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Table of Contents
Double and triple integrals
History of integrals
Application in engineering
Application and real life.
Integration is about chopping things up, and adding the pieces. This “chopping and adding” viewpoint is often helpful in setting up problems involving integration, and will be especially useful later for multiple integrals.
Figure 1: Chopping a line into small pieces.
For example, consider finding the total amount of chocolate on a straight piece of wafer (like a stick of Pocky), given the density of chocolate on the wafer. What does “density” mean? In this case, the amount of chocolate, which could be measured in grams, per unit distance along the wafer, which could be measured in centimeters. We call this quantity the (linear) mass density of chocolate on the wafer, which we will denote by λ. Using x to measure distance along the wafer, chop the wafer into small pieces of length dx, as indicated symbolically in Figure 1.
http://math.oregonstate.edu/BridgeBook/book/math/single#fn__11) What is the mass of each piece? Clearly, λdx. The total mass M of the chocolate on the wafer is given by adding up the mass of each piece. Since in general λ depends on position, it can be thought of as a function of x, that is, λ=λ(x). Thus, “adding” really means “integrating”, and we obtain
You may have learned that integration is antidifferentiation, and that integrals are areas. Yes, the total amount of chocolate, thought of as a function of the distance from one end, is indeed an antiderivative of the function λ(x). And yes, this integral represents the “area” under the graph of the function λ(x), although the dimensions (mass) are not those of geometric area (length squared). But the “infinitesmal mass” λdx better represents the relevant physical process, and should therefore be regarded as fundamental. Note the importance of dx (and its units) to this argument!
From this point of view, the fundamental theorem of calculus is easy. If you add up little bits of choclate (λdx), you get the total amount of chocolate. Similarly, if you add up the small changes in some quantity, say dq, the charge on a small piece of wire, you get its total change, so that
which might look more familiar if you write q=f(x), so that (
Yes, this means that the integral of a derivative is the function you started with. In the chocolate example, that function is the “total chocolate” function, whose derivative is λ(x). Nontheless, what you're adding up is chocolate, that is, λdx, not merely λ.
The use of differentials has other advantages. In the language of differentials, substitution is easy: If u=x2, then du=2xdx, so that e.g.
And integration by parts is just the product rule: Start with
and integrate both sides (and rearrange terms as needed).
All of the above integrals are indefinite; both sides of the equality are functions. Definite integrals are obtained simply by evaluating both sides at the endpoints of some interval. For example:
and both sides of this equality are now numbers.
In practice, (
http://math.oregonstate.edu/BridgeBook/book/math/single#mjx-eqn-intdiff2) tells us that (single) integration is nothing more than antidifferentation — just run the derivative rules backwards. Here are the basic derivative rules, in differential form:
To obtain the corresponding integration rules, simply integrate both sides, and use (
http://math.oregonstate.edu/BridgeBook/book/math/single#mjx-eqn-intdiff2). For example, the first rule becomes
which is more commonly written in the form
where of course m=n−1≠−1.
A definite integral is an
with upper and lower limits. If is restricted to lie on the
http://mathworld.wolfram.com/RealLine.htmlreal line, the definite integral is known as a
http://mathworld.wolfram.com/RiemannIntegral.htmlRiemann integral (which is the usual definition encountered in elementary textbooks). However, a general definite integral is taken in the complex plane, resulting in the
with , , and in general being complex numbers and the path of integration from to known as a
http://mathworld.wolfram.com/FirstFundamentalTheoremofCalculus.htmlfirst fundamental theorem of calculus allows definite integrals to be computed in terms of
http://mathworld.wolfram.com/IndefiniteIntegral.htmlindefinite integrals, since if is the
http://mathworld.wolfram.com/IndefiniteIntegral.htmlindefinite integral for a
http://mathworld.wolfram.com/ContinuousFunction.htmlcontinuous function , then
This result, while taught early in elementary
http://mathworld.wolfram.com/Calculus.htmlcalculus courses, is actually a very deep result connecting the purely algebraic
http://mathworld.wolfram.com/IndefiniteIntegral.htmlindefinite integral and the purely analytic (or geometric) definite integral. Definite integrals may be evaluated in the
http://www.wolfram.com/language/Wolfram Language using
http://reference.wolfram.com/language/ref/Integrate.htmlIntegrate[f, x, a, b].
