In the previous section, we restricted our attention to rectangular domains. Now we shall treat the more general case of domains \({\mathcal{D}}\) whose boundaries are simple closed curves (a curve is simple if it does not intersect itself). We assume that the boundary of \({\mathcal{D}}\) is smooth as in Figure 15.15 or consists of finitely many smooth curves, joined together with possible corners, as in Figure 15.15. A boundary curve of this type is called piecewise smooth. We also assume that \({\mathcal{D}}\) is a closed domain; that is, \({\mathcal{D}}\) contains its boundary.
Fortunately, we do not need to start from the beginning to define the double integral over a domain \({\mathcal{D}}\) of this type. Given a function \(f(x,y)\) on \({\mathcal{D}}\), we choose a rectangle \({\mathcal{R}} = [a,b]\times[c,d]\) containing \({\mathcal{D}}\) and define a new function \(\tilde f (x,y)\) that agrees with \(f(x,y)\) on \({\mathcal{D}}\) and is zero outside of \({\mathcal{D}}\) Figure 15.16: \[ \tilde{f}(x,y) = \begin{cases} f(x,y) & \textrm{if }(x,y)\in{\mathcal{D}} \\ 0 & \textrm{if }(x,y)\notin{\mathcal{D}} \end{cases} \]
The double integral of \(f\) over \({\mathcal{D}}\) is defined as the integral of \(\tilde{f}\) over \({\mathcal{R}}\): \begin{equation*} \boxed{ \iint_{{\mathcal{D}}\,} f(x,y)\,dA = \iint_{{\mathcal{R}}\,} \tilde{f}(x,y)\,dA}\tag{1} \end{equation*}
We say that \(f\) is integrable over \({\mathcal{D}}\) if the integral of \(\tilde{f}\) over \({\mathcal{R}}\) exists. The value of the integral does not depend on the particular choice of \({\mathcal{R}}\) because \(\tilde{f}\) is zero outside of \({\mathcal{D}}\).
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This definition seems reasonable because the integral of \(\tilde{f}\) only “picks up” the values of \(f\) on \({\mathcal{D}}\). However, \(\tilde{f}\) is likely to be discontinuous because its values jump suddenly to zero beyond the boundary. Despite this possible discontinuity, the next theorem guarantees that the integral of \(\tilde{f}\) over \({\mathcal{R}}\) exists if our original function \(f\) is continuous.
If \(f(x,y)\) is continuous on a closed domain \({\mathcal{D}}\) whose boundary is a closed, simple, piecewise smooth curve, then \(\iint_{{\mathcal{D}}\,} f(x,y)\,dA\) exists.
In Theorem 1, we define continuity on \({\mathcal{D}}\) to mean that \(f\) is defined and continuous on some open set containing \({\mathcal{D}}\).
As in the previous section, the double integral defines the signed volume between the graph of \(f(x,y)\) and the \(xy\)-plane, where regions below the \(xy\)-plane are assigned negative volume.
We can approximate the double integral by Riemann sums for the function \(\tilde{f}\) on a rectangle \({\mathcal{R}}\) containing \({\mathcal{D}}\). Because \(\tilde{f}(P) = 0\) for points \(P\) in \({\mathcal{R}}\) that do not belong to \({\mathcal{D}}\), any such Riemann sum reduces to a sum over those sample points that lie in \({\mathcal{D}}\): \begin{equation*} \iint_{{\mathcal{D}}\,} f(x,y)\,dA \approx \sum_{i=1}^N\sum_{j=1}^M \tilde{f}(P_{ij})\,\Delta x_i\,\Delta y_j = \underbrace{\sum f(P_{ij})\,\Delta x_i\,\Delta y_j}_{\mbox{Sum only over points} \atop {P_{ij}\mbox{ that lie in }{\mathcal{D}}}}\tag{2} \end{equation*}
Compute \(S_{4,4}\) for the integral \(\iint_{{\mathcal{D}}\,} (x+y)\,dA\), where \({\mathcal{D}}\) is the shaded domain in Figure 15.17. Use the upper right-hand corners of the squares as sample points.
