cálculo i, aplicación de la integral

Preface Here are my online notes for my Calculus I course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus I or needing a refresher in some of the early topics in calculus. I’ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from an Algebra or Trig class or contained in other sections of the notes. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class.

2. Because I want these notes to provide some more examples for you to read through, I don’t always work the same problems in class as those given in the notes. Likewise, even if I do work some of the problems in here I may work fewer problems in class than are presented here.

3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible when writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are.

4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class.

Area Between Curves In this section we are going to look at finding the area between two curves. There are actually two cases that we are going to be looking at. In the first case we want to determine the area between ( )y f x= and ( )y g x= on the interval

[a,b]. We are also going to assume that ( ) ( )f x g x≥ . Take a look at the following sketch to

get an idea of what we’re initially going to look at.

In the Area and Volume Formulas section of the Extras chapter we derived the following formula for the area in this case.

( ) ( )b

aA f x g x dx= −∫ (1)

The second case is almost identical to the first case. Here we are going to determine the area between ( )x f y= and ( )x g y= on the interval [c,d] with ( ) ( )f y g y≥ .

In this case the formula is,

( ) ( )d

cA f y g y dy= −∫ (2)

Now (1) and (2) are perfectly serviceable formulas, however, it is sometimes easy to forget that these always require the first function to be the larger of the two functions. So, instead of these formulas we will instead use the following “word” formulas to make sure that we remember that the area is always the “larger” function minus the “smaller” function. In the first case we will use,

upper lower

,function function

b

a

A dx a x b = − ≤ ≤

⌠⌡

(3)

In the second case we will use,

right left

,function function

d

c

A dy c y d = − ≤ ≤

⌠⌡

(4)

Using these formulas will always force us to think about what is going on with each problem and to make sure that we’ve got the correct order of functions when we go to use the formula. Let’s work an example.

Example 1 Determine the area of the region enclosed by 2y x= and y x= . Solution First of all, just what do we mean by “area enclosed by”. This means that the region we’re interested in must have one of the two curves on every boundary of the region. So, here is a graph of the two functions with the enclosed region shaded.

Note that we don’t take any part of the region to the right of the intersection point of these two graphs. In this region there is no boundary on the right side and so is not part of the enclosed area. Remember that one of the given functions must be on the each boundary of the enclosed region. Also from this graph it’s clear that the upper function will be dependent on the range of x’s that we use. Because of this you should always sketch of a graph of the region. Without a sketch it’s often easy to mistake which of the two functions is the larger. In this case most would probably say that 2y x= is the upper function and they would be right for the vast majority of the x’s. However, in this case it is the lower of the two functions. The limits of integration for this will be the intersection points of the two curves. In this case it’s pretty easy to see that they will intersect at 0x = and 1x = so these are the limits of integration. So, the integral that we’ll need to compute to find the area is,

1 2

0

1332

0

upper lowerfunction function

2 13 3

13

b

a

A dx

x x dx

x x

= −

= −

= −

=

⌠⌡

∫

Before moving on to the next example, there are a couple of important things to note. First, in almost all of these problems a graph is pretty much required. Often the bounding region, which will give the limits of integration, is difficult to determine without a graph. Also, it can often be difficult to determine which of the functions is the upper function and which is the lower function without a graph. This is especially true in cases like the last example where the answer to that question actually depended upon the range of x’s that we were using. Finally, unlike the area under a curve that we looked at in the previous chapter the area between two curves will always be positive. If we get a negative number or zero we can be sure that we’ve made a mistake somewhere and will need to go back and find it. Note as well that sometimes instead of saying region enclosed by we will say region bounded by. They mean the same thing. Let’s work some more examples.

Example 2 Determine the area of the region bounded by 2xy x −= e , 1y x= + , 2x = , and the

y-axis. Solution In this case the last two pieces of information, 2x = and the y-axis, tell us the right and left boundaries of the region. Also, recall that the y-axis is given by the line 0x = . Here is the graph with the enclosed region shaded in.

Here, unlike the first example, the two curves don’t meet. Instead we rely on two vertical lines to bound the left and right sides of the region as we noted above Here is the integral that will give the area.

