Iterated integral on a bounded closed set

We saw the method of iterative integral for a rectangular region. We now extend the method to the case of non-rectangular regions.

See also: Iterated integral on a rectangular region

Consider two continuous functions \(y = \varphi(x)\) and  \(y = \psi(x)\) on \([a,b]\) such that \(\psi(x) < \varphi(x)\) for all \(x \in [a,b]\). We can define the following region \(D\) in \(\mathbb{R}^2\) sandwiched by these functions:

\[D = \{ (x,y) \mid a \leq x \leq b, \psi(x) \leq y \leq \varphi(x)\}.\tag{Eq:Dsand}\]

Clearly, \(D\) is a closed Jordan region. Let \(f(x,y)\) be a continuous function on \(D\). Then, \(f(x,y)\) is integrable on \(D\). We can calculate the integral of \(f(x,y)\) on \(D\) by the method of iterated integral.

See also: Multiple integral on a bounded closed set for the explanation of closed Jordan regions.

First, fix an \(x \in [a,b]\), and define the univariate function \(F_1(x)\) by 

\[F_1(x) = \int_{\psi(x)}^{\varphi(x)}f(x,y)dy.\tag{Eq:F1}\]

Lemma 

The function \(F_1(x)\) defined by (Eq:F1) is continuous on \([a,b]\).

Proof. We show that \(F_1(x)\) is continuous at an arbitrary \(x_0 \in [a,b]\). Let \(\varepsilon\) be any positive real number. Since \(\varphi(x)\) and \(\psi(x)\) are continuous, there exists a \(\delta > 0\) such that, if \(|x - x_0|< \delta\), then \(|\psi(x) - \psi(x_0)|\) and \(|\varphi(x) - \varphi(x_0)|\) are arbitrarily small. Therefore, with \(c = \max\{\psi(x), \psi(x_0)\}\) and \(d = \min \{\varphi(x), \varphi(x_0)\}\), we may assume that the following inequalities hold:

\[\begin{eqnarray} \left|F_1(x) - \int_c^df(x,y)dy\right| & < & \frac{\varepsilon}{3},\\ \left|F_1(x_0) - \int_c^df(x_0,y)dy\right| & < & \frac{\varepsilon}{3}. \end{eqnarray}\]

Noting that \(f(x,y)\) is uniformly continuous on \(D\), we can redefine \(\delta\) to be sufficiently small so that, for all \((x,y), (x',y')\in D\), if \(|x-x'|< \delta\) and \(|y - y'|< \delta\), then

\[|f(x,y) - f(x',y')| < \frac{\varepsilon}{3(d-c)}.\]

Then, for all \(x \in [a,b]\), if \(|x - x_0|< \delta\), then

\[\begin{eqnarray*} |F_1(x) - F_1(x_0)| &=&\left|F_1(x) - \int_c^df(x,y)dy + \int_c^df(x,y)dy\right.\\ && \left. -\int_c^df(x_0,y)dy +\int_c^df(x_0,y)dy - F_1(x_0)\right|\\ &\leq&\left|F_1(x) - \int_c^df(x,y)dy\right|\\ && + \int_c^d\left|f(x,y)dy -f(x_0,y)\right|dy +\left|\int_c^df(x_0,y)dy - F_1(x_0)\right|\\ &<& \frac{\varepsilon}{3} + \int_c^d\frac{\varepsilon}{3(d-c)}dy + \frac{\varepsilon}{3}\\ &=& \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{eqnarray*}\]

Therefore, \(F_1(x)\) is continuous at \(x_0\). ■

Since \(F_1(x)\) is continuous, it is integrable. The integral

\[\int_a^bF_1(x)dx = \int_a^b\left(\int_{\psi(x)}^{\varphi(x)}f(x,y)dy\right)dx\]

is equal to the double integral \(\iint_Df(x,y)dxdy\). That is,

Theorem

With the region \(D\) as defined in (Eq:Dsand) and \(f(x,y)\) a continuous function on \(D\), we have

\[\iint_Df(x,y)dxdy = \int_a^b\left(\int_{\psi(x)}^{\varphi(x)}f(x,y)dy\right)dx.\]

Proof. Let \(R = [a,b]\times[c,d]\) be a region such that \(D \subset R\). Define the following bounded function on \(R\):

\[\tilde{f}(x,y) = \left\{ \begin{array}{cc} f(x,y) & ((x,y) \in D),\\ 0 & ((x,y) \not\in D). \end{array} \right.\]

