# Applications of Multiple Integrals in Engineering

2012 words (8 pages) Essay

18th Jul 2017 Engineering Reference this

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Integrals are used to consider the full body on the basis of analysis done on a small part of 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, this is where Multiple Integrals into application. Multiple integrals are there for multiple dimensions of a body. Now for taking a cuboid into consideration we need to be working 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 coming from the interval a ≤ x ≤ b . For these integrals we can say that we 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 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 a ≤ x ≤ b into n subintervals of width Δx and choose a point, , from each interval as

shown below,

Each of the rectangles has height of f() and we could then use the area of each of these

rectangles to approximate the area as follows,

A ≈ f(Δx + f(Δx – – – + f() Δx – – – f() Δx

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 a single dimension in a single variable, but we 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) 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).

## DOUBLE INTEGRALS

We will start out by assuming that the region in is a rectangle which we will denote as follows,

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 initially assume that

f ( x, y) ≥ 0

although this doesn’t really have to be the case. Let’s start out with the graph of the surface S give by graphing f (x, y) over the rectangle R.

Now, just like with functions of one variable let’s first talk about what the volume of the region under S (and above the xy-plane) is.

We will first approximate the volume much as we approximated the area 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 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 f (,).

Here is a sketch of that.

Each of the rectangles has a base area of ΔA and a height of f (,) so the volume of each of

these boxes is f (,).ΔA. The volume under the surface S is then approximately,

V ≈ [f (,).ΔA]

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,

V = [f (,).ΔA]

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.

= [f (,).ΔA]

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.

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Find out moreNote 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 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 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,

≈ [f ( , ).ΔA]

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 there are 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 the limits on the inner integral must be y limits of integration and if the outer differential is dy then the limits on the outer integral must be y limits of integration.

Now, on some level this is just notation and doesn’t really tell us how to compute the double

integral. Let’s just take the first possibility above and change the notation a little.

We will compute the double integral by first computing

and we compute this by holding x constant and integrating with respect to y as if 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 a function with respect to y we treated the x’s as constants and differentiated with respect to y as if it was a function of a single 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 with respect to y first. So, the iterated integral that we need to

compute is,

When setting these up make sure the limits match up to the differentials. Since the dy is the inner

differential (i.e. we are integrating with respect to y first) the inner integral needs to have y limits.

To compute this we will 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 constant 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 think about which one you should 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 easier to compute on occasion.

## Fact

If f (x, y) = g (x)h( y) and we are integrating over the rectangle R = [a,b]*[c, d] then,

dA = ().()

So, if we can break 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 using this integral.

Example- Evaluate , R= [-2,3]*[0,]

Solution-

Since the integrand is a function of x times a function of y we can use the fact.

## ().() = . (]

= + [y+]

## =

## DOUBLE INTEGRALS OVER GENERAL REGIONS

In the previous section we looked at double integrals over rectangular regions. The problem with

this is that most of the regions are not rectangular so we need to now look at the following double

integral

òò f(x,y) dA

Such figures show us the area that we 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 understand these type of figures and problems well,

Example- dA, where D is the region bounded by y = y = .

Solution- In this case we need to determine the two inequalities for x and y that we need to 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 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 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,

dV

Let’s start simple by integrating over the box,

B = [a,b]*[c, d]*[r, s]

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 example of this type of triple integral.

Example-Determine the volume of the region that lies behind the plane x + y + z = 8 and in front of the region in the yz-plane that is bounded by z = √y and z = y .

## Solution-

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.

0 ≤ y ≤ 4

≤ z ≤

The volume is then,

V = = [

## = [

= dz

=dy

=(8-) =

So this is it about the triple integrals as well as the multiple integrals, there are yet some details which are not covered like double integrals in polar coordinates, triple integrals in cylindrical coordinates, and triple integrals in spherical coordinates etc.

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View our services## APPLICATIONS IN MECHANICAL ENGINEERING

Now the applications of multiple integrals in mechanical engineering are the basic applications of them i.e. to find areas and volumes of various bodies just by taking a little part of them into consideration.

