_j), y(a_i, c_j))\). The integral of \(f(x,y)\) over \(D\) is calculated as follows. Let \(\mu(D_{ij})\) be the area of the region \(D_{ij}\). 

Remark. In the following, \(\mu(X)\) indicates the area of the set \(X\in\mathbb{R}^2\). □

The integral \(\iint_Df(x,y)dxdy\) is approximated by the sum
\[\iint_Df(x,y)dxdy \approx  \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x(a_i,c_j),y(a_i,c_j))\mu(D_{ij})\tag{Eq:intD}\]
where each \(f(x(a_i,c_j),y(a_i,c_j))\mu(D_{ij})\) represents the signed volume of the shape with base \(D_{ij}\) and (signed) height \(f(x(a_i,c_j),y(a_i,c_j))\). As the partition is refined, the sum on the right-hand side converges to the integral on the left-hand side.
On the other hand, the right-hand side of the formula (Eq:chvar) can be approximated as
\[\iint_{E}f(x(u,v), y(u,v))|J(u,v)|dudv \approx \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x(a_i,c_j),y(a_i,c_j))|J(a_i,c_j)|\mu(E_{ij})\tag{Eq:intE}\]
To compare (Eq:intD) and (Eq:intE), we need to compare the areas of \(D_{ij}\) and \(E_{ij}\). Note that it suffices to show that
\[\mu(D_{ij}) = |J(a_i,c_j)|\mu(E_{ij}).\]
The difficulty here is we don't know how to measure the area of \(D_{ij}\), \(\mu(D_{ij})\). On the other hand, we know that \(\mu(E_{ij}) = (a_{i+1}-a_{i})(c_{j+1} - c_{j})\) (as \(E_{ij} = [a_i, a_{i+1}]\times [c_j, c_{j+1}]\) is rectangular). To simplify the problem, we approximate the region \(D_{ij}\) by a parallelogram spanned by the two vectors
\[\begin{eqnarray} \overrightarrow{q_{i,j}q_{i+1,j}} &=& (x(a_{i+1},c_j) - x(a_{i},c_j), y(a_{i+1},c_j) - y(a_{i},c_j)),\\ \overrightarrow{q_{i,j}q_{i,j+1}} &=& (x(a_{i},c_{j+1}) - x(a_{i},c_j), y(a_{i},c_{j+1}) - y(a_{i},c_j)). \end{eqnarray}\]
We further approximate each component of these vectors by using the first-order Taylor expansion (Figure 4):
\[\begin{eqnarray} x(a_{i+1},c_j) - x(a_{i},c_j) &\approx& x_u(a_i,c_j)(a_{i+1} - a_{i}),\\ y(a_{i+1},c_j) - y(a_{i},c_j) &\approx& y_u(a_i,c_j)(a_{i+1} - a_{i}),\\ x(a_{i},c_{j+1}) - x(a_{i},c_j) &\approx& x_v(a_i,c_j)(c_{j+1} - c_{j}),\\ y(a_{i},c_{j+1}) - y(a_{i},c_j) &\approx& y_v(a_i,c_j)(c_{j+1} - c_{j}) \end{eqnarray}\]
where \(x_u = \frac{\partial x}{\partial u}\), \(y_u = \frac{\partial y}{\partial u}\), \(x_v = \frac{\partial x}{\partial v}\), \(y_v = \frac{\partial y}{\partial v}\).
Figure 4. Approximating the region \(D_{ij}\) by the parallelogram \(P_{ij}\).

Thus, consider the parallelogram \(P_{ij}\) spanned by the following two vectors:
\[\begin{eqnarray} \mathbf{u} &=& (x_u(a_i,c_i)(a_{i+1} - a_{i}),y_u(a_i,c_i)(a_{i+1} - a_{i})),\\ \mathbf{v} &=& (x_v(a_i,c_i)(c_{j+1} - c_{j}),y_v(a_i,c_i)(c_{j+1} - c_{j})). \end{eqnarray}\]
We can approximate the area of \(D_{ij}\) by that of \(P_{ij}\) (Fig. 4), that is,
\[\mu(D_{ij}) \approx \mu(P_{ij}).\]
Now, recall the fact that the area of the parallelogram spanned by \(\mathbf{u} = (a, b)\) and \(\mathbf{v} = (c, d)\) is given by \(|ad - bc|\). Therefore, the area of \(P_{ij}\) is given by
\[\begin{eqnarray} \mu(P_{ij}) &=& |x_u(a_i,c_i)(a_{i+1} - a_{i})\cdot y_v(a_i,c_i)(c_{j+1} - c_{j}) \nonumber\\ && - y_u(a_i,c_i)(a_{i+1} - a_{i})\cdot x_v(a_i,c_i)(c_{j+1} - c_{j})|\\ &=& |x_u(a_i,c_i)y_v(a_i,c_j) - y_u(a_i,c_j)x_v(a_i,c_j)|(a_{i+1} - a_{i})(c_{j+1} - c_{j})\nonumber\\ \\ &=& |J(a_i,c_j)|\mu(E_{ij}). \end{eqnarray}\]
Thus, if we take the limit of infinitesimally fine partitions of \(E_{ij}\), the sums in (Eq:intD) and (Eq:intE) will converge to the same value that is \(\iint_Df(x,y)dxdy\).