Solution Let \(f(x,y) = x+y\). The subrectangles in Figure 15.17 have sides of length \(\Delta x = \Delta y = \frac12\) and area \(\Delta A = \frac14\). Only \(7\) of the 16 sample points lie in \({\mathcal{D}}\), so \[ \begin{array}{rcl} S_{4,4} &=& \sum_{i=1}^4\sum_{j=1}^4 \tilde{f}(P_{ij})\,\Delta x\,\Delta y=\frac14\big(f(0.5,0.5)+f(1,0.5) +f(0.5,1)+f(1,1)\\ &&+f(1.5,1)+f(1,1.5)+f(1.5,1.5)\big)\\ &=&\frac14\bigl(1+1.5 +1.5+2 +2.5 +2.5+3\bigr) =\frac72 \end{array} \]
The linearity properties of the double integral carry over to general domains: If \(f(x,y)\) and \(g(x,y)\) are integrable and \(C\) is a constant, then \[ \begin{array}{rl} \iint_{{\mathcal{D}}\,} (f(x,y)+g(x,y))\,dA &=\iint_{{\mathcal{D}}\,} f(x,y) \,dA + \iint_{{\mathcal{D}}\,} g(x,y) \,dA\\ \iint_{{\mathcal{D}}\,} Cf(x,y) \,dA &= C\iint_{{\mathcal{D}}\,} f(x,y) \,dA \end{array} \]
Although we usually think of double integrals as representing volumes, it is worth noting that we can express the area of a domain \({\mathcal{D}}\) in the plane as the double integral of the constant function \(f(x,y)=1\): \begin{equation*} \boxed{\textrm{Area}({\mathcal{D}}) = \iint_{{\mathcal{D}}\,} 1\,dA}\tag{3} \end{equation*}
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Indeed, as we see in Figure 15.18, the the area of \({\mathcal{D}}\) is equal to the volume of the “cylinder” of height \(1\) with \({\mathcal{D}}\) as base. More generally, for any constant \(C\), \begin{equation*} \iint_{{\mathcal{D}}\,} C\,dA = C \,\textrm{Area}({\mathcal{D}})\tag{4} \end{equation*}
Eq. (3) tells us that we can approximate the area of a domain \({\mathcal{D}}\) by a Riemann sum for \(\iint_{{\mathcal{D}}\,} 1\,dA\). In this case, \(f(x,y)=1\), and we obtain a Riemann sum by adding up the areas \(\Delta x_i\,\Delta y_j\) of those rectangles in a grid that are contained in \({\mathcal{D}}\) or that intersects the boundary of \({\mathcal{D}}\) Figure 15.19. The finer the grid, the better the approximation. The exact area is the limit as the sides of the rectangles tend to zero.
When \({\mathcal{D}}\) is a region between two graphs in the \(xy\)-plane, we can evaluate double integrals over \({\mathcal{D}}\) as iterated integrals. We say that \({\mathcal{D}}\) is vertically simple if it is the region between the graphs of two continuous functions \(y={g_1}(x)\) and \(y = g_2(x)\) (Figure 15.20): \[ {\mathcal{D}} = \{(x,y): a\le x \le b,\quad {g_1}(x) \le y \le g_2(x)\} \]
Similarly, \({\mathcal{D}}\) is horizontally simple if \[ {\mathcal{D}} = \{(x,y): c\le y \le d,\quad {g_1}(y) \le x \le g_2(y)\} \]
When you write a double integral over a vertically simple region as an iterated integral, the inner integral is an integral over the dashed segment shown in Figure 15.20. For a horizontally simple region, the inner integral is an integral over the dashed segment shown in Figure 15.20.
