Iterated Integrals

ITERATED INTEGRALS

The method of evaluation of double integrals depend upon the nature of the curves binding the region R. Let the region R be bounded by the curves x = x₁, x = x₂, y = y₁ and y = y₂ is shown in Figure 5.2.

Case (i) When x₁, x₂ are functions of y and y₁, y₂ are constants. (The variable x has variable limits, y has constant limits) Let AC and BD be the curves x₁ = φ₁(y) and x₂ = φ₂(y).

Here the double integral is evaluated first with respect to x (treating y as a constant temporarily). The resulting expression is a function of y which is integrated with respect to y between the limits y = y₁ and y = y₂.

The integration being carried from the inner to the outer rectangle. Geometrically the integral in the inner rectangle indicates that the integration is performed along the horizontal strip PQ (keeping y constant) while the outer rectangle corresponds to the sliding of strip PQ from AC to BD thus covering the entire region ABDC of integration.

Case (ii) When y₁, y₂ are functions of x and x₁, x₂ are constants (the variable y has variable limits and x has constant limits shown in Figure 5.3. Let AB and CD be the curves y₁ = φ₁(x) and y₂ = φ₂(x).

Here the double integration is evaluated first with respect to y (treating x as a constant temporarily). The resulting expression is a function of x is integrated with respect to x between the limits x = x₁ and x = x₂.

The integration being carried from the inner to the outer rectangle. Geometrically, the integral in the inner rectangle indicates that the integration is performed along the vertical strip PQ (keeping x constant) while the outer rectangle correspondence to the sliding of the strip from AB to CD thus covering the entire region ABDC of the integration.

Case (iii) When x₁, x₂, y₁ and y₂ are constants. Here the region of integration is a rectangle ABDC. It is immaterial whether we integrate first with respect to x and then with respect to y or first to y and then with respect to x. Hence the order of integration is immaterial provided the limits of integration are changed accordingly,

Properties of Double Integrals

Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications.

If f(x, y) and g(x, y) are continuous on the bounded region R, then the following properties hold.

1. Constant Multiple:

2. Sum and Difference:

3. Domination:

4. Additivity:

if R is the union of two nonoverlapping regions R₁ and R₂.

Property 4 assumes that the region of integration R is decomposed into nonoverlapping regions R₁ and R₂ with boundaries consisting of a finite number of line segments or smooth curves. Figure 15.17 illustrates an example of this property.

The idea behind these properties is that integrals behave like sums. If the function f(x, y) is replaced by its constant multiple cf(x, y), then a Riemann sum for f

is replaced by a Riemann sum for cf

Taking limits as n → ∞ shows that c lim_n→∞ S_n = c ∫∫_R f(x,y) dA and lim_n→∞ cS_n = ∫∫_R cf dA are equal. It follows that the Constant Multiple Property carries over from sums to double integrals.

The other properties are also easy to verify for Riemann sums, and carry over to double integrals for the same reason. While this discussion gives the idea, an actual proof that these properties hold requires a more careful analysis of how Riemann sums converge.

FUBINI'S THEOREM

If is continuous on the rectangle

 R = [(x,y) |a≤x≤b, c≤y≤d}, then

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 1. Evaluate the iterated integrals.

(a) ₀∫³₁∫² x²y dydx

(b) ₁∫²₀∫³ x²y dxdy

Solution:

(a) Here we first integrate with respect to y:

(b) Here we first integrate with respect to x:

Example 2. Evaluate the iterated integral ₁∫²₁∫³ xy² dxdy.

Solution:

 = 28/3

Example 3. Evaluate the iterated integral ₁∫²₂∫⁵ [xy] dxdy

Solution:

Example 4. Evaluate the iterated integral ₀∫¹₀∫² xy(x + y) dxdy

Solution:

 I=2

Example 5. Evaluate the iterated integral ₁∫²₀∫^x2 x dydx

Solution:

Example 6. Evaluate the iterated integral ₀∫^a₀∫^√[a2‒x2] dydx

Solution:

EXERCISE

Evaluate the iterated integral