614 Views 8 Replies Latest reply: Jun 21, 2006 12:00 AM by Wave-disabled
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Jun 5, 2003
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## Triple Integrals

the problem says The region common to the interiors of the cylinders x^2 + y^2 = 1 and x^2 +z^2 = 1, one-eight of which is shown in the accompanying figure.

I set up the integral in terms of dzdydx
where the domain of x is between 0 and 1, and the domain of y and z both are between 0 and sqrt[1-x^2]. my deficulty on this problem is actually finding the integral. I integrated in term of dz. it has the sqrt[1-x^2] which is really complicated when I used trigonometry function as substitution. By the way, the definition of volume in space is define to be the triple integral of f(x,y,z)=1 over the domain D. Any hints on this problem?
• 10,537 posts since
Oct 22, 2006
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Apr 14, 2006 12:00 AM (in response to Wave-disabled)
Triple Integrals
You have to remember that we get to integrals from two very different directions, and that the way we calculate them has little to do with their underlying meaning.

The basic concept of an integral is an infinite sum of infinitesimals. We have some figure. It can be a line (straight or curved), an area (with simple or complicated boundary, a volume (with any sort of boundary), or even higher dimensional analogs of a volume. We divide the figure into an infinite number of infinitesimals, with an infinetesimal value for each, and add up all the values to get the integral (the whole).

There are a few minor technical problems with this. There are no such things as infinitesimals, and we can't do infinite numbers of operations. So for rigor we divide our figure into a large number of small (but still finite) pieces with a small (but still finite) value for each and consider the limit of the sum as the number of pieces is allowed to grow without limit. But while this may be the formal definition, and the basis for numerical methods, the underlying concept is still the infinite sum of infinitesimals.

Note that with this concept of an integral there is no such thing as a double or triple integral. There are line integrals, area integrals, volume integrals, etc. And the integrals are the infinite sums of infinitesimal lengths, areas, or volumes. Problems generally lead to unique integrals, as specified here. The decomposition of a multidimensional integral into a series of one dimensional integrals is something that is done strictly for the purpose of doing the calculations. There are generally many different ways of decomposing an integral, and much of the practical consideration is finding the decomposition that makes the evaluations the simplest.

The other direction comes from the differential calculus. There we define the derivative of a function, the linear approximation to a function (over some suitably small interval). A natural development from this is the anti-derivative, given a function find another function whose derivative is the original function.

The amazing part is that for the simple case, real functions of one variable, the two concepts become the same. The area under a curve is the sum of all the infinitesimal areas for each of the points and so is an integral. The fundamental theorem of the calculus states that this integral is equal to the difference of an antiderivative at the two end points. So for exact solutions we use antiderivatives, since these can often be calculated. And one can end up thinking of integrals as antiderivatives. But that is not what they are, and that concept does not extend well to higher dimensions (although there are some interesting analogues of the fundamental theorem in higher dimensions).

YOu need to get used to thinking of the integrals as sums. To get a double integral from an area integral we break down the summation into two steps. We divide the area into a infinite set of lines (which, with respect to area, are infinitesimals) and sum those lines. To calculate each line we sum the points (the infinitesimal lengths) on that line. In three dimensions we just extend that a bit. We divide the volume into a set of areas, then divide the areas into lines, and then the lines into points. Then each line is the integral (sum) of its points, each area is an integral of its lines, and the volume is the integral of its areas.

Another way of looking at this is a bit more geometric. A line is something that is swept out by a moving point. An area is something that is swept out by a moving line. And a volume is something that is swept out by a moving area. So we calculate the value of a line by integrating the values of the moving point, the value of an area by integrating the values of the moving line, and the value of a volume by integrating the values of the moving area. There's a technical issue here that only perpendicular motion counts, typically one arranges the problem so that there is only perpendicular motion.

� � � � Tom Gutman
• 10,537 posts since
Oct 22, 2006
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Jun 21, 2006 12:00 AM (in response to Wave-disabled)
Triple Integrals
Oh, we get exact answers (occasionally, mostly in school) all right. But we don't do it calculating with infinitesimals. First we replace the intuitively clear but mathematically questionable concept of an infinitesimal with the mathematically sound concept of a limit. Using that we prove the fundamental theorem of the calculus. Now we can do one dimensional integrals exactly by finding an exact anti-derivative. Higher dimensional integrals are decomposed into multiple integrals and anti-derivatives are found for each.

But in practice we find that most integrals do not have expressible antiderivatives, and we tend to run out of patience with even the ones that do. So we typically do numerical integrations.

� � � � Tom Gutman
• 10,537 posts since
Oct 22, 2006
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Apr 14, 2006 12:00 AM (in response to Wave-disabled)
Triple Integrals
The key to this problem is doing the integrals in the right order. If you set it up in the order you state, dzdydx you should have no problems. What do you get for the two inner integrals.

Note that when you are integrating with respect to y, x is a constant. Also remember that if you take the integrals from zero, you end up calculating the volume of one eighth of the figure, the part that is shown in the diagram. And if you get really lazy you can skip the two inner integrals altogether, by recognizing that the intersection of the figure and a plane perpendicular to the x axis is a square, a square whose area is easy to compute, and you need only integrate that square along the x axis.

� � � � Tom Gutman
• 7,759 posts since
Jun 1, 2007
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Jun 20, 2006 12:00 AM (in response to Wave-disabled)
Triple Integrals
On 6/20/2006 9:47:23 PM, Wave wrote:
>On 4/14/2006 3:36:50 PM, Tom_Gutman
>wrote:
>>There are a few minor
>>technical problems with this.
>>There are no such things as
>>infinitesimals, and we can't
>>do infinite numbers of
>>operations.
>
>If that is the case, how do we actually
>get the "exact" answer to the problem or
>is it really "exact"? another
>definition?

The integrals are exact. They are proven using induction on the infinitesimals expression.

TTFN,
Eden

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