Section 15.1: Double Integrals over rectangles

First let’s recall the basic facts concerning definite integrals of functions of a single variable. If \(f(x)\) is defined for \(a \le x \le b\), we start by dividing the interval \([a, b]\) into \(n\) subintervals \([x_{i-1}, x_i]\) of equal width \(\Delta x = (b - a)/n\) and we choose sample points \(x_i^*\) in these subintervals. Then we form the Riemann sum \[ \sum_{i=1}^{n} f(x_i^*) \Delta x \tag{1} \] and take the limit of such sums as \(n \to \infty\) to obtain the definite integral of \(f\) from \(a\) to \(b\):

\[ \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \tag{2} \] In the special case where \(f(x) \ge 0\), the Riemann sum can be interpreted as the sum of the areas of the approximating rectangles in Figure 1, and \(\int_a^b f(x) dx\) represents the area under the curve \(y = f(x)\) from \(a\) to \(b\).

We consider a function \(f\) of two variables defined on a closed rectangle \[ R = [a, b] \times [c, d] = \{(x, y) \in \mathbb{R}^2 | a \le x \le b, c \le y \le d\} \] and we first suppose that \(f(x, y) \ge 0\). The graph of \(f\) is a surface with equation \(z = f(x, y)\).

Let \(S\) be the solid that lies above \(R\) and under the graph of \(f\), that is, \[ S = \{(x, y, z) \in \mathbb{R}^3 | 0 \le z \le f(x, y), (x, y) \in R\} \] (See Figure 2.) Our goal is to find the volume of \(S\).

The first step is to divide the rectangle \(R\) into subrectangles. We accomplish this by dividing the interval \([a, b]\) into \(m\) subintervals \([x_{i-1}, x_i]\) of equal width \(\Delta x = (b - a)/m\) and dividing \([c, d]\) into \(n\) subintervals \([y_{j-1}, y_j]\) of equal width \(\Delta y = (d - c)/n\). By drawing lines parallel to the coordinate axes through the endpoints of these subintervals, as in Figure 3, we form the subrectangles

\[ R_{ij} = [x_{i-1}, x_i] \times [y_{j-1}, y_j] = \{(x, y) | x_{i-1} \le x \le x_i, y_{j-1} \le y \le y_j\} \] each with area \(\Delta A = \Delta x \Delta y\).

If we choose a sample point \((x_{ij}^*, y_{ij}^*)\) in each \(R_{ij}\), then we can approximate the part of \(S\) that lies above each \(R_{ij}\) by a thin rectangular box (or “column”) with base \(R_{ij}\) and height \(f(x_{ij}^*, y_{ij}^*)\) as shown in Figure 4. The volume of this box is the height of the box times the area of the base rectangle: \[ f(x_{ij}^*, y_{ij}^*) \Delta A \]

If we follow this procedure for all the rectangles and add the volumes of the corresponding boxes, we get an approximation to the total volume of \(S\):

\[ \tag{3} V \approx \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A \]

(See Figure 5.) This double sum means that for each subrectangle we evaluate \(f\) at the chosen point and multiply by the area of the subrectangle, and then we add the results.

Our intuition tells us that the approximation given in (Eq 3) becomes better as \(m\) and \(n\) become larger and so we would expect that

\[ \tag{4} V = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A \]

We use the expression in Equation (4) to define the volume of the solid \(S\) that lies under the graph of \(f\) and above the rectangle \(R\).

This dicussion leads to the following definition.

Definition The double integral of \(f\) over the rectangle \(R\) is \[ \tag{5} \iint_R f(x, y) dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A \] if this limit exists.

The precise meaning of the limit in Definition 5 is that for every number \(\varepsilon > 0\) there is an integer \(N\) such that \[ \left| \iint_R f(x, y) dA - \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A \right| < \varepsilon \] for all integers \(m\) and \(n\) greater than \(N\) and for any choice of sample points \((x_{ij}^*, y_{ij}^*)\) in \(R_{ij}\).

A function \(f\) is called integrable if the limit in Definition exists.

