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DOUBLE INTEGRAL, TRIPLE INTEGRAL, ITERATED INTEGRAL

Def. Double integral. Let z = f(x, y) be a function defined and continuous over a finite
region R of the xy-plane. Let this region be
subdivided into n subregions R_{1}, R_{2}, ... ,R_{n} of
respective areas ΔA_{1}, ΔA_{2}, ... ,ΔA_{n}. See Fig. 1.
The manner of subdivision is arbitrary. In each
subregion R_{k} we select a point P_{k}(x_{k}, y_{k}) and form
the sum

Next we define the diameter r_{i} of a subregion R_{i} to
be the greatest distance between any two points
within or on its boundary. The double integral of
the function f(x, y) over the region R is then defined
as

where we stipulate that the maximum r_{i} → 0 as n → 0.

When z = f(x, y) is non-negative over the region R the
double integral 2) may be interpreted as a volume. Any
term f_{k}(x_{k}, y_{k})ΔA_{k} of 1) gives the volume of a vertical
column standing on ΔA_{k} as a base and of height z_{k}. See
Fig. 2. The sum of all these terms gives an approximation
to the volume of the cylindrical column standing on a base
consisting of the region R, sides generated by a line
parallel to the z axis moving along the boundary of R, and
top, the cut-off section of surface z = f(x, y).

Thus we see that in exact analogy to the way we defined the definite integral

as the limit of a sum and interpreted it as an area, we define the double integral as the limit of a sum and interpret it as a volume.

The iterated integral. Consider the function z = f(x, y) defined over a region R as shown in Fig. 3. The double integral of this function over region R is defined by 2) above. We will now demonstrate a method for computing this integral. We know how to find the volume of various solids by the method of slicing. The volume is expressed as the integral

where A(x) represents the area of a variable cross-section of the solid, all the sections being made by planes perpendicular to a fixed line. We shall now employ this method to evaluate the double integral.

Consider the cylindrical column shown in Fig. 4. Label
the cylindrical column “Q”. Let a plane x = x', parallel
to the yz-plane, be passed through column Q as shown.
The cross-section of column Q created by this plane
corresponds to the area under the curve z = f(x', y)
between points (x', Y_{1}) and (x', Y_{2}), where (x', Y_{1}) and
(x', Y_{2}) are the points where the intersecting plane crosses
the boundary of region R. The area of this cross-sectional
slice is given by the integral

Consider the drawing of region R shown in Fig. 5. Let a
be the smallest value which x can have in R, and b the
largest value. R is assumed to be a region such that the
line x = x', for each value of x' between a and b, cuts the
boundary of R just twice, no more. The corresponding values
of y are Y_{1} and Y_{2}, where Y_{1} < Y_{2}. For a given value of x', the
values of Y_{1} and Y_{2} can be found from the equations of the
boundary of R where y = Y_{1}(x) defines one part of the boundary
and y = Y_{2}(x) defines the other part.

Upon dropping the primes, 4) can be written as

The expression A_{s}(x) plays the role of A(x) in formula 3), representing a slice of cylinder Q. Hence
the volume V of cylinder Q is given by the integral

or

The inner integration, with respect to y, is to be performed first. The result is a function of x, which
is then integrated between the limits x = a and x = b. *In the first integration x is regarded as a
constant.*

The expression 5) for V is called an iterated integral. It is usually written as

we have just shown the technique for evaluating double integrals. They are evaluated by an iterated integral. The relationship between iterated integrals and double integrals is so important that we state it in a theorem:

Theorem 1. The double integral of a continuous function f(x, y) over a region A of the plane can be evaluated by either of the two iterated integrals:

where the limits of integration are determined with reference to the boundary of area A.

Def. Iterated integral. An indicated succession of integrals in which integration is to be performed first with respect of one variable, the others being held constant, then with respect to a second, the remaining ones being held constant, etc.

Indefinite iterated integrals. The integral

is an indefinite iterated integral. It is interpreted as

where the inner integral, ∫ f(x, y) dy, is integrated first, with respect to y, regarding x as a constant, and then the outer integration is performed with respect to x, regarding y as a constant. Each integration thus performed is called partial integration, and is the inverse of partial differentiation.

Example 1. Find

Solution.

where the arbitrary function
was introduced in order to provide the most general function
whose partial derivative with respect to x is x^{3} + y^{3}. Now performing the remaining integration with
respect to y, we obtain

where Ψ(x) and Φ(y) are arbitrary functions.

One can confirm the same result will be obtained by reversing the order of integration.

The above integration is verified by observing that

Definite iterated integrals. The integral

is a definite iterated integral where a and b are constants and u_{1} and u_{2} are either constants or
functions of x. The inner integral is integrated first with respect to y, regarding x as constant, and
then the outer integral is integrated with respect to x, regarding y as constant.

This notation can be extended to functions of any number of variables. Thus

represents a definite iterated integral in three variables.

Example 1. Evaluate

Solution.

Area computation using double integrals.

Problem. Compute the area of PQRS bounded by the curves y = f(x) and y = φ(x) and the ordinates at x = a and x = b shown in Fig. 6 using double integrals.

Solution.

If we were to compute this area using single integrals we would get

Note that this integral is exactly the same as

Example. Compute the area enclosed by the parabola y^{2} = 2x + 4 and the line y = x - 2. See Fig.
7.

Solution. We will integrate first with respect to x, summing the little area elements in a horizontal
strip extending from the parabola to the line. The limits of integration are from x = ½ (y^{2} - 4) to x =
y + 2. We then sum these strips by integrating with respect to y between the limits y = -2 and y = 4.
Thus

Volume computation by double integration.

Problem. Compute the volume between the xy-plane and the surface z = f(x, y), bounded on the sides by two arbitrary cylindrical surfaces y = α(x) and y = β(x), and the planes x = a and x = b, as shown in Fig. 8 using double integrals

Solution. The formula is

Example.

Express the volume of one octant of the sphere x^{2} +
y^{2} + z^{2} = r^{2} as a double integral.

See Fig. 9.

Solution.

It is important to interpret this integration as a double summation process. The quantity z Δx Δy is the volume of a rectangular column with base Δy by Δx and height z, as indicated in Fig. 10. One can see that the first integration (with respect to y) sums these columns into a slice extending from y = 0 to y = The second integration, then, sums all the slices from x = 0 to x = r.

****************************

Triple integrals.

****************************

Def. Triple integral. Let f(x, y, z) be a function defined and continuous over a finite region
R of xyz-space. Suppose that R is divided into n subregions R_{1}, R_{2}, ... ,R_{n} of volumes ΔV_{1}, ΔV_{2, .... ,
}ΔV_{n }, respectively. In each subregion R_{k} we select a
point P_{k}(x_{k}, y_{k}, z_{k}) and form the sum

Next we define the diameter r_{i} of a subregion R_{i} to be
the greatest distance between any two points within or
on its boundary. The triple integral of the function f(x,
y) over the region R is then defined as

where we stipulate that the maximum r_{i} → 0 as n → 0.

Evaluation of the triple integral.

Rectangular coordinates.

where the limits of integration are chosen to cover the region R.

Cylindrical coordinates.

where the limits of integration are chosen to cover the region R.

Spherical coordinates.

where the limits of integration are chosen to cover the region R.

Centroids and moments of inertia.

Centroids. The coordinates ( ) of the centroid of a volume satisfy the relations

Moments of inertia. The moments of inertia of a volume with respect to the coordinate axes are given by

References.

Middlemiss. Differential and Integral Calculus.

Ayres. Calculus. (Schaum)

Sherwood, Taylor. Calculus.

Smith, Salkover, Justice. Calculus.

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