2

0

22

04

2

2


1

1 12 2

7 3.50922 2

b

a

x

x

A dx

x x dx

x x

−

−

−

= −

= + −

= + +

= + =

⌠⌡

∫ e

e

e

Example 3 Determine the area of the region bounded by 22 10y x= + and 4 16y x= + . Solution In this case the intersection points (which we’ll need eventually) are not going to be easily identified from the graph so let’s go ahead and get them now. Note that for most of these problems you’ll not be able to accurately identify the intersection points from the graph and so you’ll need to be able to determine them by hand. In this case we can get the intersection points

by setting the two equations equal.

( ) ( )

2

2

2 10 4 162 4 6 0

2 1 3 0

x xx x

x x

+ = +

− − =

+ − =

So it looks like the two curves will intersect at 1x = − and 3x = . If we need them we can get the y values corresponding to each of these by plugging the values back into either of the equations. We’ll leave it to you to verify that the coordinates of the two intersection points on the graph are (-1,12) and (3,28). Note as well that if you aren’t good at graphing knowing the intersection points can help in at least getting the graph started. Here is a graph of the region.

With the graph we can now identify the upper and lower function and so we can now find the enclosed area.

( )3 2

1

3 2

1

33 2

1


4 16 2 10

2 4 6

2 2 63

643

b

a

A dx

x x dx

x x dx

x x x

−

−

−

= −

= + − +

= − + +

= − + +

=

⌠⌡

∫

∫

Be careful with parenthesis in these problems. One of the more common mistakes students make with these problems is to neglect parenthesis on the second term.

Example 4 Determine the area of the region bounded by 22 10y x= + , 4 16y x= + , 2x = − and 5x = Solution So, the functions used in this problem are identical to the functions from the first problem. The difference is that we’ve extended the bounded region out from the intersection points. Since these are the same functions we used in the previous example we won’t bother finding the intersection points again. Here is a graph of this region.

Okay, we have a small problem here. Our formula requires that one function always be the upper function and the other function always be the lower function and we clearly do not have that here. However, this actually isn’t the problem that it might at first appear to be. There are three regions in which one function is always the upper function and the other is always the lower function. So, all that we need to do is find the area of each of the three regions, which we can do, and then add them all up. Here is the area.

( ) ( ) ( )1 3 52 2 2

2 1 3

1 3 52 2 2

2 1 3

1 3 53 2 3 2 3 2

2 1 3

2 10 4 16 4 16 2 10 2 10 4 16

2 4 6 2 4 6 2 4 6

2 2 22 6 2 6 2 63 3 3

14 64 643 3 3

1423

A x x dx x x dx x x dx

x x dx x x dx x x dx

x x x x x x x x x

−

− −

−

− −

−

− −

= + − + + + − + + + − +

= − − + − + + + − −

= − − + − + + + − −

= + +

=

∫ ∫ ∫

∫ ∫ ∫

Example 5 Determine the area of the region enclosed by siny x= , cosy x= , 2

x π= , and the

y-axis. Solution First let’s get a graph of the region.

So, we have another situation where we will need to do two integrals to get the area. The intersection point will be where sin cosx x=

in the interval. We’ll leave it to you to verify that this will be 4

x π= . The area is then,

( ) ( )

( )

24

0 4

240 4

cos sin sin cos

sin cos cos sin

2 1 2 1

2 2 2 0.828427

A x x dx x x dx

x x x x

π π

π

π π

π

= − + −

= + + − −

= − + −

= − =

∫ ∫

We will need to be careful with this next example.

Example 6 Determine the area of the region enclosed by 21 32

x y= − and 1y x= − .

Solution Don’t let the first equation get you upset. We will have to deal with these kinds of equations occasionally so we’ll need to get used to dealing with them. As always, it will help if we have the intersection points for the two curves. In this case we’ll get

the intersection points by solving the second equation for x and then setting them equal. Here is that work,

( ) ( )

2

2

2

11 32

2 2 60 2 80 4 2

y y

y yy yy y

+ = −

+ = −

= − −

= − +

So, it looks like the two curves will intersect at 2y = − and 4y = or if we need the full coordinates they will be : (-1,-2) and (5,4). Here is a sketch of the two curves.