• It is shown in courses on advanced calculus that all continuous functions are integrable. In fact, the double integral of \(f\) exists provided that \(f\) is “not too discontinuous.” In particular, if \(f\) is bounded on \(R\), [that is, there is a constant \(M\) such that \(|f(x, y)| \le M\) for all \((x, y)\) in \(R\)], and \(f\) is continuous there, except on a finite number of smooth curves, then \(f\) is integrable over \(R\).

The sample point \((x_{ij}^*, y_{ij}^*)\) can be chosen to be any point in the subrectangle \(R_{ij}\), but if we choose it to be the upper right-hand corner of \(R_{ij}\) [namely \((x_i, y_j)\), see Figure 3], then the expression for the double integral looks simpler:

\[ \tag{6} \iint_R f(x, y) dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i, y_j) \Delta A \]

We see that a volume can be written as a double integral:

If \(f(x, y) \ge 0\), then the volume \(V\) of the solid that lies above the rectangle \(R\) and below the surface \(z = f(x, y)\) is \[ V = \iint_R f(x, y) dA \]

The sum in Definition 5, \[ \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \Delta A \] is called a double Riemann sum and is used as an approximation to the value of the double integral. If \(f\) happens to be a positive function, then the double Riemann sum represents the sum of volumes of columns, as in Figure 5, and is an approximation to the volume under the graph of \(f\).

EXAMPLE 2 If \(R = \{(x, y) | -1 \le x \le 1, -2 \le y \le 2\}\), evaluate the integral \(\iint_R \sqrt{1 - x^2} dA\).

The methods that we used for approximating single integrals (the Midpoint Rule, the Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals. Here we consider only the Midpoint Rule for double integrals. This means that we use a double Riemann sum to approximate the double integral, where the sample point \((x_{ij}^*, y_{ij}^*)\) in \(R_{ij}\) is chosen to be the center \((\bar{x}_i, \bar{y}_j)\) of \(R_{ij}\). In other words, \(\bar{x}_i\) is the midpoint of \([x_{i-1}, x_i]\) and \(\bar{y}_j\) is the midpoint of \([y_{j-1}, y_j]\).

Midpoint Rule for Double Integrals \[ \iint_R f(x, y) dA \approx \sum_{i=1}^{m} \sum_{j=1}^{n} f(\bar{x}_i, \bar{y}_j) \Delta A \] where \(\bar{x}_i\) is the midpoint of \([x_{i-1}, x_i]\) and \(\bar{y}_j\) is the midpoint of \([y_{j-1}, y_j]\).

EXAMPLE 3 Use the Midpoint Rule with \(m = n = 2\) to estimate the value of the integral \(\iint_R (x - 3y^2) dA\), where \(R = \{(x, y) | 0 \le x \le 2, 1 \le y \le 2\}\).

Recall that it is usually difficult to evaluate single integrals directly from the definition of an integral, but the Fundamental Theorem of Calculus provides a much easier method. The evaluation of double integrals from first principles is even more difficult, but here we see how to express a double integral as an iterated integral, which can then be evaluated by calculating two single integrals.

Suppose that \(f\) is a function of two variables that is integrable on the rectangle \(R = [a, b] \times [c, d]\). We use the notation \(\int_c^d f(x, y) dy\) to mean that \(x\) is held fixed and \(f(x, y)\) is integrated with respect to \(y\) from \(y = c\) to \(y = d\). This procedure is called partial integration with respect to y. Now \(\int_c^d f(x, y) dy\) is a number that depends on the value of \(x\), so it defines a function of \(x\): \[ A(x) = \int_c^d f(x, y) dy \] If we now integrate the function \(A\) with respect to \(x\) from \(x = a\) to \(x = b\), we get

\[ \tag{7} \int_a^b A(x) dx = \int_a^b \left[ \int_c^d f(x, y) dy \right] dx \]

The integral on the right side of Equation (7)) is called an iterated integral. Usually the brackets are omitted. Thus

\[ \tag{8} \int_a^b \int_c^d f(x, y) dy dx = \int_a^b \left[ \int_c^d f(x, y) dy \right] dx \] means that we first integrate with respect to \(y\) from \(c\) to \(d\) and then with respect to \(x\) from \(a\) to \(b\).

