Integrals are being used to consider the full body based on analysis done on a small part from it, but these analysis are just on a single dimension of any body, for e. g. if we take a cuboid it has three dimensions i. e. its length, breadth and the height. But by to analyse it we have consider all of its dimensions, that's where Multiple Integrals into application. Multiple integrals is there for multiple dimensions of the body. Now when planning on taking a cuboid into consideration we have to be employed in Triple Integration.
The definition of a definite integrals for functions of single variable, while working with the integral of single variable is as below,
f(x) dx
we think of x's as from the interval a ‰ x ‰ b. For these integrals we can say that people are
integrating over the interval a ‰ x ‰ b. Note that this does assume that a < b, however, if we
have b < a then we can just use the interval b ‰ x ‰ a.
Now, when we derived the definition of the definite integral we first considered this as an area
problem. We first asked what the area under the curve was and to do that we broke up the
interval a ‰ x ‰ b into n subintervals of width Оx and choose a spot, , from each interval as
shown below,
Each of the rectangles has height of f() and we're able to then use the area of each of these
rectangles to approximate the area as follows,
A ‰ f(Оx + f(Оx - - - + f() Оx - - - f() Оx
To get the exact area then we take the limits as 'n' that go till infinity, to fulfil the definition of definite integrals
=
This was how we integrate for an individual dimension within a variable, but you want to integrate a function of two variables, f (x, y). With functions of one variable we integrated over an interval (i. e. a one-dimensional space) therefore it creates some sense then that whenever integrating a function of two variables we will integrate over a region of (two dimensional space).
DOUBLE INTEGRALS
We will start out by assuming that the spot in is a rectangle which we will denote the following,
R = [a, b]*[c, d]
This means that the ranges for x and y are a ‰ x ‰ b and c ‰ y ‰ d.
Also, we will primarily assume that
f ( x, y) ‰Ґ 0
although this won't really have to be the situation. Let's start out with the graph of the surface S give by graphing f (x, y) within the rectangle R.
Now, exactly like with functions of 1 variable let's first speak about what the quantity of the region under S (and above the xy-plane) is.
We will first approximate the volume much as we approximated the region above. We will first divide up a ‰ x ‰ b into n subintervals and divide up c ‰ y ‰ d into m subintervals.
This will divide up R into a series of smaller rectangles and from each one of these we will choose a point (, ). Here is a sketch of the setup.
Now, over each one of these smaller rectangles we will construct a box whose height is given by f (, ).
Here is a sketch of that.
Each of the rectangles has basics section of ОA and a height of f (, ) therefore the level of each of
these boxes is f (, ). ОA. The volume under the surface S is then approximately,
V ‰ [f (, ). ОA]
We will have a double sum since we will need to add up volumes in both x and y directions.
To get a better estimation of the quantity we will need n and m larger and larger also to get the
exact volume we should take the limit as both n and m go to infinity. In other words,
V = [f (, ). ОA]
Now, this will look familiar. This looks nearly the same as the definition of the integral of any function of
single variable. Actually this is also the definition of any double integral, or even more exactly an integral
of a function of two variables on the rectangle.
Here is the state definition of a double integral of an function of two variables over the rectangular
region R as well as the notation that we'll use for it.
= [f (, ). ОA]
Note the similarities and dissimilarities in the notation to single integrals. We've two integrals to denote the actual 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 both integrals to denote the spot that people are integrating over. Note that one interpretation of the double integral of f (x, y) over the rectangle R is the volume under the function f (x, y) (and above the xy-plane).
V =
We can use this double sum in the definition to estimate the worthiness of a double integral if we need to. We can do that by choosing (, ) to be the midpoint of every rectangle. When we do this we usually denote the idea as (, ). This leads to the Midpoint Rule,
‰ [f (, ). ОA]
Now we be looking at how to actually compute double integrals.
Fubini's Theorem
If f (x, y) is continuous on R = [a, b]*[c, d] then,
=
These integrals are called iterated integrals.
Note that we now have in fact two ways of computing a double integral and also notice that the inner differential matches up with the limits on the inner integral and similarly for the out differential and limits. In other words, if the inner differential is dy then your limits on the inner integral must be y limits of integration and when the outer differential is dy then your limits on the outer integral must be y limits of integration.