If \({\mathcal{D}}\) is vertically simple with description \[ a\le x \le b,\qquad {g_1}(x) \le y \le g_2(x) \] then \[ \boxed{\iint_{{\mathcal{D}}\,} f(x,y)\, dA = \int_a^b\int_{{g_1}(x)}^{g_2(x)} f(x,y)\,dy\,dx} \]
If \({\mathcal{D}}\) is a horizontally simple region with description \[ c\le y \le d,\qquad {g_1}(y) \le x \le g_2(y) \] then \[ \boxed{\iint_{{\mathcal{D}}\,} f(x,y)\, dA = \int_c^d\int_{{g_1}(y)}^{g_2(y)} f(x,y)\,dx\,dy} \]
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Proof We sketch the proof, assuming that \({\mathcal{D}}\) is vertically simple (the horizontally simple case is similar). Choose a rectangle \({\mathcal{R}} = [a,b]\times[c,d]\) containing \({\mathcal{D}}\). Then \begin{equation*} \iint_{{\mathcal{D}}\,} f(x,y)\, dA = \int_a^b \int_c^d \tilde f(x,y)\,dy\,dx\tag{5} \end{equation*}
Although \(\tilde f\) need not be continuous, the use of Fubini’s Theorem in Eq. (5) can be justified. In particular, the integral \(\int_c^d \tilde f(x,y)\,dy\) exists and is a continuous function of \(x\).
By definition, \(\tilde f(x,y)\) is zero outside \({\mathcal{D}}\), so for fixed \(x\), \(\tilde f(x,y)\) is zero unless \(y\) satisfies \({g_1}(x)\le y \le g_2(x)\). Therefore, \[ \int_c^d \tilde f(x,y)\,dy = \int_{{g_1}(x)}^{g_2(x)} f(x,y)\,dy \]
Substituting in Eq. (5), we obtain the desired equality: \[ \iint_{{\mathcal{D}}\,} f(x,y)\, dA = \int_a^b\int_{{g_1}(x)}^{g_2(x)} f(x,y)\,dy\,dx \]
Integration over a simple region is similar to integration over a rectangle with one difference: The limits of the inner integral may be functions instead of constants.
Evaluate \(\iint_{{\mathcal{D}}\,}x^2y\,dA\), where \({\mathcal{D}}\) is the region in Figure 15.21.
Solution
Step 1. Describe \({\mathcal{D}}\) as a vertically simple region.
\[ \underbrace{\,\,1\le x \le 3\,\,}_{\mbox{ Limits of outer}\atop \mbox{ integral}},\quad\qquad \underbrace{\,\,\frac1x\le y \le \sqrt{x}\,\,}_{\mbox{Limits of inner}\atop \mbox{ integral}} \] In this case, \(g_1(x)=1/x\) and \(g_2(x)=\sqrt{x}\).
Step 2. Set up the iterated integral.
\[ \iint_{{\mathcal{D}}\,} x^2y\,dA = \int_1^3\int_{y=1/x}^{\sqrt{x}} x^2y\,dy\,dx \] Notice that the inner integral is an integral over a vertical segment between the graphs of \(y={1}/{x}\) and \(y=\sqrt{x}\).
Step 3. Compute the iterated integral.
As usual, we evaluate the inner integral by treating \(x\) as a constant, but now the upper and lower limits depend on \(x\): \[ \int_{y=1/x}^{\sqrt{x}}x^2y\,dy = \frac12x^2y^2 \bigg|_{y=1/x}^{\sqrt{x}} =\frac12x^2(\sqrt{x})^2 - \frac12x^2\left(\frac1{x}\right)^2 = \frac12x^3-\frac12 \]
We complete the calculation by integrating with respect to \(x\): \[ \begin{array}{rl} \iint_{{\mathcal{D}}\,} x^2y\,dA = \int_1^3\left(\frac12x^3-\frac12\right) \,dx = \left(\frac18x^4-\frac12x\right)\bigg|_1^3 \\ &= \frac{69}8 - \left(-\frac38\right)=9 \end{array} \]
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Find the volume \(V\) of the region between the plane \(z=2x+3y\) and the triangle \({\mathcal{D}}\) in Figure 15.22.