The question of which definite integrals can be expressed in terms of
http://mathworld.wolfram.com/ElementaryFunction.htmlelementary functions is not susceptible to any established theory. In fact, the problem belongs to transcendence theory, which appears to be "infinitely hard." For example, there are definite integrals that are equal to the
http://mathworld.wolfram.com/Euler-MascheroniConstant.htmlEuler-Mascheroni constant . However, the problem of deciding whether can be expressed in terms of the values at rational values of
http://mathworld.wolfram.com/ElementaryFunction.htmlelementary functionsinvolves the decision as to whether is rational or algebraic, which is not known.
Integration rules of definite integration include
If is continuous on and is continuous and has an antiderivative on an
http://mathworld.wolfram.com/Interval.htmlinterval containing the values of for , then
http://mathworld.wolfram.com/WatsonsTripleIntegrals.htmlWatson's triple integrals are examples of (very) challenging
http://mathworld.wolfram.com/MultipleIntegral.htmlmultiple integrals. Other challenging integrals include
http://mathworld.wolfram.com/AhmedsIntegral.htmlAhmed's integral and
Definite integration for general input is a tricky problem for computer mathematics packages, and some care is needed in their application to definite integrals. Consider the definite integral of the form
which can be done trivially by taking advantage of the trigonometric identity
Many computer mathematics packages, however, are able to compute this integral only for specific values of , or not at all. Another example that is difficult for computer software packages is
which is nontrivially equal to 0.
Some definite integrals, the first two of which are due to Bailey and Plouffe (1997) and the third of which is due to Guénard and Lemberg (2001), which were identified by Borwein and Bailey (2003, p. 61) and Bailey et al. (2007, p. 62) to be "technically correct" but "not useful" as computed by Mathematica Version 4.2 are reproduced below. More recent versions of
http://www.wolfram.com/language/Wolfram Language return them directly in the same simple form given by Borwein and Bailey without even the need for additional simplification:
http://oeis.org/A091476A091476), where is
http://mathworld.wolfram.com/CatalansConstant.htmlCatalan's constant. A fourth integral proposed by a challenge is also trivially computable in modern versions of the
http://oeis.org/A091477A091477), where is
A pretty definite integral due to L. Glasser and O. Oloa (L. Glasser, pers. comm., Jan. 6, 2007) is given by
http://oeis.org/A127196A127196), where is the
http://mathworld.wolfram.com/Euler-MascheroniConstant.htmlEuler-Mascheroni constant. This integral (in the form considered originally by Oloa) is the case of the class of integrals
previously studied by Glasser. The closed form given above was independently found by Glasser and Oloa (L. Glasser, pers. comm., Feb. 2, 2010; O. Oloa, pers. comm., Feb. 2, 2010), and proofs of the result were subsequently published by Glasser and Manna (2008) and Oloa (2008). Generalizations of this integral have subsequently been studied by Oloa and others; see also Bailey and Borwein (2008).
An interesting class of integrals is
which have the special values
(Bailey et al. 2007, pp. 42 and 60).
An amazing integral determined empirically is
(Bailey et al. 2007, p. 61).
A complicated-looking definite integral of a
http://mathworld.wolfram.com/RationalFunction.htmlrational function with a simple solution is given by
(Bailey et al. 2007, p. 258).
Another challenging integral is that for the volume of the
Integrands that look alike could provide very different results, as illustrated by the beautiful pair
due to V. Adamchik (OEIS
http://oeis.org/A115287A115287; Moll 2006; typo corrected), where is the
http://mathworld.wolfram.com/OmegaConstant.htmlomega constant and is the
http://mathworld.wolfram.com/LambertW-Function.htmlLambert W-function. These can be computed using contour integration.