Now, on some level this is just notation and doesn't really reveal how to compute the double
integral. Let's simply take the first possibility above and change the notation just a little.
We will compute the double integral by first computing
and we calculate this by holding x regular and integrating regarding y as though this were an single integral. This will give a function involving only x's which we can in turn integrate. We've done a similar process with partial derivatives. To take the derivative of your function regarding y we treated the x's as constants and differentiated regarding y as if it was a function of an individual variable.
Double integrals work in the same manner. We think of all the x's as constants and integrate with
respect to y or we think of all y's as constants and integrate with respect to x.
In this case we will integrate regarding y first. So, the iterated integral that people need to
compute is,
When setting these up make sure the limits match to the differentials. Since the dy is the inner
differential (i. e. we live integrating regarding y first) the inner integral needs to have y limits.
To compute this we can do the inner integral first and we typically keep the outer integral around
as follows,
(2x) dx
= dx
= dx
Remember that we treat the x as a regular when doing the first integral and we don't do any
integration with it yet. Now, we have a normal single integral so let's finish the integral by
computing this.
= 7 = 84
we can do the integral in either direction. However, sometimes one direction of integration is significantly easier than the other so make sure that you see which one you must do first before actually doing the integral.
The next topic of this section is a quick fact that can be used to make some iterated integrals
somewhat much easier to compute sometimes.
Fact
If f (x, y) = g (x)h( y) and we are integrating within the rectangle R = [a, b]*[c, d] then,
dA = (). ()
So, if we can split up the function into a function only of x times a function of y then we can do
the two integrals individually and multiply them together.
Let's do a example applying this integral.
Example- Evaluate, R= [-2, 3]*[0, ]
Since the integrand is a function of x times a function of y we may use the fact.
(). () =. (]
= + [y+]
=
DOUBLE INTEGRALS OVER GENERAL REGIONS
In the prior section we looked at double integrals over rectangular regions. The challenge with
this is the fact that the majority of the regions are not rectangular so we need to now look at the following double
integral
f(x, y) dA
Such figures show us the region that people have to consider and the details about that area i. e. the points from which its starting or finishing at etc.
Let's do an example to comprehend these kind of figures and problems well,
Example- dA, where D is the spot bounded by y = y =.
Solution- In cases like this we need to determine both inequalities for x and y that we should do the integral.
The best way to do this is the graph the two curves. Here is a sketch.
So, from the sketch we can easily see that that two inequalities are,
0 ‰ x ‰ 1, ‰ y ‰ x
We can now do the integral,
dA = dydx
= [2xy - ]dx
= - + =
TRIPLE INTEGRAL
Now that people learn how to integrate over a two-dimensional region we have to move on to integrating more than 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 the 3d region. The notation for the overall triple integrals is,
dV
Let's start simple by integrating on the box,
B = [a, b]*[c, d]*[r, s]
Note that when by using notation we list the x's first, the y's second and the z's third.
The triple integral in cases like this is,
=
Note that people 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. You will discover 6 different possible orders to do the integral in and which order you are doing the integral in will depend after the function and the order that you feel would be the easiest. We will get the same answer regardless of the order however.
Let's execute a example of this type of triple integral.
Example-Determine the volume of the spot that lies behind the plane x + y + z = 8 and before the region in the yz-plane that is bounded by z = y and z = y.
Solution-
In this case we have been given D therefore we won't have to really work to realize 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 a concept of what the spot we're dealing with appears like.
Now, the graph of the spot above is all okay, but it doesn't really show us what the spot is.
So, here is a sketch of the region itself.
Here are the limits for each and every of the variables.
0 ‰ y ‰ 4
‰ z ‰
The volume is then,
V = = [
= [
= dz
=dy
=(8-) =
So this could it be about the triple integrals as well as the multiple integrals, there are yet some details that are not covered like double integrals in polar coordinates, triple integrals in cylindrical coordinates, and triple integrals in spherical coordinates etc.
APPLICATIONS IN MECHANICAL ENGINEERING
Now the applications of multiple integrals in mechanical engineering are the basic applications of these i. e. to find areas and volumes of various bodies just by taking a little part of these under consideration.
And this does apply in various fields like while preparing a machine, or the parts to built in any machine its size and volume etc. are very important.