Solution The triangle \({\mathcal{D}}\) is bounded by the lines \(y=x/2\), \(y=x\), and \(y=2\). We see in Figure 15.23 that \({\mathcal{D}}\) is vertically simple, but the upper curve is not given by a single formula: The formula switches from \(y=x\) to \(y=2\). Therefore, it is more convenient to describe \({\mathcal{D}}\) as a horizontally simple region Figure 15.23: \[ {\mathcal{D}}: 0\le y \le 2,\quad y \le x \le 2y \]
The volume is equal to the double integral of \(f(x,y)=2x+3y\) over \({\mathcal{D}}\), \[ \begin{array}{rl} V &=\iint_{{\mathcal{D}}\,} f(x,y)\,dA = \int_0^2\int_{x=y}^{2y} (2x+3y)\,dx\,dy \\ &=\int_0^2\big( x^2+3yx\big)\bigg|_{x=y}^{2y}\,dy = \int_0^2\big( (4y^2+6y^2) - (y^2+3y^2) \big)\,dy\\ & =\int_0^2 6y^2 \, dy = 2y^3\bigg|_0^2=16 \end{array} \]
The next example shows that in some cases, one iterated integral is easier to evaluate than the other.
Evaluate \(\iint_{{\mathcal{D}}\,} e^{y^2}dA\) for \({\mathcal{D}}\) in Figure 15.24.
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Solution First, let’s try describing \({\mathcal{D}}\) as a vertically simple domain. Referring to Figure 15.24, we have \[ {\mathcal{D}}: 0\le x \le 4,\quad \dfrac12 x \leq y \leq 2 \quad\Rightarrow\quad \iint_{{\mathcal{D}}\,} e^{y^2}\,dA=\int_{x=0}^4 \int_{y=x/2}^2 e^{y^2}\,dy\,dx \]
The inner integral cannot be evaluated because we have no explicit antiderivative for \(e^{y^2}\). Therefore, we try describing \({\mathcal{D}}\) as horizontally simple [Figure 15.24]: \[ {\mathcal{D}}: 0\le y \le 2,\quad 0 \le x \le 2y \]
This leads to an iterated integral that can be evaluated: \[ \begin{array}{rcl} \int_{0}^2 \int_{x=0}^{2y} e^{y^2}\,dx\,dy &=& \int_{0}^2 \Big( xe^{y^2}\Big|_{x=0}^{2y}\Big) \,dy = \int_{0}^2 2ye^{y^2} \,dy\\ &=& e^{y^2}\Big|_0^2 = e^4-1 \end{array} \]
Sketch the domain of integration \({\mathcal{D}}\) corresponding to \[ \int_1^9\int_{\sqrt y}^3 xe^y\,dx\,dy \]
Then change the order of integration and evaluate.
Solution The limits of integration give us inequalities that describe the domain \({\mathcal{D}}\) (as a horizontally simple region since \(dx\) precedes \(dy\)): \[ 1\le y \le 9,\qquad \sqrt{y} \le x \le 3 \]
We sketch the region in Figure 15.25. Now observe that \({\mathcal{D}}\) is also vertically simple: \[ 1\le x \le 3,\qquad 1 \le y \le x^2 \] so we can rewrite our integral and evaluate: \[ \begin{array}{rl} \int_1^9\int_{x=\sqrt y}^3 xe^y\,dx\,dy&=\int_1^3 \int_{y=1}^{x^2} xe^{y}\,dy\,dx =\int_1^3\left( \int_{y=1}^{x^2} xe^{y}\,dy \right)\,dx\\ &= \int_1^3 \Big(xe^y\Big|_{y=1}^{x^2}\Big)\,dx =\int_1^3 (xe^{x^2}-ex)\,dx = \frac12(e^{x^2}-ex^2)\Big|_1^3\\ & = \frac12(e^9-9e)-0 = \frac12 (e^9 - 9e) \end{array} \]
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In the next theorem, part (a) is a formal statement of the fact that larger functions have larger integrals, a fact that we also noted in the single-variable case. Part (b) is useful for estimating integrals.
Let \(f(x,y)\) and \(g(x,y)\) be integrable functions on \({\mathcal{D}}\).
Proof If \(f(x,y)\le g(x,y)\), then every Riemann sum for \(f(x,y)\) is less than or equal to the corresponding Riemann sum for \(g\): \[ \sum f(P_{ij})\,\Delta x_i\,\Delta y_j \le \sum g(P_{ij})\,\Delta x_i\,\Delta y_j \]
We obtain (6) by taking the limit. Now suppose that \(f(x,y)\le M\) and apply (6) with \(g(x,y)=M\): \[ \iint_{{\mathcal{D}}\,} f(x,y)\,dA \le \iint_{{\mathcal{D}}\,} M\,dA = M\,\textrm{Area}({\mathcal{D}}) \]
This proves half of (7). The other half follows similarly.