Now, we will have a serious problem at this point if we aren’t careful. To this point we’ve been using an upper function and a lower function. To do that here notice that there are actually two portions of the region that will have different lower functions. In the range [-2,-1] the parabola is actually both the upper and the lower function. To use the formula that we’ve been using to this point we need to solve the parabola for y. This gives,

2 6y x= ± + where the “+” gives the upper portion of the parabola and the “-” gives the lower portion. Here is a sketch of the complete area with each region shaded that we’d need if we were going to use the first formula.

The integrals for the area would then be,

( ) ( )1 5

3 1

1 5

3 1

1 5 5

3 1 1

4 16 53 322 2

10 4

2 6 2 6 2 6 1

2 2 6 2 6 1

2 2 6 2 6 1

2 1 13 3 2

18

A x x dx x x dx

x dx x x dx

x dx x dx x dx

u u x x

−

− −

−

− −

−

− − −

−

= + − − + + + − −

= + + + − +

= + + + + − +

= + + − +

=

∫ ∫

∫ ∫

∫ ∫ ∫

While these integrals aren’t terribly difficult they are more difficult than they need to be. Recall that there is another formula for determining the area. It is,

right left

,function function

d

c

A dy c y d = − ≤ ≤

⌠⌡

and in our case we do have one function that is always on the left and the other is always on the right. So, in this case this is definitely the way to go. Note that we will need to rewrite the equation of the line since it will need to be in the form ( )x f y= but that is easy enough to do.

Here is the graph for using this formula.

The area is,

( )4

2

2

42

2

43 2

2

right leftfunction function

11 32

1 42

1 1 46 2

18

d

c

A dy

y y dy

y y dy

y y y

−

−

−

= −

= + − −

= − + +

= − + +

=

⌠⌡

⌠⌡

⌠⌡

This is the same that we got using the first formula and this was definitely easier than the first method. So, in this last example we’ve seen a case where we could use either formula to find the area. However, the second was definitely easier. Students often come into a calculus class with the idea that the only easy way to work with functions is to use them in the form ( )y f x= . However, as we’ve seen in this previous

example there are definitely times when it will be easier to work with functions in the form

( )x f y= . In fact, there are going to be occasions when this will be the only way in which a

problem can be worked so make sure that you can deal with functions in this form.

Let’s take a look at one more example to make sure we can deal with functions in this form.

Example 7 Determine the area of the region bounded by 2 10x y= − + and ( )22x y= − . Solution First, we will need intersection points.

( )

( )( )

22

2 2

2

10 2

10 4 40 2 4 60 2 1 3

y y

y y yy yy y

− + = −

− + = − +

= − −

= + −

The intersection points are 1y = − and 3y = . Here is a sketch of the region.

This is definitely a region where the second area formula will be easier. If we used the first formula there would be three different regions that we’d have to look at. The area in this case is,

( )

3 22

1

3 2

1

33 2

1

right leftfunction function

10 2

2 4 6

2 42 63 3

d

c

A dy

y y dy

y y dy

y y y

−

−

−

= −

= − + − −

= − + +

6 = − + + =

⌠⌡

∫

∫

Volumes of Solids of Revolution / Method of Rings In this section we will start looking at the volume of a solid of revolution. We should first define just what a solid of revolution is. To get a solid of revolution we start out with a function,

( )y f x= , on an interval [a,b].

We then rotate this curve about a given axis to get the surface of the solid of revolution. For purposes of this discussion let’s rotate the curve about the x-axis, although it could be any vertical or horizontal axis. Doing this for the curve above gives the following three dimensional region.

What we want to do over the course of the next two sections is to determine the volume of this object. In the final the Area and Volume Formulas section of the Extras chapter we derived the following formulas for the volume of this solid.