Computer mathematics packages also often return results much more complicated than necessary. An example of this type is provided by the integral
for and which follows from a simple application of the
http://mathworld.wolfram.com/LeibnizIntegralRule.htmlLeibniz integral rule (Woods 1926, pp. 143-144).
There are a wide range of methods available for
http://mathworld.wolfram.com/NumericalIntegration.htmlnumerical integration. Good sources for such techniques include Press et al. (1992) and Hildebrand (1956). The most straightforward numerical integration technique uses the
http://mathworld.wolfram.com/Newton-CotesFormulas.htmlNewton-Cotes formulas (also called quadrature formulas), which approximate a function tabulated at a sequence of regularly spaced
http://mathworld.wolfram.com/Interval.htmlintervals by various degree
http://mathworld.wolfram.com/Polynomial.htmlpolynomials. If the endpoints are tabulated, then the 2- and 3-point formulas are called the
http://mathworld.wolfram.com/TrapezoidalRule.htmltrapezoidal rule and
http://mathworld.wolfram.com/SimpsonsRule.htmlSimpson's rule, respectively. The 5-point formula is called
http://mathworld.wolfram.com/BoolesRule.htmlBoole's rule. A generalization of the
http://mathworld.wolfram.com/TrapezoidalRule.htmltrapezoidal rule is
http://mathworld.wolfram.com/RombergIntegration.htmlromberg integration, which can yield accurate results for many fewer function evaluations.
If the analytic form of a function is known (instead of its values merely being tabulated at a fixed number of points), the best numerical method of integration is called
http://mathworld.wolfram.com/GaussianQuadrature.htmlGaussian quadrature. By picking the optimal
http://mathworld.wolfram.com/Abscissa.htmlabscissas at which to compute the function, Gaussian quadrature produces the most accurate approximations possible. However, given the speed of modern computers, the additional complication of the
http://mathworld.wolfram.com/GaussianQuadrature.htmlGaussian quadrature formalism often makes it less desirable than the brute-force method of simply repeatedly calculating twice as many points on a regular grid until convergence is obtained. An excellent reference for
http://mathworld.wolfram.com/GaussianQuadrature.htmlGaussian quadrature is Hildebrand (1956).
The June 2, 1996 comic strip FoxTrot by Bill Amend (Amend 1998, p. 19; Mitchell 2006/2007) featured the following definite integral as a "hard" exam problem intended for a remedial math class but accidentally handed out to the normal class:
The integral corresponds to integration over a
http://mathworld.wolfram.com/SphericalCone.htmlspherical cone with opening angle and radius 4. However, it is not clear what the integrand physically represents (it resembles computation of a moment of inertia, but that would give a factor rather than the given ).
Before starting on double integrals let’s do a quick review of the definition of a definite integrals for functions of single variables. First, when working with the integral,
we think of x’s as coming from the interval . For these integrals we can say that we are integrating over the interval . Note that this does assume that , however, if we have then we can just use the interval .
Now, when we
http://tutorial.math.lamar.edu/Classes/CalcI/AreaProblem.aspxderived the definition of the definite integral we first thought of this as an area problem. We first asked what the area under the curve was and to do this we broke up the interval into n subintervals of width and choose a point, , from each interval as shown below,
Each of the rectangles has height of and we could then use the area of each of these rectangles to approximate the area as follows.
To get the exact area we then took the limit as n goes to infinity and this was also the definition of the definite integral.
In this section we want to integrate a function of two variables, . With functions of one variable we integrated over an interval (i.e. a one-dimensional space) and so it makes some sense then that when integrating a function of two variables we will integrate over a region of (two-dimensional space).
We will start out by assuming that the region in is a rectangle which we will denote as follows,
This means that the ranges for x and y are and .
Also, we will initially assume that although this doesn’t really have to be the case. Let’s start out with the graph of the surface S given by graphing over the rectangle R.
Now, just like with functions of one variable let’s not worry about integrals quite yet. Let’s first ask what the volume of the region under S (and above the xy-plane of course) is.