Estimate \(\iint_{\mathcal{D}}\,\frac{dA}{\sqrt{x^2+(y-2)^2}}\) where \({\mathcal{D}}\) is the disk of radius 1 centered at the origin.
Solution The quantity \(\sqrt{x^2+(y-2)^2}\) is the distance \(d\) from \((x,y)\) to \((0,2)\), and we see from Figure 15.26 that \(1\le d\le 3\). Taking reciprocals, we have \[ \frac13\le \frac1{\sqrt{x^2+(y-2)^2}}\le 1 \]
We apply (7) with \(m=\frac13\) and \(M=1\), using the fact that \(\textrm{Area}({\mathcal{D}})=\pi\), to obtain \[ \frac{\pi}3 \le \iint_{{\mathcal{D}}}\,\frac{dA}{\sqrt{x^2+(y-2)^2}}\le \pi \]
Equation (8) is similar to the definition of an average value in one variable: \[ \overline{f} = \frac1{b-a}\int_a^b\,f(x)\,dx = \frac{\int_a^b\,f(x)\,dx}{\int_a^b\,1\,dx} \]
The average value (or mean value) of a function \(f(x,y)\) on a domain \({\mathcal{D}}\), which we denote by \(\overline{f}\), is the quantity \begin{equation*} \boxed{\overline{f} = \frac1{\textrm{Area}({\mathcal{D}})}\iint_{{\mathcal{D}}\,} f(x,y)\,dA = \frac{\iint_{{\mathcal{D}}\,} f(x,y)\,dA}{\iint_{{\mathcal{D}}\,} 1\,dA}}\tag{8} \end{equation*}
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Equivalently, \(\overline{f}\) is the value satisfying the relation \[ \boxed{\iint_{{\mathcal{D}}\,} f(x,y)\,dA = \overline{f}\cdot\textrm{Area}({\mathcal{D}})} \]
The solid region under the graph has the same (signed) volume as the cylinder with base \({\mathcal{D}}\) of height \(\overline{f}\) Figure 15.27.
An architect needs to know the average height \(\overline{H}\) of the ceiling of a pagoda whose base \({\mathcal{D}}\) is the square \([-4,4]\times [-4,4]\) and roof is the graph of \[ H(x,y)=32-x^2-y^2 \] where distances are in feet Figure 15.28. Calculate \(\overline{H}\).
Solution First, we compute the integral of \(H(x,y)\) over \({\mathcal{D}}\): \[ \begin{array}{rl} \!\!\iint_{{\mathcal{D}}\,}(32-x^2-y^2)\,dA &=\int_{-4}^4\int_{-4}^4(32-x^2-y^2)\,dy\,dx \\ &= \int_{-4}^4 \left(\left(32y-x^2y-\frac13y^3\right)\bigg|_{-4}^4\right) dx = \int_{-4}^4 \left(\frac{640}3-8x^2\right) dx\\ &=\left(\frac{640}3x-\frac83x^3\right)\bigg|_{-4}^4 = \frac{4096}3 \end{array} \]
The area of \({\mathcal{D}}\) is \(8\times 8=64\), so the average height of the pagoda’s ceiling is \[ \overline{H} = \frac1{\textrm{Area}({\mathcal{D}})}\iint_{{\mathcal{D}}\,} H(x,y)\,dA = \frac1{64}\left(\frac{4096}3\right)=\frac{64}3\approx 21.3~\textrm{ft} \]
The Mean Value Theorem states that a continuous function on a domain \({\mathcal{D}}\) must take on its average value at some point \(P\) in \({\mathcal{D}}\), provided that \({\mathcal{D}}\) is closed, bounded, and also connected (see Exercise 63 for a proof). By definition, \({\mathcal{D}}\) is connected if any two points in \({\mathcal{D}}\) can be joined by a curve in \({\mathcal{D}}\) Figure 15.29.