( ) ( )b d

a cV A x dx V A y dy= =∫ ∫

where, ( )A x and ( )A y is the cross-sectional area of the solid. There are many ways to get the

cross-sectional area and we’ll see two (or three depending on how you look at it) over the next two sections. Whether we will use ( )A x or ( )A y will depend upon the method and the axis of

rotation used for each problem. One of the easier methods for getting the cross-sectional area is to cut the object perpendicular to the axis of rotation. Doing this the cross section will be either a solid disk if the object is solid (as our above example is) or a ring if we’ve hollowed out a portion of the solid (we will see this eventually). In the case that we get a solid disk the area is,

( )2radiusA π= where the radius will depend upon the function and the axis of rotation. In the case that we get a ring the area is,

2 2outer inner

radius radiusA π

= −

where again both of the radii will depend on the functions given and the axis of rotation. Note as well that in the case of a solid disk we can think of the inner radius as zero and we’ll arrive at the correct formula for a solid disk and so this is a much more general formula to use. Also, in both cases, whether the area is a function of x or a function of y will depend upon the axis of rotation as we will see. This method is often called the method of disks or the method of rings. Let’s do an example. Example 1 Determine the volume of the solid obtained by rotating the region bounded by

2 4 5y x x= − + , 1x = , 4x = , and the x-axis about the x-axis. Solution The first thing to do is get a sketch of the bounding region and the solid obtained by rotating the region about the x-axis. Here are both of these sketches.

Okay, to get a cross section we cut the solid at any x. Below are a couple of sketches showing a typical cross section. The sketch on the right shows a cut away of the object with a typical cross section without the caps. The sketch on the left shows just the curve we’re rotating as well as its mirror image along the bottom of the solid.

In this case the radius is simply the distance from the x-axis to the curve and this is nothing more than the function value at that particular x as shown above. The cross-sectional area is then,

( ) ( ) ( )22 4 3 24 5 8 26 40 25A x x x x x x xπ π= − + = − + − + Next we need to determine the limits of integration. Working from left to right the first cross section will occur at 1x = and the last cross section will occur at 4x = . These are the limits of integration.

The volume of this solid is then,

( )4 4 3 2

1

45 4 3 2

1

8 26 40 25

1 262 20 255 3

785

b

aV A x dx

x x x x dx

x x x x x

π

π

π

=

= − + − +

= − + − +

=

∫

∫

In the above example the object was a solid object, but the more interesting objects are those that are not solid so let’s take a look at one of those. Example 2 Determine the volume of the solid obtained by rotating the portion of the region

bounded by 3y x= and 4xy = that lies in the first quadrant about the y-axis.

Solution First, let’s get a graph of the bounding region and a graph of the object. Remember that we only want the portion of the bounding region that lies in the first quadrant. There is a portion of the bounding region that is in the third quadrant as well, but we don't want that for this problem.

There are a couple of things to note with this problem. First, we are only looking for the volume of the “walls” of this solid, not the complete interior as we did in the last example. Next, we will get our cross section by cutting the object perpendicular to the axis of rotation. The cross section will be a ring (remember we are only looking at the walls) for this example and it will be horizontal at some y. This means that the inner and outer radius for the ring will be x values and so we will need to rewrite our functions into the form ( )x f y= . Here are the

functions written in the correct form for this example.

33

44

y x x yxy x y

= ⇒ =

= ⇒ =

Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right.

The inner radius in this case is the distance from the y-axis to the inner curve while the outer radius is the distance from the y-axis to the outer curve. Both of these are then x distances and so are given by the equations of the curves as shown above. The cross-sectional area is then,

( ) ( ) ( )( ) ( )22 3 2 64 16A y y y y yπ π= − = −

Working from the bottom of the solid to the top we can see that the first cross-section will occur at 0y = and the last cross-section will occur at 2y = . These will be the limits of integration. The volume is then,

( )2 2 6

0

23 7

0

16

16 13 7

51221

d

cV A y dy

y y dy

y y

π

π

π

=

= −

= −

=

∫

∫

With these two examples out of the way we can now make a generalization about this method. If we rotate about a horizontal axis (the x-axis for example) then the cross sectional area will be a

function of x. Likewise, if we rotate about a vertical axis (the y-axis for example) then the cross sectional area will be a function of y. The remaining two examples in this section will make sure that we don’t get too used to the idea of always rotating about the x or y-axis. Example 3 Determine the volume of the solid obtained by rotating the region bounded by

2 2y x x= − and y x= about the line 4y = . Solution First let’s get the bounding region and the solid graphed.

Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about a horizontal axis we know that the cross-sectional area will be a function of x. Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right.