And this is applicable in various fields like while preparing a machine,or the parts to fitted in any machine its size and volume etc. are very important.

Integrals are used to consider the full body on the basis of analysis done on a small part of 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, this is where Multiple Integrals into application. Multiple integrals are there for multiple dimensions of a body. Now for taking a cuboid into consideration we need to be working 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 coming from the interval a ≤ x ≤ b . For these integrals we can say that we 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 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 a ≤ x ≤ b into n subintervals of width Δx and choose a point, , from each interval as

shown below,

Each of the rectangles has height of f() and we could then use the area of each of these

rectangles to approximate the area as follows,

A ≈ f(Δx + f(Δx – – – + f() Δx – – – f() Δx

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 a single dimension in a single variable, but we 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) 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).

## DOUBLE INTEGRALS

We will start out by assuming that the region in is a rectangle which we will denote as follows,

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 initially assume that

f ( x, y) ≥ 0

although this doesn’t really have to be the case. Let’s start out with the graph of the surface S give by graphing f (x, y) over the rectangle R.

Now, just like with functions of one variable let’s first talk about what the volume of the region under S (and above the xy-plane) is.

We will first approximate the volume much as we approximated the area 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 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 f (,).

Here is a sketch of that.

Each of the rectangles has a base area of ΔA and a height of f (,) so the volume of each of

these boxes is f (,).ΔA. The volume under the surface S is then approximately,

V ≈ [f (,).ΔA]

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,

V = [f (,).ΔA]

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.

= [f (,).ΔA]

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 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 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,

≈ [f ( , ).ΔA]

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 there are 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 the limits on the inner integral must be y limits of integration and if the outer differential is dy then the limits on the outer integral must be y limits of integration.

Now, on some level this is just notation and doesn’t really tell us how to compute the double

integral. Let’s just take the first possibility above and change the notation a little.

We will compute the double integral by first computing

and we compute this by holding x constant and integrating with respect to y as if 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 a function with respect to y we treated the x’s as constants and differentiated with respect to y as if it was a function of a single 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 with respect to y first. So, the iterated integral that we need to

compute is,

When setting these up make sure the limits match up to the differentials. Since the dy is the inner

differential (i.e. we are integrating with respect to y first) the inner integral needs to have y limits.

To compute this we will 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 constant 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 think about which one you should 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 easier to compute on occasion.

## Fact

If f (x, y) = g (x)h( y) and we are integrating over the rectangle R = [a,b]*[c, d] then,

dA = ().()

So, if we can break 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 using this integral.

Example- Evaluate , R= [-2,3]*[0,]

Solution-

Since the integrand is a function of x times a function of y we can use the fact.

## ().() = . (]

= + [y+]

## =

## DOUBLE INTEGRALS OVER GENERAL REGIONS

In the previous section we looked at double integrals over rectangular regions. The problem with

this is that most of the regions are not rectangular so we need to now look at the following double

integral

òò f(x,y) dA

Such figures show us the area that we 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 understand these type of figures and problems well,

Example- dA, where D is the region bounded by y = y = .

Solution- In this case we need to determine the two inequalities for x and y that we need to 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 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 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,

dV

Let’s start simple by integrating over the box,

B = [a,b]*[c, d]*[r, s]

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 example of this type of triple integral.

Example-Determine the volume of the region that lies behind the plane x + y + z = 8 and in front of the region in the yz-plane that is bounded by z = √y and z = y .

## Solution-

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.

0 ≤ y ≤ 4

≤ z ≤

The volume is then,

V = = [

## = [

= dz

=dy

=(8-) =

So this is it about the triple integrals as well as the multiple integrals, there are yet some details which 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 them i.e. to find areas and volumes of various bodies just by taking a little part of them into consideration.

And this is applicable in various fields like while preparing a machine,or the parts to fitted in any machine its size and volume etc. are very important.

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