If \(f(x,y)\) is continuous and \({\mathcal{D}}\) is closed, bounded, and connected, then there exists a point \(P\in{\mathcal{D}}\) such that \begin{equation*} \iint_{{\mathcal{D}}\,} f(x,y)\,dA = f(P)\,\textrm{Area}({\mathcal{D}})\tag{9} \end{equation*}
Equivalently, \(f(P)=\overline{f}\), where \(\overline{f}\) is the average value of \(f\) on \({\mathcal{D}}\).
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Double integrals are additive with respect to the domain: If \({\mathcal{D}}\) is the union of domains \({\mathcal{D}}_1, {\mathcal{D}}_2,\dots,{\mathcal{D}}_N\) that do not overlap except possibly on boundary curves Figure 15.30, then \[ \boxed{\iint_{{\mathcal{D}}\,} f(x,y)\, dA = \iint_{{\mathcal{D}}_1\,} f(x,y)\, dA + \cdots + \iint_{{\mathcal{D}}_N\,} f(x,y)\, dA} \]
Additivity may be used to evaluate double integrals over domains \({\mathcal{D}}\) that are not simple but can be decomposed into finitely many simple domains.
In general, the approximation (10) is useful only if \({\mathcal{D}}\) is small in both width and length, that is, if \({\mathcal{D}}\) is contained in a circle of small radius. If \({\mathcal{D}}\) has small area but is very long and thin, then \(f\) may be far from constant on \({\mathcal{D}}\).
We close this section with a simple but useful remark. If \(f(x,y)\) is a continuous function on a small domain \({\mathcal{D}}\), then \begin{equation*} \iint_{{\mathcal{D}}\,} f(x,y)\,dA \approx \underbrace{f(P)\,\textrm{Area}({\mathcal{D}})}_{\textrm{Function value }\times\textrm{ area}}\tag{10} \end{equation*} where \(P\) is any sample point in \({\mathcal{D}}\). In fact, we can choose \(P\) so that (10) is an equality by Theorem 4. But if \({\mathcal{D}}\) is small, then \(f\) is nearly constant on \({\mathcal{D}}\), and (10) holds as a good approximation for all \(P \in {\mathcal{D}}\).
If the domain \({\mathcal{D}}\) is not small, we may partition it into \(N\) smaller subdomains \({\mathcal{D}}_1, \dots, {\mathcal{D}}_N\) and choose sample points \(P_j\) in \({\mathcal{D}}_j\). By additivity, \[ \iint_{{\mathcal{D}}\,} f(x,y)\,dA = \sum_{j=1}^N \iint_{{\mathcal{D}}_j\,} f(x,y)\,dA \approx \sum_{j=1}^N f(P_j)\,\textrm{Area}({\mathcal{D}}_j) \] and thus we have the approximation \begin{equation*} \boxed{\iint_{{\mathcal{D}}\,} f(x,y)\,dA \approx \sum_{j=1}^N f(P_j)\,\textrm{Area}({\mathcal{D}}_j)}\tag{11} \end{equation*}
We can think of Eq. (11) as a generalization of the Riemann sum approximation. In a Riemann sum, \({\mathcal{D}}\) is partitioned by rectangles \({\mathcal{R}}_{ij}\) of area \(\Delta A_{ij} = \Delta x_i\,\Delta y_j\).
Estimate \(\iint_{{\mathcal{D}}\,} f(x,y)\,dA\) for the domain \({\mathcal{D}}\) in Figure 15.31, using the areas and function values given there and the accompanying table.
\(j\) | \(1\) | \(2\) | \(3\) | \(4\) |
\(\textrm{Area}({\mathcal{D}}_j)\) | \(1\) | \(1\) | \(0.9\) | \(1.2\) |
\(f(P_j)\) | \(1.8\) | \(2.2\) | \(2.1\) | \(2.4\) |
Solution \[ \begin{array}{rl} \iint_{{\mathcal{D}}\,} f(x,y)\,dA &\approx \sum_{j=1}^4 f(P_j)\,\textrm{Area}({\mathcal{D}}_j) \\ &= (1.8)(1)+(2.2)(1) +(2.1)(0.9)+(2.4)(1.2) \approx 8.8 \end{array} \]
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