Now, we’re going to have to be careful here in determining the inner and outer radius as they aren’t going to be quite as simple they were in the previous two examples. Let’s start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distance from the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis of rotation to the inner edge of the ring. So, we know that the distance from the axis of rotation to the x-axis is 4 and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference between these two. Or, inner radius 4 x= − The outer radius works the same way. The outer radius is,

( )2 2outer radius 4 2 2 4x x x x= − − = − + + Note that given the location of the typical ring in the sketch above the formula for the outer radius may not look quite right but it is in fact correct. As sketched the outer edge of the ring is below the x-axis and at this point the value of the function will be negative and so when we do the subtraction in the formula for the outer radius we’ll actually be subtracting off a negative number which has the net effect of adding this distance onto 4 and that gives the correct outer radius. Likewise, if the outer edge is above the x-axis, the function value will be positive and so we’ll be doing an honest subtraction here and again we’ll get the correct radius in this case. The cross-sectional area for this case is,

( ) ( ) ( )( ) ( )2 22 4 3 22 4 4 4 5 24A x x x x x x x xπ π= − + + − − = − − +

The first ring will occur at 0x = and the last ring will occur at 3x = and so these are our limits of integration. The volume is then,

( )3 4 3 2

0

35 4 3 2

0

4 5 24

1 5 125 3

1535

b

aV A x dx

x x x x dx

x x x x

π

π

π

=

= − − +

= − − +

=

∫

∫

Example 4 Determine the volume of the solid obtained by rotating the region bounded by

2 1y x= − and 1y x= − about the line 1x = − . Solution As with the previous examples, let’s first graph the bounded region and the solid.

Now, let’s notice that since we are rotating about a vertical axis and so the cross-sectional area will be a function of y. This also means that we are going to have to rewrite the functions to also get them in terms of y.

2

2 1 14

1 1

yy x x

y x x y

= − ⇒ = +

= − ⇒ = +

Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right.

The inner and outer radius for this case is both similar and different from the previous example. This example is similar in the sense that the radii are not just the functions. In this example the functions are the distances from the y-axis to the edges of the rings. The center of the ring however is a distance of 1 from the y-axis. This means that the distance from the center to the edges is a distance from the axis of rotation to the y-axis (a distance of 1) and then from the y-axis to the edge of the rings. So, the radii are then the functions plus 1 and that is what makes this example different from the previous example. Here we had to add the distance to the function value whereas in the previous example we needed to subtract the function from this distance. Note that without sketches the radii on these problems can be difficult to get. So, in summary, we’ve got the following for the inner and outer radius for this example.

2 2

outer radius 1 1 2

inner radius 1 1 24 4

y yy y

= + + = +

= + + = +

The cross-sectional area is then,

( ) ( )22 4

22 2 44 16y yA y y yπ π

= + − + = −

The first ring will occur at 0y = and the final ring will occur at 4y = and so these will be our limits of integration. The volume is,

( )4 4

0

42 5

0

416

1280

965

d

cV A y dy

yy dy

y y

π

π

π

=

= −

= −

=

⌠⌡

∫

Work This is the final application of integral that we’ll be looking at in this course. In this section we will be looking at the amount of work that is done by a force in moving an object. In a first course in Physics you typically look at the work that a constant force, F, does when moving an object over a distance of d. In these cases the work is, W Fd= However, most forces are not constant and will depend upon where exactly the force is acting. So, let’s suppose that the force at any x is given by F(x). Then the work done by the force in moving an object from x a= to x b= is given by,

( )b

aW F x dx= ∫

To see a justification of this formula see the Proof of Various Integral Properties section of the Extras chapter. Notice that if the force is constant we get the correct formula for a constant force.

( )

b

a

b

a

W F dx

Fx

F b a

=

=

= −

∫

where b-a is simply the distance moved, or d. So, let’s take a look at a couple of examples of non-constant forces. Example 1 A spring has a natural length of 20 cm. A 40 N force is required to stretch (and hold the spring) to a length of 30 cm. How much work is done in stretching the spring from 35 cm to 38 cm? Solution This example will require Hooke’s Law to determine the force. Hooke’s Law tells us that the force required to stretch a spring a distance of x meters from its natural length is, ( )F x k x= where 0k > is called the spring constant. The first thing that we need to do is determine the spring constant for this spring. We can do that using the initial information. A force of 40 N is required to stretch the spring 30cm-20cm = 10cm = 0.10m from its natural length. Using Hooke’s Law we have, 40 0.10 400k k= ⇒ = So, according to Hooke’s Law the force required to hold this spring x meters from its natural

length is, ( ) 400F x x= We want to know the work required to stretch the spring from 35cm to 38cm. First we need to convert these into distances from the natural length in meters. Doing that gives us x’s of 0.15m and 0.18m. The work is then,

