Section 15.2: Integration over general 2d regions

For single integrals, the region over which we integrate is always an interval. But for double integrals, we want to be able to integrate a function \(f\) not just over rectangles but also over regions \(D\) of more general shape, such as the one illustrated in Figure 1.

We suppose that \(D\) is a bounded region, which means that \(D\) can be enclosed in a rectangular region \(R\) as in Figure 2. Then we define a new function \(F\) with domain \(R\) by \[ F(x, y) = \begin{cases} f(x, y) & \text{if } (x, y) \text{ is in } D \\ 0 & \text{if } (x, y) \text{ is in } R \text{ but not in } D \end{cases} \tag{1} \]

If \(F\) is integrable over \(R\), then we define the double integral of f over D by \[ \iint_D f(x, y) dA = \iint_R F(x, y) dA \tag{2} \] where F is given by Equation 1.

In the case where \(f(x, y) \ge 0\), we can still interpret \(\iint_D f(x, y) dA\) as the volume of the solid that lies above \(D\) and under the surface \(z = f(x, y)\) (the graph of \(f\)). You can see that this is reasonable by comparing the graphs of \(f\) and \(F\) in Figures 3 and 4 and remembering that \(\iint_R F(x, y) dA\) is the volume under the graph of \(F\).

Figure 4 also shows that \(F\) is likely to have discontinuities at the boundary points of \(D\). Nonetheless, if \(f\) is continuous on \(D\) and the boundary curve of \(D\) is “well behaved” (in a sense outside the scope of this book), then it can be shown that \(\iint_R F(x, y) dA\) exists and therefore \(\iint_D f(x, y) dA\) exists. In particular, this is the case for the following two types of regions.

A plane region \(D\) is said to be of type I if it lies between the graphs of two continuous functions of \(x\), that is, \[ D = \{(x, y) | a \le x \le b, g_1(x) \le y \le g_2(x)\} \] where \(g_1\) and \(g_2\) are continuous on \([a, b]\). Some examples of type I regions are shown in Figure below.

In order to evaluate \(\iint_D f(x, y) dA\) when \(D\) is a region of type I, we choose a rectangle \(R = [a, b] \times [c, d]\) that contains \(D\), as in Figure 6, and we let \(F\) be the function given by Equation 1; that is, \(F\) agrees with \(f\) on \(D\) and \(F\) is 0 outside \(D\). Then, by Fubini’s Theorem, \[ \iint_D f(x, y) dA = \iint_R F(x, y) dA = \int_a^b \int_c^d F(x, y) dy dx \] Observe that \(F(x, y) = 0\) if \(y < g_1(x)\) or \(y > g_2(x)\) because \((x, y)\) then lies outside \(D\). Therefore \[ \int_c^d F(x, y) dy = \int_{g_1(x)}^{g_2(x)} F(x, y) dy = \int_{g_1(x)}^{g_2(x)} f(x, y) dy \] because \(F(x, y) = f(x, y)\) when \(g_1(x) \le y \le g_2(x)\). Thus we have the following formula that enables us to evaluate the double integral as an iterated integral.

If f is continuous on a type I region \(D\) such that

\[ D = \{(x, y) | a \le x \le b, g_1(x) \le y \le g_2(x)\}, \] then \[ \iint_D f(x, y) dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) dy dx \tag{3} \] The integral on the right side of (3) is an iterated integral that is similar to the ones we considered in the preceding section, except that in the inner integral we regard \(x\) as being constant not only in \(f(x, y)\) but also in the limits of integration, \(g_1(x)\) and \(g_2(x)\).

We also consider plane regions of type II, which can be expressed as \[ D = \{(x, y) | c \le y \le d, h_1(y) \le x \le h_2(y)\} \tag{4} \] where \(h_1\) and \(h_2\) are continuous. Two such regions are illustrated in Figure 7.

Using the same methods that were used in establishing (3), we can show that \[ \iint_D f(x, y) dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y) dx dy \tag{5} \] where \(D\) is a type II region given by Equation 4.

EXAMPLE 1 Evaluate \(\iint_D (x + 2y) dA\), where \(D\) is the region bounded by the parabolas \(y = 2x^2\) and \(y = 1 + x^2\).

