example 1

By changing the order of integration, evaluate \[ \int^a_0 \int_{0}^{(a^2-x^2)^{1/2}} (a^2-y^2)^{1/2} {\it dy}\,{\it dx} . \]

solution Note that \(x\) varies between 0 and \(a\), and for each such fixed \(x\), we have \(0\leq y\leq (a^2-x^2)^{1/2}\). Thus, the iterated integral is equivalent to the double integral \[ \intop\!\!\!\intop\nolimits_{D}\, (a^2-y^2)^{1/2} {\it dy}\,{\it dx} , \] where \(D\) is the set of points \((x, y)\) such that \(0\leq x\leq a\) and \(0\leq y\leq (a^2-x^2)^{1/2}\). But this is the representation of one-quarter (the positive quadrant portion) of the disk of radius \(a\); hence, \(D\) can also be described as the set of points \((x, y)\) satisfying \[ 0\leq y\leq a, 0\leq x\leq (a^2-y^2)^{1/2} \] (see Figure 15.28). Thus, \begin{eqnarray*} \int_0^a \int_0^{(a^2-x^2)^{1/2}} (a^2-y^2)^{1/2} {\it dy}\,{\it dx} &=& \int_0^a \bigg[ \int_0^{(a^2-y^2)^{1/2}} (a^2-y^2)^{1/2} {\it dx} \bigg]\, {\it dy}\\[4pt] &=& \int_0^a \big[x(a^2-y^2)^{1/2}\big]_{x=0}^{(a^2-y^2)^{1/2}} {\it dy} \\[4pt] &=& \int_0^a (a^2-y^2)\, {\it dy} =\bigg[a^2y-\frac{y^3}{3}\bigg]^a_0 =\frac{2a^3}{3}. \end{eqnarray*}

Figure 15.28: The positive-quadrant portion of a disk of radius \(a\;\).

example 2

Evaluate \[ \int_1^2 \int_0^{\log x} (x-1) \sqrt{1+e^{2y}}\ {\it dy}\, {\it dx} . \]

solution It will simplify matters if we first interchange the order of integration. First, notice that the integral is equal to \({\intop\!\!\!\intop}_D\, (x-1)\sqrt{1+e^{2y}}\ {\it dA}\), where \(D\) is the set of \((x, y)\) such that \[ 1\leq x\leq 2\qquad\hbox{and}\qquad 0\leq y\leq \log x. \]

The region \(D\) is simple (see Figure 15.29) and can also be described by \[ 0\leq y\leq \log 2\qquad\hbox{and}\qquad e^y\leq x\leq 2. \]

Thus, the given iterated integral is equal to \begin{equation*} \int_0^{\log 2}\int_{e^y}^2 (x-1)\sqrt{1+e^{2y}}\ {\it dx}\, {\it dy} = \int_0^{\log 2} \sqrt{1+e^{2y}}\, \bigg[ \int^2_{e^y} (x-1)\, {\it dx} \bigg]\, {\it dy}\\[4pt] \quad = \int_0^{\log 2}\sqrt{1+e^{2y}}\, \bigg[\frac{x^2}{2}-x\bigg]_{e^y}^2 {\it dy} \\[4pt] \quad = -\int_0^{\log 2}\bigg(\frac{e^{2y}}{2}-e^y\bigg)\sqrt{1+e^{2y}} {\it dy} \\[4pt] \quad = {-}\frac{1}{2}\int_0^{\log 2} e^{2y} \sqrt{1+e^{2y}} {\it dy}+ \int_0^{\log 2} e^y \sqrt{1+e^{2y}} {\it dy}.\tag{1} \end{equation*}

In the first integral in expression (1), we substitute \(u=e^{2y}\), and in the second, \(v=e^y\). Hence, we obtain \begin{equation*} -\frac{1}{4}\int_1^4 \sqrt{1+u} {\,d} u+\int_1^2 \sqrt{1+v^2} {\,d} v.\tag{2} \end{equation*}