We will first approximate the volume much as we approximated the area above. We will first divide up into n subintervals and divide up into m subintervals. This will divide up Rinto a series of smaller rectangles and from each of these we will choose a point . Here is a sketch of this set up.
Now, over each of these smaller rectangles we will construct a box whose height is given by . Here is a sketch of that.
Each of the rectangles has a base area of and a height of so the volume of each of these boxes is . The volume under the surface S is then approximately,
We will have a double sum since we will need to add up volumes in both the x and y directions.
To get a better estimation of the volume we will take n and m larger and larger and to get the exact volume we will need to take the limit as both n and m go to infinity. In other words,
Now, this should look familiar. This looks a lot like the definition of the integral of a function of single variable. In fact this is also the definition of a double integral, or more exactly an integral of a function of two variables over a rectangle.
Here is the official definition of a double integral of a function of two variables over a rectangular region R as well as the notation that we’ll use for it.
Note the similarities and differences in the notation to single integrals. We have two integrals to denote the fact that we are dealing with a two dimensional region and we have a differential here as well. Note that the differential is dA instead of the dx and dy that we’re used to seeing. Note as well that we don’t have limits on the integrals in this notation. Instead we have the R written below the two integrals to denote the region that we are integrating over.
Note that one interpretation of the double integral of over the rectangle R is the volume under the function (and above the xy-plane). Or,
We can use this double sum in the definition to estimate the value of a double integral if we need to. We can do this by choosing to be the midpoint of each rectangle. When we do this we usually denote the point as . This leads to the Midpoint Rule,
In the next section we start looking at how to actually compute double integrals.
Now that we know how to integrate over a two-dimensional region we need to move on to integrating over a three-dimensional region. We used a double integral to integrate over a two-dimensional region and so it shouldn’t be too surprising that we’ll use a triple integral to integrate over a three dimensional region. The notation for the general triple integrals is,
Let’s start simple by integrating over the box,
Note that when using this notation we list the x’s first, the y’s second and the z’s third.
The triple integral in this case is,
Note that we integrated with respect to x first, then y, and finally z here, but in fact there is no reason to the integrals in this order. There are 6 different possible orders to do the integral in and which order you do the integral in will depend upon the function and the order that you feel will be the easiest. We will get the same answer regardless of the order however.
Let’s do a quick example of this type of triple integral.
Example 1 Evaluate the following integral.
Just to make the point that order doesn’t matter let’s use a different order from that listed above. We’ll do the integral in the following order.
Before moving on to more general regions let’s get a nice geometric interpretation about the triple integral out of the way so we can use it in some of the examples to follow.
The volume of the three-dimensional region E is given by the integral,
Let’s now move on the more general three-dimensional regions. We have three different possibilities for a general region. Here is a sketch of the first possibility.
In this case we define the region E as follows,
where is the notation that means that the point lies in the region D from the xy-plane. In this case we will evaluate the triple integral as follows,
where the double integral can be evaluated in any of the methods that we saw in the previous couple of sections. In other words, we can integrate first with respect to x, we can integrate first with respect to y, or we can use polar coordinates as needed.
Example 2 Evaluate where E is the region under the plane that lies in the first octant.
We should first define octant. Just as the two-dimensional coordinates system can be divided into four quadrants the three-dimensional coordinate system can be divided into eight octants. The first octant is the octant in which all three of the coordinates are positive.
Here is a sketch of the plane in the first octant.
We now need to determine the region D in the xy-plane. We can get a visualization of the region by pretending to look straight down on the object from above. What we see will be the region D in the xy-plane. So D will be the triangle with vertices at , , and . Here is a sketch of D.
Now we need the limits of integration. Since we are under the plane and in the first octant (so we’re above the plane ) we have the following limits for z.
We can integrate the double integral over D using either of the following two sets of inequalities.
Since neither really holds an advantage over the other we’ll use the first one. The integral is then,
Let’s now move onto the second possible three-dimensional region we may run into for triple integrals. Here is a sketch of this region.