EXAMPLE 2 Find the volume of the solid that lies under the paraboloid \(z = x^2 + y^2\) and above the region \(D\) in the \(xy\)-plane bounded by the line \(y = 2x\) and the parabola \(y = x^2\).

Type 1 Solution

Type II Solution

EXAMPLE 3 Evaluate \(\iint_D xy dA\), where \(D\) is the region bounded by the line \(y = x - 1\) and the parabola \(y^2 = 2x + 6\).

EXAMPLE 4 Find the volume of the tetrahedron bounded by the planes \(x + 2y + z = 2, x = 2y, x = 0,\) and \(z = 0\).

EXAMPLE 5 Evaluate the iterated integral \(\int_0^1 \int_x^1 \sin(y^2) dy dx\).

General regions the properties follow from Definition.

\[ \iint_D [f(x, y) + g(x, y)] dA = \iint_D f(x, y) dA + \iint_D g(x, y) dA \tag{6} \]

\[ \iint_D cf(x, y) dA = c \iint_D f(x, y) dA \quad \text{where c is a constant} \tag{7} \]

If \(f(x, y) \ge g(x, y)\) for all \((x, y)\) in \(D\), then

\[ \iint_D f(x, y) dA \ge \iint_D g(x, y) dA \tag{8} \]

The next property of double integrals is similar to the property of single integrals given by the equation \(\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx\). If \(D = D_1 \cup D_2\), where \(D_1\) and \(D_2\) don’t overlap except perhaps on their boundaries (see Figure 17), then

\[ \iint_D f(x, y) dA = \iint_{D_1} f(x, y) dA + \iint_{D_2} f(x, y) dA \tag{9} \] Property 9 can be used to evaluate double integrals over regions \(D\) that are neither type I nor type II but can be expressed as a union of regions of type I or type II. Figure 18 illustrates this procedure.

If we integrate the constant function \(f(x, y) = 1\) over a region \(D\), we get the area of \(D\): \[ \iint_D 1 dA = A(D) \tag{10} \]

Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is \(D\) and whose height is 1 has volume \(A(D) \cdot 1 = A(D)\), but we know that we can also write its volume as \(\iint_D 1 dA\).

If \(m \le f(x, y) \le M\) for all \((x, y) \text{ in } D\) , then

\[ mA(D) \le \iint_D f(x, y) dA \le MA(D) \tag{11} \]

EXAMPLE 6 Use Property 11 to estimate the integral \(\iint_D e^{\sin x \cos y} dA\), where \(D\) is the disk with center the origin and radius 2.

Evaluate the iterated integral \(\int_0^3 \int_0^1 (8x - 2y) dy dx\).

Evaluate the iterated integral \(\int_0^2 \int_0^4 x^2y dx dy\).

Evaluate the iterated integral \(\int_0^1 \int_0^1 xe^{y^2} dx dy\).

Evaluate the iterated integral \(\int_0^{\pi/2} \int_0^{1/2} x \sin y dy dx\).

Evaluate the iterated integral \(\int_0^1 \int_0^1 \cos(s^3) dt ds\).

Evaluate the iterated integral \(\int_0^1 \int_0^1 \sqrt{1 + e^v} dw dv\).

Evaluate the double integral \(\iint_D \frac{y}{x^2 + 1} dA\), where \(D = \{(x, y) | 0 \le x \le 4, 0 \le y \le \sqrt{x}\}\).

Evaluate the double integral \(\iint_D (2x + y) dA\), where \(D = \{(x, y) | 1 \le y \le 2, y - 1 \le x \le 1\}\).

Evaluate the double integral \(\iint_D e^{-y^2} dA\), where \(D = \{(x, y) | 0 \le y \le 3, 0 \le x \le y\}\).

Evaluate the double integral \(\iint_D y\sqrt{x^2 - y^2} dA\), where \(D = \{(x, y) | 0 \le x \le 2, 0 \le y \le x\}\).

Express \(D\) as a region of type I and also as a region of type II. Then evaluate the double integral \(\iint_D x dA\) in two ways, where \(D\) is enclosed by the lines \(y = x, y = 0, x = 1\).

Express \(D\) as a region of type I and also as a region of type II. Then evaluate the double integral \(\iint_D xy dA\) in two ways, where \(D\) is enclosed by the curves \(y = \sqrt{x}, y = 3x\).

