Change of variables in multiple integrals

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 $\mathrm{\Phi }$

$$\mathrm{\Phi }(u,v)=(x(u,v),y(u,v))$$

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

$$\mathrm{\Phi }(E)=D.$$

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

We further assume the following:

1. $\mathrm{\Phi }(u,v)$ 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 $$\frac{\partial x}{\partial u}(u_0,v_0)\frac{\partial y}{\partial v}(u_0,v_0)-\frac{\partial x}{\partial v}(u_0,v_0)\frac{\partial y}{\partial u}(u_0,v_0)\ne 0.$$ That is, the determinant of the Jacobian of $\mathrm{\Phi }$ is non-zero in $E$.

See also: Differentiation of more general maps ${\mathbb{R}}^n\to {\mathbb{R}}^m$ for Jacobians.

Let $f(x,y)$ be a continuous function on $D$. By composing with $\mathrm{\Phi }$, we have the function

$$(f\circ \mathrm{\Phi })(u,v)=f(x(u,v),y(u,v))$$

on $E$.

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

The following holds:

$$\begin{array}{}\text{(Eq:chvar)}& {\iint }_{D}f(x,y)dxdy={\iint }_{E}f(x(u,v),y(u,v))|J(u,v)|dudv\end{array}$$

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 $q_{ij}=\mathrm{\Phi }(p_{ij})=(x(a_i,c_j),y(a_i,c_j))$. 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 ${\iint }_{D}f(x,y)dxdy$ is approximated by the sum

$$\begin{array}{}\text{(Eq:intD)}& {\iint }_{D}f(x,y)dxdy\approx \sum _{i=0}^{n-1}\sum _{j=0}^{m-1}f(x(a_i,c_j),y(a_i,c_j))\mu (D_{ij})\end{array}$$

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

$$\begin{array}{}\text{(Eq:intE)}& {\iint }_{E}f(x(u,v),y(u,v))|J(u,v)|dudv\approx \sum _{i=0}^{n-1}\sum _{j=0}^{m-1}f(x(a_i,c_j),y(a_i,c_j))|J(a_i,c_j)|\mu (E_{ij})\end{array}$$

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

$$\begin{array}{rcl}\overrightarrow{q_{i,j}{q}_{i+1,j}}& =& (x(a_{i+1},c_j)-x(a_i,c_j),y(a_{i+1},c_j)-y(a_i,c_j)),\\ \overrightarrow{q_{i,j}{q}_{i,j+1}}& =& (x(a_i,c_{j+1})-x(a_i,c_j),y(a_i,c_{j+1})-y(a_i,c_j)).\end{array}$$

We further approximate each component of these vectors by using the first-order Taylor expansion (Figure 4):

$$\begin{array}{rcl}x(a_{i+1},c_j)-x(a_i,c_j)& \approx & x_u(a_i,c_j)(a_{i+1}-a_i),\\ y(a_{i+1},c_j)-y(a_i,c_j)& \approx & y_u(a_i,c_j)(a_{i+1}-a_i),\\ x(a_i,c_{j+1})-x(a_i,c_j)& \approx & x_v(a_i,c_j)(c_{j+1}-c_j),\\ y(a_i,c_{j+1})-y(a_i,c_j)& \approx & y_v(a_i,c_j)(c_{j+1}-c_j)\end{array}$$

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:

$$\begin{array}{rcl}\mathbf{u}& =& (x_u(a_i,c_i)(a_{i+1}-a_i),y_u(a_i,c_i)(a_{i+1}-a_i)),\\ \mathbf{v}& =& (x_v(a_i,c_i)(c_{j+1}-c_j),y_v(a_i,c_i)(c_{j+1}-c_j)).\end{array}$$

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

$$\begin{array}{rcl}\mu (P_{ij})& =& |x_u(a_i,c_i)(a_{i+1}-a_i)\cdot y_v(a_i,c_i)(c_{j+1}-c_j)\\ & & -y_u(a_i,c_i)(a_{i+1}-a_i)\cdot x_v(a_i,c_i)(c_{j+1}-c_j)|\\ & =& |x_u(a_i,c_i)y_v(a_i,c_j)-y_u(a_i,c_j)x_v(a_i,c_j)|(a_{i+1}-a_i)(c_{j+1}-c_j)\\ \\ & =& |J(a_i,c_j)|\mu (E_{ij}).\end{array}$$

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 ${\iint }_{D}f(x,y)dxdy$.

Example. Let us calculate ${\iint }_D(x+y)dxdy$ where$D$ is the square region surrounded by the following four lines

$$\begin{array}{rcl}y& =& \frac{1}{2}x,\\ y& =& \frac{1}{2}x+\frac{5}{2},\\ y& =& -2x,\\ y& =& -2x+5.\end{array}$$

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 $(-1,2)$:

$$(x,y)=u(2,1)+v(-1,2)=(2u-v,u+2v),0\le u,v\le 1.$$

Thus, consider the change of variables

$$\begin{array}{rcl}x(u,v)& =& 2u-v,\\ y(u,v)& =& u+2v.\end{array}$$

Then, the square region $E=[0,1]\times [0,1]$ on the $(u,v)$ plane is mapped to $D$.

The Jacobian determinant is

$$\begin{array}{r}J(u,v)=\left|\begin{array}{cc}{x}_u(u,v)& x_v(u,v)\\ y_u(u,v)& y_v(u,v)\end{array}\right|=\left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|=5.\end{array}$$

Thus, we have

$$\begin{array}{rcl}{\iint }_D(x+y)dxdy& =& {\iint }_E\{(2u-v)+(u+2v)\}5dudv\\ & =& 5{\int }_0^1({\int }_0^1(3u+v)dv)du\\ & =& 5{\int }_0^1{[3uv+\frac{1}{2}{v}^2]}_{v=0}^{v=1}du\\ & =& 5{\int }_0^1(3u+\frac{1}{2})du\\ & =& 5{[\frac{3}{2}{u}^2+\frac{1}{2}u]}_{u=0}^{u=1}\\ & =& 10.\end{array}$$

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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

$$\begin{array}{rcl}x(u,v)& =& au+cv,\\ y(u,v)& =& bu+dv.\end{array}$$

In such a case, we have

$${\iint }_{D}f(x,y)dxdy={\iint }_{E}f(au+cv,bu+dv)|ad-bc|dudv$$

Another important change of variables is the polar coordinates:

$$\begin{array}{rcl}x(r,\theta )& =& r\cos \theta ,\\ y(r,\theta )& =& r\sin \theta .\end{array}$$

Then, the Jacobian determinant is

$$\begin{array}{rcl}J(r,\theta )& =& \left|\begin{array}{cc}\cos \theta & \sin \theta \\ -r\sin \theta & r\cos \theta \end{array}\right|=r({\cos }^2\theta +{\sin }^2\theta )=r.\end{array}$$

Example. Given

$$D=\{(x,y)\mid x\ge 0,y\ge 0,x^2+y^2\le a^2\}$$

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 $[0,\frac{\pi }{2}]$, respectively. Noting that $x^2+y^2=r^2$, we have

$$\begin{array}{rcl}{\iint }_D(x^2+y^2)dxdy& =& {\int }_0^a({\int }_0^{\frac{\pi }{2}}{r}^2\cdot rd\theta )dr\\ & =& \frac{\pi }{2}{\int }_0^a{r}^{3}dr\\ & =& \frac{\pi }{2}{\left[\frac{r^4}{4}\right]}_0^a\\ & =& \frac{a^4\pi }{8}.\end{array}$$

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