For this possibility we define the region E as follows,
So, the region D will be a region in the yz-plane. Here is how we will evaluate these integrals.
As with the first possibility we will have two options for doing the double integral in the yz-plane as well as the option of using polar coordinates if needed.
Example 3 Determine the volume of the region that lies behind the plane and in front of the region in the yz-plane that is bounded by and .
In this case we’ve been given D and so we won’t have to really work to find that. Here is a sketch of the region D as well as a quick sketch of the plane and the curves defining D projected out past the plane so we can get an idea of what the region we’re dealing with looks like.
Now, the graph of the region above is all okay, but it doesn’t really show us what the region is. So, here is a sketch of the region itself.
Here are the limits for each of the variables.
The volume is then,
We now need to look at the third (and final) possible three-dimensional region we may run into for triple integrals. Here is a sketch of this region.
In this final case E is defined as,
and here the region D will be a region in the xz-plane. Here is how we will evaluate these integrals.
where we will can use either of the two possible orders for integrating D in the xz-plane or we can use polar coordinates if needed.
Example 4 Evaluate where E is the solid bounded by and the plane .
Here is a sketch of the solid E.
The region D in the xz-plane can be found by “standing” in front of this solid and we can see that D will be a disk in the xz-plane. This disk will come from the front of the solid and we can determine the equation of the disk by setting the elliptic paraboloid and the plane equal.
This region, as well as the integrand, both seems to suggest that we should use something like polar coordinates. However we are in the xz-plane and we’ve only seen polar coordinates in the xy-plane. This is not a problem. We can always “translate” them over to the xz-plane with the following definition.
Since the region doesn’t have y’s we will let z take the place of y in all the formulas. Note that these definitions also lead to the formula,
With this in hand we can arrive at the limits of the variables that we’ll need for this integral.
The integral is then,
Now, since we are going to do the double integral in polar coordinates let’s get everything converted over to polar coordinates. The integrand is,
HISTORY OF INTEGRATION
http://www.wolfram.com/products/mathematica/Mathematica combines the recent decades of the computer revolution with the previous few centuries of mathematical research, fulfilling the original goal of the early computer pioneers: to do mathematics by computer...
Over 2000 years ago, Archimedes (287-212 BC) found formulas for the surface areas and volumes of solids such as the sphere, the cone, and the paraboloid. His method of integration was remarkably modern considering that he did not have algebra, the function concept, or even the decimal representation of numbers.
Leibniz (1646-1716) and Newton (1642-1727) independently discovered calculus. Their key idea was that differentiation and integration undo each other. Using this symbolic connection, they were able to solve an enormous number of important problems in mathematics, physics, and astronomy.
Fourier (1768-1830) studied heat conduction with a series of trigonometric terms to represent functions. Fourier series and integral transforms have applications today in fields as far apart as medicine, linguistics, and music.
Gauss (1777-1855) made the first table of integrals, and with many others continued to apply integrals in the mathematical and physical sciences. Cauchy (1789-1857) took integrals to the complex domain. Riemann (1826-1866) and Lebesgue (1875-1941) put definite integration on a firm logical foundation.
Liouville (1809-1882) created a framework for constructive integration by finding out when indefinite integrals of elementary functions are again elementary functions. Hermite (1822-1901) found an algorithm for integrating rational functions. In the 1940s Ostrowski extended this algorithm to rational expressions involving the logarithm.
In the 20th century before computers, mathematicians developed the theory of integration and applied it to write tables of integrals and integral transforms. Among these mathematicians were Watson, Titchmarsh, Barnes, Mellin, Meijer, Grobner, Hofreiter, Erdelyi, Lewin, Luke, Magnus, Apelblat, Oberhettinger, Gradshteyn, Ryzhik, Exton, Srivastava, Prudnikov, Brychkov, and Marichev.