Set up iterated integrals for both orders of integration. Then evaluate the double integral \(\iint_D y dA\) using the easier order and explain why it’s easier, where \(D\) is bounded by \(y = x - 2, x = y^2\).

Set up iterated integrals for both orders of integration. Then evaluate the double integral \(\iint_D y^2e^{xy} dA\) using the easier order and explain why it’s easier, where \(D\) is bounded by \(y = x, y = 4, x = 0\).

Evaluate the double integral \(\iint_D x \cos y dA\), where \(D\) is bounded by \(y = 0, y = x^2, x = 1\).

Evaluate the double integral \(\iint_D (x^2 + 2y) dA\), where \(D\) is bounded by \(y = x, y = x^3, x \ge 0\).

Evaluate the double integral \(\iint_D y^2 dA\), where \(D\) is the triangular region with vertices \((0, 1), (1, 2), (4, 1)\).

Evaluate the double integral \(\iint_D xy dA\), where \(D\) is enclosed by the quarter-circle \(y = \sqrt{1 - x^2}, x \ge 0\), and the axes.

Evaluate the double integral \(\iint_D (2x - y) dA\), where \(D\) is bounded by the circle with center the origin and radius 2.

Evaluate the double integral \(\iint_D y dA\), where \(D\) is the triangular region with vertices \((0, 0), (1, 1),\) and \((4, 0)\).

Find the volume of the solid under the plane \(3x + 2y - z = 0\) and above the region enclosed by the parabolas \(y = x^2\) and \(x = y^2\).

Find the volume of the solid under the surface \(z = 1 + x^2y^2\) and above the region enclosed by \(x = y^2\) and \(x = 4\).

Find the volume of the solid under the surface \(z = xy\) and above the triangle with vertices \((1, 1), (4, 1),\) and \((1, 2)\).

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

Find the volume of the tetrahedron enclosed by the coordinate planes and the plane \(2x + y + z = 4\).

Find the volume of the solid bounded by the planes \(z = x, y = x, x + y = 2,\) and \(z = 0\).

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

Find the volume of the solid bounded by the cylinder \(y^2 + z^2 = 4\) and the planes \(x = 2y, x = 0, z = 0\) in the first octant.

Find the volume of the solid bounded by the cylinder \(x^2 + y^2 = 1\) and the planes \(y = z, x = 0, z = 0\) in the first octant.

Find the volume of the solid bounded by the cylinders \(x^2 + y^2 = r^2\) and \(y^2 + z^2 = r^2\).

Use a graphing calculator or computer to estimate the x-coordinates of the points of intersection of the curves \(y = x^4\) and \(y = 3x - x^2\). If \(D\) is the region bounded by these curves, estimate \(\iint_D x dA\).

Find the approximate volume of the solid in the first octant that is bounded by the planes \(y = x, z = 0,\) and \(z = x\) and the cylinder \(y = \cos x\). (Use a graphing device to estimate the points of intersection.)

Find the volume of the solid by subtracting two volumes. The solid enclosed by the parabolic cylinders \(y = 1 - x^2, y = x^2 - 1\) and the planes \(x + y + z = 2, 2x + 2y - z + 10 = 0\).

Find the volume of the solid by subtracting two volumes. The solid enclosed by the parabolic cylinder \(y = x^2\) and the planes \(z = 3y, z = 2 + y\).

Find the volume of the solid by subtracting two volumes. The solid under the plane \(z = 3\), above the plane \(z = y\), and between the parabolic cylinders \(y = x^2\) and \(y = 1 - x^2\).

Find the volume of the solid by subtracting two volumes. The solid in the first octant under the plane \(z = x + y\), above the surface \(z = xy\), and enclosed by the surfaces \(x = 0, y = 0,\) and \(x^2 + y^2 = 4\).

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

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

Use a computer algebra system to find the exact volume of the solid under the surface \(z = x^3y^4 + xy^2\) and above the region bounded by the curves \(y = x^3 - x\) and \(y = x^2 + x\) for \(x \ge 0\).

Use a computer algebra system to find the exact volume of the solid between the paraboloids \(z = 2x^2 + y^2\) and \(z = 8 - x^2 - 2y^2\) and inside the cylinder \(x^2 + y^2 = 1\).

Use a computer algebra system to find the exact volume of the solid enclosed by \(z = 1 - x^2 - y^2\) and \(z = 0\).