Similarly, the iterated integral \[ \tag{eq:9} \int_c^d \int_a^b f(x, y) dx dy = \int_c^d \left[ \int_a^b f(x, y) dx \right] dy \] means that we first integrate with respect to \(x\) (holding \(y\) fixed) from \(x = a\) to \(x = b\) and then we integrate the resulting function of \(y\) with respect to \(y\) from \(y = c\) to \(y = d\).

Notice that in both Equations (8) and (9) we work from the inside out.

EXAMPLE 4 Evaluate the iterated integrals. (a) \(\int_0^3 \int_1^2 x^2y dy dx\) (b) \(\int_1^2 \int_0^3 x^2y dx dy\)

The following theorem gives a practical method for evaluating a double integral by expressing it as an iterated integral (in either order).

10 Fubini’s Theorem If \(f\) is continuous on the rectangle \(R = \{(x, y) | a \le x \le b, c \le y \le d\}\), then \[ \iint_R f(x, y) dA = \int_a^b \int_c^d f(x, y) dy dx = \int_c^d \int_a^b f(x, y) dx dy \] More generally, this is true if we assume that \(f\) is bounded on \(R\), \(f\) is discontinuous only on a finite number of smooth curves, and the iterated integrals exist.

EXAMPLE 5 Evaluate the double integral \(\iint_R (x - 3y^2) dA\), where \(R = \{(x, y) | 0 \le x \le 2, 1 \le y \le 2\}\).

EXAMPLE 6 Evaluate \(\iint_R y \sin(xy) dA\), where \(R = [1, 2] \times [0, \pi]\).

EXAMPLE 7 Find the volume of the solid \(S\) that is bounded by the elliptic paraboloid \(x^2 + 2y^2 + z = 16\), the planes \(x = 2\) and \(y = 2\), and the three coordinate planes.

Recall from one variable calculs that the average value of a functio \(f\) of one variable defined on an interval \([a, b]\) is \[ f_{ave} = \frac{1}{b-a} \int_a^b f(x) dx \] In a similar fashion we define the average value of a function \(f\) of two variables defined on a rectangle \(R\) to be \[ f_{ave} = \frac{1}{A(R)} \iint_R f(x, y) dA \] where \(A(R)\) is the area of \(R\). If \(f(x, y) \ge 0\), the equation \[ A(R) \times f_{ave} = \iint_R f(x, y) dA \] says that the box with base \(R\) and height \(f_{ave}\) has the same volume as the solid that lies under the graph of \(f\).

1. Estimate the volume of the solid that lies below the surface \(z = xy\) and above the rectangle \(R = \{(x, y) | 0 \le x \le 6, 0 \le y \le 4\}\). Use a Riemann sum with \(m = 3\), \(n = 2\), and take the sample point to be the upper right corner of each square.
2. Use the Midpoint Rule to estimate the volume of the solid in part (a).

If \(R = [0, 4] \times [-1, 2]\), use a Riemann sum with \(m = 2\), \(n = 3\) to estimate the value of \(\iint_R (1 - xy^2) dA\). Take the sample points to be (a) the lower right corners and (b) the upper left corners of the rectangles.

1. Use a Riemann sum with \(m = n = 2\) to estimate the value of \(\iint_R xe^{-xy} dA\), where \(R = [0, 2] \times [0, 1]\). Take the sample points to be upper right corners.
2. Use the Midpoint Rule to estimate the integral in part (a).

1. Estimate the volume of the solid that lies below the surface \(z = 1 + x^2 + 3y\) and above the rectangle \(R = [1, 2] \times [0, 3]\). Use a Riemann sum with \(m = n = 2\) and choose the sample points to be lower left corners.
2. Use the Midpoint Rule to estimate the volume in part (a).