In 1969 Risch made the major breakthrough in algorithmic indefinite integration when he published his work on the general theory and practice of integrating elementary functions. His algorithm does not automatically apply to all classes of elementary functions because at the heart of it there is a hard differential equation that needs to be solved. Efforts since then have been directed at handling this equation algorithmically for various sets of elementary functions. These efforts have led to an increasingly complete algorithmization of the Risch scheme. In the 1980s some progress was also made in extending his method to certain classes of special functions.
The capability for definite integration gained substantial power in Mathematica, first released in 1988. Comprehensiveness and accuracy have been given strong consideration in the development of Mathematica and have been successfully accomplished in its integration code. Besides being able to replicate most of the results from well-known collections of integrals (and to find scores of mistakes and typographical errors in them), Mathematica makes it possible to calculate countless new integrals not included in any published handbook.
Gottfried Wilhelm Leibniz: Father of Calculus. Gottfried Leibniz was indeed a remarkable man. During his lifetime between 1646 and 1716, he discovered and developed monumental mathematical theories. In addition, he speculated on the universe, physics and other human philosoph
Newton and Leibniz[
The major advance in integration came in the 17th century with the independent discovery of the
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculusfundamental theorem of calculus by
https://en.wikipedia.org/wiki/Gottfried_LeibnizLeibniz. The theorem demonstrates a connection between integration and differentiation. This connection, combined with the comparative ease of differentiation, can be exploited to calculate integrals. In particular, the fundamental theorem of calculus allows one to solve a much broader class of problems. Equal in importance is the comprehensive mathematical framework that both Newton and Leibniz developed. Given the name infinitesimal calculus, it allowed for precise analysis of functions within continuous domains. This framework eventually became modern
https://en.wikipedia.org/wiki/Calculuscalculus, whose notation for integrals is drawn directly from the work of Leibniz
APPLICATION OF INTEGRATION IN ENGINEERING AND REAL LIFE
In primary school, we learned how to find areas of shapes with straight sides (e.g. area of a triangle or rectangle). But how do you find areas when the sides are curved? We'll find out how in:
http://www.intmath.com/applications-integration/2-area-under-curve.php2. Area Under a Curve and
http://www.intmath.com/applications-integration/3-area-between-curves.php3. Area Between 2 Curves
http://www.intmath.com/applications-integration/4-volume-solid-revolution.php4. Volume of Solid of Revolution explains how to use integration to find the volume of an object with curved sides, e.g. wine barrels.
http://www.intmath.com/applications-integration/5-centroid-area.php5. Centroid of an Area means the centre of mass. We see how to use integration to find the centroid of an area with curved sides.
http://www.intmath.com/applications-integration/6-moments-inertia.php6. Moments of Inertia explains how to find the resistance of a rotating body. We use integration when the shape has curved sides.
http://www.intmath.com/applications-integration/7-work-variable-force.php7. Work by a Variable Force shows how to find the work done on an object when the force is not constant. This section includes Hooke's Law for springs.
(This chapter is easier if you can draw curves confidently.)
You may also wish to see the
http://www.intmath.com/calculus/calculus-intro.phpIntroduction to Calculus.
http://www.intmath.com/applications-integration/8-electric-charges.php8. Electric Charges have a force between them that varies depending on the amount of charge and the distance between the charges. We use integration to calculate the work done when charges are separated.
http://www.intmath.com/applications-integration/9-average-value-function.php9. Average Value of a curve can be calculated using integration.
http://www.intmath.com/applications-integration/hic-head-injury-criterion.phpHead Injury Criterion is an application of average value and used in road safety research.
http://www.intmath.com/applications-integration/10-force-liquid.php10. Force by Liquid Pressure varies depending on the shape of the object and its depth. We use integration to find the force.
In each case, we solve the problem by considering the simple case first. Usually this means the area or volume has straight sides. Then we extend the straight-sided case to consider curved sides. We need to use integration because we have curved sides and cannot use the simple formulas any more. collections of integrals (and to find scores of mistakes and typographical errors in them), Mathematica makes it possible to calculate countless new integrals not included in any published handbook.
In primary school, we learned how to find areas of shapes with straig