Use a computer algebra system to find the exact volume of the solid enclosed by \(z = x^2 + y^2\) and \(z = 2y\).

Sketch the region of integration and change the order of integration. \(\int_0^4 \int_0^{\sqrt{x}} f(x, y) dy dx\).

Sketch the region of integration and change the order of integration. \(\int_0^1 \int_{x^2}^1 f(x, y) dy dx\).

Sketch the region of integration and change the order of integration. \(\int_0^{\pi/2} \int_0^{\cos x} f(x, y) dy dx\).

Sketch the region of integration and change the order of integration. \(\int_{-2}^2 \int_0^{\sqrt{4-y^2}} f(x, y) dx dy\).

Sketch the region of integration and change the order of integration. \(\int_1^2 \int_0^{\ln x} f(x, y) dy dx\).

Sketch the region of integration and change the order of integration. \(\int_0^1 \int_{\arctan x}^{\pi/4} f(x, y) dy dx\).

Evaluate the integral by reversing the order of integration. \(\int_0^1 \int_{3y}^3 e^{x^2} dx dy\).

Evaluate the integral by reversing the order of integration. \(\int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3 + 1} dx dy\).

Evaluate the integral by reversing the order of integration. \(\int_0^3 \int_{y^2}^9 y \cos(x^2) dx dy\).

Evaluate the integral by reversing the order of integration. \(\int_0^1 \int_{\arcsin y}^{\pi/2} \cos x \sqrt{1 + \cos^2 x} dx dy\).

Evaluate the integral by reversing the order of integration. \(\int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4} dx dy\).

Evaluate the integral by reversing the order of integration. \(\int_0^1 \int_x^1 e^{y^2} dy dx\).

Express \(D\) as a union of regions of type I or type II and evaluate the integral \(\iint_D x^2 dA\).

Express \(D\) as a union of regions of type I or type II and evaluate the integral \(\iint_D y dA\).

Use Property 11 to estimate the value of the integral \(\iint_S \sqrt{4 - x^2y^2} dA\), where \(S = \{(x, y) | x^2 + y^2 \le 1, x \ge 0\}\).

Use Property 11 to estimate the value of the integral \(\iint_T \sin^4(x + y) dA\), where \(T\) is the triangle enclosed by the lines \(y = 0, y = 2x,\) and \(x = 1\).

Find the average value of \(f(x, y) = xy\) over the region \(D\), where \(D\) is the triangle with vertices \((0, 0), (1, 0),\) and \((1, 3)\).

Find the average value of \(f(x, y) = x \sin y\) over the region \(D\), where \(D\) is enclosed by the curves \(y = 0, y = x^2,\) and \(x = 1\).

In evaluating a double integral over a region \(D\), a sum of iterated integrals was obtained as follows: \(\iint_D f(x, y) dA = \int_0^1 \int_0^{2y} f(x, y) dx dy + \int_1^3 \int_0^{3-y} f(x, y) dx dy\). Sketch the region \(D\) and express the double integral as an iterated integral with reversed order of integration.

Use geometry or symmetry, or both, to evaluate the double integral \(\iint_D (x + 2) dA\), where \(D = \{(x, y) | 0 \le y \le \sqrt{9 - x^2}\}\).

Use geometry or symmetry, or both, to evaluate the double integral \(\iint_D \sqrt{R^2 - x^2 - y^2} dA\), where \(D\) is the disk with center the origin and radius \(R\).

Use geometry or symmetry, or both, to evaluate the double integral \(\iint_D (2x + 3y) dA\), where \(D\) is the rectangle \(0 \le x \le a, 0 \le y \le b\).

Use geometry or symmetry, or both, to evaluate the double integral \(\iint_D (2 + x^2y^3 - y^2\sin x) dA\), where \(D = \{(x, y) | |x| + |y| \le 1\}\).

Use geometry or symmetry, or both, to evaluate the double integral \(\iint_D (ax^3 + by^3 + \sqrt{a^2 - x^2}) dA\), where \(D = [-a, a] \times [-b, b]\).

Graph the solid bounded by the plane \(x + y + z = 1\) and the paraboloid \(z = 4 - x^2 - y^2\) and find its exact volume. (Use your CAS to do the graphing, to find the equations of the boundary curves of the region of integration, and to evaluate the double integral.)