Let V be the volume of the solid that lies under the graph of \(f(x, y) = \sqrt{52 - x^2 - y^2}\) and above the rectangle given by \(2 \le x \le 4\), \(2 \le y \le 6\). Use the lines \(x = 3\) and \(y = 4\) to divide R into subrectangles. Let L and U be the Riemann sums computed using lower left corners and upper right corners, respectively. Without calculating the numbers V, L, and U, arrange them in increasing order and explain your reasoning.

A contour map is shown for a function \(f\) on the square \(R = [0, 4] \times [0, 4]\). (a) Use the Midpoint Rule with \(m = n = 2\) to estimate the value of \(\iint_R f(x, y) dA\). (b) Estimate the average value of \(f\).

The contour map shows the temperature, in degrees Fahrenheit, at 4:00 PM on February 26, 2007, in Colorado. (The state measures 388 mi west to east and 276 mi south to north.) Use the Midpoint Rule with \(m = n = 4\) to estimate the average temperature in Colorado at that time.

Evaluate the double integral \(\iint_R \sqrt{2} dA\), where \(R = \{(x, y) | 2 \le x \le 6, -1 \le y \le 5\}\) by first identifying it as the volume of a solid.

Evaluate the double integral \(\iint_R (2x + 1) dA\), where \(R = \{(x, y) | 0 \le x \le 2, 0 \le y \le 4\}\) by first identifying it as the volume of a solid.

Evaluate the double integral \(\iint_R (4 - 2y) dA\), where \(R = [0, 1] \times [0, 1]\) by first identifying it as the volume of a solid.

The integral \(\iint_R \sqrt{9 - y^2} dA\), where \(R = [0, 4] \times [0, 2]\), represents the volume of a solid. Sketch the solid.

Find \(\int_0^5 f(x, y) dx\) and \(\int_0^1 f(x, y) dy\) for \(f(x, y) = x + 3x^2y^2\).

Find \(\int_0^5 f(x, y) dx\) and \(\int_0^1 f(x, y) dy\) for \(f(x, y) = y\sqrt{x} + 2\).

Calculate the iterated integral \(\int_1^4 \int_0^2 (6x^2y - 2x) dy dx\).

Calculate the iterated integral \(\int_0^1 \int_1^2 (x + e^{-y})^2 dx dy\).

Calculate the iterated integral \(\int_0^1 \int_0^1 (x + e^y) dx dy\).

Calculate the iterated integral \(\int_0^{\pi/6} \int_0^{\pi/2} (\sin x + \sin y) dy dx\).

Calculate the iterated integral \(\int_0^3 \int_{-2}^0 (y^2 + y^3 \cos x) dx dy\).

Calculate the iterated integral \(\int_1^4 \int_1^2 (\frac{x}{y} + \frac{y}{x}) dy dx\).

Calculate the iterated integral \(\int_1^2 \int_0^1 \frac{x}{y} dy dx\).

Calculate the iterated integral \(\int_0^1 \int_0^1 ye^{xy} dx dy\).

Calculate the iterated integral \(\int_0^{\pi/2} \int_0^{\pi/2} \sin \theta \cos \phi d\theta d\phi\).

Calculate the iterated integral \(\int_0^1 \int_0^1 xy\sqrt{x^2 + y^2} dy dx\).

Calculate the iterated integral \(\int_0^1 \int_0^1 \sqrt{u + v^2} du dv\).

Calculate the iterated integral \(\int_0^1 \int_0^1 \frac{1}{\sqrt{s+t}} ds dt\).

Calculate the double integral \(\iint_R x \sec^2 y dA\), where \(R = \{(x, y) | 0 \le x \le 2, 0 \le y \le \pi/4\}\).

Calculate the double integral \(\iint_R (y + xy^{-2}) dA\), where \(R = \{(x, y) | 0 \le x \le 2, 1 \le y \le 2\}\).

Calculate the double integral \(\iint_R \frac{xy^2}{x^2 + 1} dA\), where \(R = \{(x, y) | 0 \le x \le 1, -3 \le y \le 3\}\).

Calculate the double integral \(\iint_R \frac{\tan \theta}{\sqrt{1 - t^2}} dA\), where \(R = \{(\theta, t) | 0 \le \theta \le \pi/3, 0 \le t \le 1/2\}\).