Both integrals in expression (2) are easily found with techniques of one-variable calculus (or by consulting the table of integrals at the back of the book). For the first integral, we get \begin{equation*} \frac{1}{4}\int_1^4\sqrt{1+u} {\,d} u=\bigg[ \frac{1}{6}(1+u)^{3/2}\bigg]^4_1 = \frac{1}{6} [(1+4)^{3/2}-2^{3/2} ] = \frac{1}{6} [5^{3/2}-2^{3/2} ].\tag{3} \end{equation*}

The second integral is \begin{equation*} \begin{array}{lll} \int_1^2 \sqrt{1+v^2} {\,d} v &=& \frac{1}{2}\left[v\sqrt{1+v^2}+\log\ (\sqrt{1+v^2}+v)\right]^2_1\\[6pt] &=& \frac{1}{2}\left[2\sqrt{5}+\log\ (\sqrt{5}+2)\right]- \frac{1}{2}\left[\sqrt{2}+\log\ (\sqrt{2}+1)\right] \end{array}\tag{4} \end{equation*} (see formula 43 in the table of integrals at the back of the book). Finally, we subtract equation (3) from equation (4) to obtain the answer \[ \frac{1}{2}\bigg(2\sqrt{5}-\sqrt{2} +\log \frac{\sqrt{5}+2}{\sqrt{2}+1}\bigg) -\frac{1}{6}\big[5^{3/2}-2^{3/2}\big]. \]

example 3

Consider the integral \[ \intop\!\!\!\intop\nolimits_{D}\frac{1}{\sqrt{1+x^6+y^8}}\,{\it dx}\,{\it dy}, \] where \(D\) is the unit square \([0, 1] \times [0, 1]\). Because the integrand satisfies, for \(x\) and \(y\) between 0 and 1, \[ \frac{1}{\sqrt{3}}\leq\frac{1}{\sqrt{1+x^6+y^8}}\leq 1, \] and because the square has area 1, we get: \[ \frac{1}{\sqrt{3}}\leq\intop\!\!\!\intop\nolimits_{D}\frac{1}{\sqrt{1+x^6+y^8}}\,{\it dx}\,{\it dy}\leq 1. \]

Theorem 5 Mean-Value Theorem: Double Integrals

Suppose \(f\colon D\to {\mathbb R}\) is continuous and \(D\) is an elementary region. Then for some point \((x_0, y_0)\) in \(D\) we have \[ \intop\!\!\!\intop\nolimits_{D} \, f(x,y)\, {\it dA}=f(x_0,y_0) A(D), \] where \(A(D)\) denotes the area of \(D\).

proof

We cannot prove this theorem with complete rigor, because it requires some concepts about continuous functions not proved in this course; but we can sketch the main ideas that underlie the proof.

Because \(f\) is continuous on \(D\), it has a maximum value \(M\) and a minimum value \(m\). Thus, \(m\leq f(x,y) \leq M\) for all \((x,y)\in D\). Furthermore, \(f(x_1,y_1)=m\) and \(f (x_2,y_2)=M\) for some pairs \((x_1, y_1)\) and \((x_2, y_2)\) in \(D\).

Dividing through inequality (5) by \(A(D)\), we get \begin{equation*} m\leq \frac{1}{A(D)}\intop\!\!\!\intop\nolimits_{D}\,f(x,y)\, {\it dA}\leq M.\tag{6} \end{equation*}

Because a continuous function on \(D\) takes on every value between its maximum and minimum values (this is the two-variable intermediate-value theorem proved in advanced calculus; see also Review Exercise 32), and because the number \([1/A(D)]{\intop\!\!\!\intop}_D f(x,y)\ {\it dA}\) is, by inequality (6), between these values, there must be a point \((x_0,y_0)\in D\) with \[ f(x_0,y_0)=\frac{1}{A(D)}\intop\!\!\!\intop\nolimits_{D}f(x,y) \,{\it dA}, \] which is precisely the conclusion of Theorem 5.