Example. Let us calculate \(\iint_{D}(x+y)dxdy\) where\(D\) is the square region surrounded by the following four lines
\[\begin{eqnarray*} y &=& \frac{1}{2}x,\\ y &=& \frac{1}{2}x + \frac{5}{2},\\ y &=& -2x,\\ y &=& -2x + 5. \end{eqnarray*}\]
Observe that the four vertices of this square are \((0,0)\), \((2,1)\), \((1, 3)\), and \((-1, 2)\) (Draw a figure!). Based on this,  any point \((x,y)\) in the square can be represented as a linear combination of the two vectors \((2, 1)\) and \((-1, 2)\):
\[(x,y) = u(2,1) + v(-1, 2) = (2u -v, u + 2v), ~ 0\leq u, v \leq 1.\]
Thus, consider the change of variables
\[\begin{eqnarray*}x(u,v) &=& 2u - v,\\y(u,v) &=& u + 2v.\end{eqnarray*}\]
Then, the square region \(E = [0,1]\times [0, 1]\) on the \((u,v)\) plane is mapped to \(D\).
The Jacobian determinant is
\[\begin{eqnarray*} J(u,v) = \begin{vmatrix} x_u(u,v) & x_v(u,v)\\ y_u(u,v) & y_v(u,v) \end{vmatrix} = \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} = 5. \end{eqnarray*}\]
Thus, we have
\[\begin{eqnarray*} \iint_D(x+y)dxdy &=& \iint_{E}\{(2u-v) + (u + 2v)\}5dudv\\ &=& 5\int_0^1\left(\int_0^1(3u +v)dv\right)du\\ &=&5\int_0^1\left[3uv + \frac{1}{2}v^2\right]_{v=0}^{v=1}du\\ &=&5\int_0^1\left(3u + \frac{1}{2}\right)du\\ &=&5\left[\frac{3}{2}u^2 + \frac{1}{2}u\right]_{u=0}^{u=1}\\ &=& 10. \end{eqnarray*}\]
□
In general, an integral over the parallelogram region \(D\) with vertices \((0,0)\), \((a,b)\), \((c,d)\) and \((a + c, b + d)\) can be transformed to an integral over \(E = [0,1]\times [0,1]\) by the linear transformation
\[\begin{eqnarray} x(u,v) &=& au + cv,\\ y(u,v) &=& bu + dv. \end{eqnarray}\]
In such a case, we have
\[\iint_Df(x,y)dxdy = \iint_Ef(au+cv, bu + dv)|ad - bc|du dv\]

Another important change of variables is the polar coordinates:
\[\begin{eqnarray} x(r, \theta) &=& r\cos\theta,\\ y(r, \theta) &=& r\sin\theta. \end{eqnarray}\]
Then, the Jacobian determinant is
\[\begin{eqnarray} J(r, \theta) &= & \begin{vmatrix} \cos\theta & \sin\theta\\ -r\sin\theta & r\cos\theta \end{vmatrix} = r(\cos^2\theta + \sin^2\theta) = r. \end{eqnarray}\]

Example. Given
\[D = \{(x,y)\mid x\geq 0, y\geq 0, x^2 + y^2 \leq a^2\}\]
where \(a > 0\), let us find
\[\iint_D(x^2 + y^2)dxdy.\]
Using the polar coordinates \(x = r\cos\theta, y = r\sin \theta\), we have \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \(\left[0, \frac{\pi}{2}\right]\), respectively. Noting that \(x^2 + y^2 = r^2\), we have
\[\begin{eqnarray*} \iint_D(x^2 + y^2)dxdy &=& \int_0^a\left(\int_0^{\frac{\pi}{2}}r^2\cdot r d\theta\right)dr\\ &=&\frac{\pi}{2}\int_0^ar^3dr\\ &=& \frac{\pi}{2}\left[\frac{r^4}{4}\right]_0^a\\ &=&\frac{a^4\pi}{8}. \end{eqnarray*}\]
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