CALCULUS III DOUBLE & TRIPLE INTEGRALS STEP-BY-STEP

CALCULUS III DOUBLE & TRIPLE INTEGRALS STEP-BY-STEP

A Manual For Self-Study prepared by

Antony Foster Department of Mathematics (office: NAC 6/273) The City College of The City University of New York 160 Convent Avenue At 138th Street New York, NY 10031 afoster1955@

Contents

i

Introduction

In this discussion we will extend the concept of a definite integral to functions of two and three variables. Whereas functions of a single variable are usually integrated over intervals, functions of two variables are usually integrated over regions in 2-space and functions of three variables are usually integrated over regions in 3-space. Calculating such integrals will require some new techniques that will be a central focus in our discussion. Once we have developed the basic methods for integrating functions of two and three variables, we will show how such integrals can be used to calculate surface areas and volumes of solids; and we will aso show how they can be used to find masses and centers of gravity of flat plates and three dimensional solids. In addition to our study of integration, we will generalize the concept of a parametric curve in 2-space to a parametric surface in 3-space. This will allow us to work with a wider variety of surfaces than previously possible and will provide a powerful tool for generating surfaces using computers an other graphing utilities such as matlab.

ii

Section 1

DOUBLE INTEGRALS

The notion of a definite integral can be extended to functions of two or more variables. In our discussion we will discuss the double integral, which is the extension to functions of two variables.

Recall that definite integral of a function of any single variable say x, arose from the area problem which we state below.

the area problem. Given a function f : [a, b] R R of any single variable, say x that is continuous and nonnegative on a closed bounded interval [a, b] on the x-axis, find the area of the plane region enclosed between the graph of or the curve y = f (x) and the interval [a, b].

b

N

N

a

f (x) dx

=

lim

max xk0

f (xk) xk

k=1

=

lim

N

f (xk) xk

k=1

(1)

[In the rightmost expression in (1), we use the "limit as N " to encapsulate the process by which we increase the

number of subintervals of [a, b] in such a way that the lengths of the subintervals approach zero.] Integrals of functions

of two variables arise from the problem of finding volumes under surfaces.

volume calculated as a double integral

the volume problem: Let f : D R2 R be a function of any two variables, say x and y. Let it be that f is continuous and is nonnegative on a bounded region D in the xy-plane, find the volume of the solid E in space enclosed between the graph of the surface z = f (x, y) and the xy-planar region D.

The restriction to a bounded region ensures that D does not extend indefinitely in any direction, thus the region D can can be enclosed within some suitably large rectangle R whose sides are parallel to the coordinate axes. The procedure for finding the volume of a solid E in space will be similar to the limiting process used for finding areas, except that now the approximating elements will be rectangular parallelepipeds rather than rectangles. We proceed as follows:

1. Using lines parallel to the coordinate axes, divide the rectangle R enclosing the region D into subrectangles, and exclude from consideration all those subrectangles that contain any points outside of D. This leaves only rectangles that are subsets of D. Assume that there are N such subrectangles, and denote the area of the kth subrectangle by Ak.

2. Choose any arbitrary point in each of the N subrectangles, and denote the point in the kth subrectangle by (xk, yk). The product f (xk, yk) Ak is the volume of a rectangular parallelepiped with base area Ak and height f (xk, yk), so the sum N f (xk, yk) Ak

k=1

1

Muliple Integration

Section 1: DOUBLE INTEGRALS

can be viewed as an approximation to the volume V (E) of the entire solid E.

3. There are two sources of error in the volume approximation: First, the parallelepipeds have flat tops, whereas the surface z = f (x, y) may be curved; the second, the rectangles that form the bases of the parallelepipeds may not completely cover the region D. However, if we repeat the above process with more and more subdivisions in such a way that both the lengths and the widths of the parallelepipeds approach zero, then it is plausible that the errors of both types approach zero, and the exact volume of the solid will be

This suggests the following definition.

N

lim

N

f (xk, yk) Ak.

k=1

2

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