Let \(\Delta\) be a partition

\[\Delta: \left\{ \begin{array}{ccc} a =& a_0 < a_1 < \cdots < a_{n-1} < a_{n} &= b,\\ c =& c_0 < c_1 < \cdots < c_{m-1} < c_{m} &= d. \end{array} \right.\]

Consider the \(nm\) rectangular regions \(D_{ij} = [a_i,a_{i+1}]\times [c_j, c_{j+1}]\) \((i = 0, 1, \cdots, n-1; j = 0, 1, \cdots, m-1)\) and let

\[\begin{eqnarray*} M_{ij}&=& \sup\{\tilde{f}(x,y) \mid (x,y) \in D_{ij}\},\\ m_{ij}&=& \inf\{\tilde{f}(x,y) \mid (x,y) \in D_{ij}\}. \end{eqnarray*}\]

Thus, if \((x,y)\in D_{ij}\), then \(m_{ij} \leq \tilde{f}(x,y) \leq M_{ij}\). Integrating these by \(y\) from \(c_j\) to \(c_{j+1}\), we have

\[m_{ij}(c_{j+1} - c_{j}) \leq \int_{c_{j}}^{c_{j+1}}\tilde{f}(x,y)dy \leq M_{ij}(c_{j+1} - c_{j}).\]

Integrating these by \(x\) from \(a_{i}\) to \(a_{i+1}\), we have

\[m_{ij}(a_{i+1} - a_{i})(c_{j+1} - c_{j}) \leq \int_{a_{i}}^{a_{i+1}}\left(\int_{c_{j}}^{c_{j+1}}\tilde{f}(x,y)dy\right)dx \leq M_{ij}(a_{i+1} - a_{i})(c_{j+1} - c_{j}).\]

Summing these over \(i\) and \(j\), we have

\[\begin{eqnarray} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}m_{ij}(a_{i+1} - a_{i})(c_{j+1} - c_{j}) &\leq & \int_{a}^{b}\left(\int_{c}^{d}\tilde{f}(x,y)dy\right)dx\nonumber\\ &\leq& \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}M_{ij}(a_{i+1} - a_{i})(c_{j+1} - c_{j}).\tag{Eq:mMij} \end{eqnarray}\]

Since \(f(x,y)\) is integrable on \(D\), the left-most-hand side (lower Riemann sum) and the right-most-hand side (upper Riemann sum) of (Eq:mMij) converge to the same value, namely, \(\iint_Df(x,y)dxdy\). On the other hand, by the definition of \(\tilde{f}(x,y)\), we have

\[\int_{c}^{d}\tilde{f}(x,y)dy = \int_{\psi(x)}^{\varphi(x)}f(x,y)dy\tag{Eq:iinty}\]

for all \(x \in [a,b]\). Substituting this into (Eq:mMij) and applying the Squeeze Theorem, we have

\[\iint_Df(x,y)dxdy = \int_{a}^{b}\left(\int_{c}^{d}\tilde{f}(x,y)dy\right)dx = \int_{a}^{b}\left(\int_{\psi(x)}^{\varphi(x)}f(x,y)dy\right)dx.\]

■

Example. Let us calculate \(\iint_D(x+y)dxdy\) where \(D = \{(x,y) \mid x \geq 0, y \geq 0, x + y \leq 2\}\). (Draw a picture!)

\[\begin{eqnarray*} \iint_D(x+y)dxdy &=& \int_0^2\left(\int_0^{2 - x}(x+y)dy\right)dx\\ &=& \int_0^2\left[xy + \frac{y^2}{2}\right]_{y = 0}^{y=2 - x}dx\\ &=&\int_0^2\left(2 - \frac{x^2}{2}\right)dx\\ &=&\left[2x - \frac{x^3}{6}\right]_{x=0}^{x=2} = \frac{8}{3}. \end{eqnarray*}\]

□

Example. Let \(D = \{(x,y) \mid 0 \leq x \leq 1, 0 \leq y \leq x^2\}\) (Draw a picture!) and \(f(x,y)\) be a continuous function on \(D\). Then,

\[\begin{eqnarray*} \iint_Df(x,y)dxdy &=& \int_0^1\left(\int_0^{x^2}f(x,y)dy\right)dx\\ &=&\int_0^1\left(\int_{\sqrt{y}}^{1}f(x,y)dx\right)dy. \end{eqnarray*}\]

□