Calculate the double integral \(\iint_R x \sin(x + y) dA\), where \(R = [0, \pi/6] \times [0, \pi/3]\).

Calculate the double integral \(\iint_R \frac{x}{1 + xy} dA\), where \(R = [0, 1] \times [0, 1]\).

Calculate the double integral \(\iint_R ye^{xy} dA\), where \(R = [0, 2] \times [0, 3]\).

Calculate the double integral \(\iint_R \frac{1}{1 + x + y} dA\), where \(R = [1, 3] \times [1, 2]\).

Sketch the solid whose volume is given by the iterated integral \(\int_0^1 \int_0^1 (4 - x - 2y) dx dy\).

Sketch the solid whose volume is given by the iterated integral \(\int_0^1 \int_0^1 (2 - x^2 - y^2) dy dx\).

Find the volume of the solid that lies under the plane \(4x + 6y - 2z + 15 = 0\) and above the rectangle \(R = \{(x, y) | -1 \le x \le 2, -1 \le y \le 1\}\).

Find the volume of the solid that lies under the hyperbolic paraboloid \(z = 3y^2 - x^2 + 2\) and above the rectangle \(R = [-1, 1] \times [1, 2]\).

Find the volume of the solid lying under the elliptic paraboloid \(x^2/4 + y^2/9 + z = 1\) and above the rectangle \(R = [-1, 1] \times [-2, 2]\).

Find the volume of the solid enclosed by the surface \(z = x^2 + xy^2\) and the planes \(z = 0, x = 0, x = 5,\) and \(y = \pm 2\).

Find the volume of the solid enclosed by the surface \(z = 1 + e^x \sin y\) and the planes \(z = 0, x = \pm 1, y = 0,\) and \(y = \pi\).

Find the volume of the solid in the first octant bounded by the cylinder \(z = 16 - x^2\) and the plane \(y = 5\).

Find the volume of the solid enclosed by the paraboloid \(z = 2 + x^2 + (y - 2)^2\) and the planes \(z = 1, x = 1, x = -1, y = 0,\) and \(y = 4\).

Graph the solid that lies between the surface \(z = 2xy/(x^2 + 1)\) and the plane \(z = x + 2y\) and is bounded by the planes \(x = 0, x = 2, y = 0,\) and \(y = 4\). Then find its volume.

Use a computer algebra system to find the exact value of the integral \(\iint_R x^5y^3e^{xy} dA\), where \(R = [0, 1] \times [0, 1]\). Then use the CAS to draw the solid whose volume is given by the integral.

Graph the solid that lies between the surfaces \(z = e^{-x^2} \cos(x^2 + y^2)\) and \(z = 2 - x^2 - y^2\) for \(|x| \le 1, |y| \le 1\). Use a computer algebra system to approximate the volume of this solid correct to four decimal places.

Find the average value of \(f(x, y) = x^2y\) over the rectangle R with vertices \((-1, 0), (-1, 5), (1, 5), (1, 0)\).

Find the average value of \(f(x, y) = e^y\sqrt{x + e^y}\) over the rectangle \(R = [0, 4] \times [0, 1]\).

Use symmetry to evaluate the double integral \(\iint_R \frac{xy}{1 + x^4} dA\), where \(R = \{(x, y) | -1 \le x \le 1, 0 \le y \le 1\}\).

Use symmetry to evaluate the double integral \(\iint_R (1 + x^2 \sin y + y^2 \sin x) dA\), where \(R = [-\pi, \pi] \times [-\pi, \pi]\).

Use a CAS to compute the iterated integrals \(\int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} dy dx\) and \(\int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} dx dy\). Do the answers contradict Fubini’s Theorem? Explain what is happening.

1. In what way are the theorems of Fubini and Clairaut similar?
2. If \(f(x, y)\) is continuous on \([a, b] \times [c, d]\) and \(g(x, y) = \int_a^x \int_c^y f(s, t) dt ds\) for \(a < x < b, c < y < d\), show that \(g_{xy} = g_{yx} = f(x, y)\).