0.18

0.15

0.182

0.15

400

200

1.98J

W x dx

x

=

=

=

∫

Example 2 We have a cable that weighs 2 lbs/ft attached to a bucket filled with coal that weighs 800 lbs. The bucket is initially at the bottom of a 500 ft mine shaft. Answer each of the following about this.

(a) Determine the amount of work required to lift the bucket to the midpoint of the shaft. (b) Determine the amount of work required to lift the bucket from the midpoint of the shaft to the top of the shaft. (c) Determine the amount of work required to lift the bucket all the way up the

shaft. Solution Before answering either part we first need to determine the force. In this case the force will be the weight of the bucket and cable at any point in the shaft. To determine a formula for this we will first need to set a convention for x. For this problem we will set x to be the amount of cable that has been pulled up. So at the bottom of the shaft 0x = , at the midpoint of the shaft 250x = and at the top of the shaft 500x = . Also at any point in the shaft there is 500 x− feet of cable still in the shaft. The force then for any x is then nothing more than the weight of the cable and bucket at that point. This is,

( )

( )weight of cable weight of bucket/coal

2 500 8001800 2

F x

xx

= +

= − +

= −

We can now answer the questions. (a) In this case we want to know the work required to move the cable and bucket/coal from 0x = to 250x = . The work required is,

( )

( )

250

0

250

0

2502

0

1800 2

1800

387500 ft-lb

W F x dx

x dx

x x

=

= −

= −

=

∫

∫

(b) In this case we want to move the cable and bucket/coal from 250x = to 500x = . The work required is,

( )

( )

500

250

500

250

5002

250

1800 2

1800

262500 ft-lb

W F x dx

x dx

x x

=

= −

= −

=

∫

∫

(c) In this case the work is,

( )

( )

500

0

500

0

5002

0

1800 2

1800

650000 ft-lb

W F x dx

x dx

x x

=

= −

= −

=

∫

∫

Note that we could have instead just added the results from the first two parts and we would have gotten the same answer to the third part. Example 3 A 20 ft cable weighs 80 lbs and hangs from the ceiling of a building without touching the floor. Determine the work that must be done to lift the bottom end of the chain all the way up until it touches the ceiling. Solution First we need to determine the weight per foot of the cable. This is easy enough to get,

80 lbs 4 lb/ft20 ft

=

Next, let x be the distance from the ceiling to any point on the cable. Using this convention we can see that the portion of the cable in the range 10 20x< ≤ will actually be lifted. The portion of the cable in the range 0 10x≤ ≤ will not be lifted at all since once the bottom of the cable has been lifted up to the ceiling the cable will be doubled up and each portion will have a length of 10 ft. So, the upper 10 foot portion of the cable will never be lifted while the lower 10 ft portion will be lifted.

Now, the very bottom of the cable, 20x = , will be lifted 10 feet to get to the midpoint and then a further 10 feet to get to the ceiling. A point 2 feet from the bottom of the cable, 18x = will lift 8 feet to get to the midpoint and then a further 8 feet until it reaches its final position (if it is 2 feet from the bottom then its final position will be 2 feet from the ceiling). Continuing on in this fashion we can see that for any point on the lower half of the cable, i.e. 10 20x≤ ≤ it will be lifted a total of ( )2 10x − .

As with the previous example the force required to lift any point of the cable in this range is simply the distance that point will be lifted times the weight/foot of the cable. So, the force is then,

( ) ( )( )( )( )( )

distance lifted weight per foot of cable

2 10 4

8 10

F x

x

x

=

= −

= −

The work required is now,

( )

( )

20

10

202

10

8 10

4 80

400 ft-lb

W x dx

x x

= −

= −

=

∫

Provided we can find the force, ( )F x , for a given problem then using the above method for

determining the work is (generally) pretty simple. However, there are some problems where this approach won’t easily work. Let’s take a look at one of those kinds of problems. Example 4 A tank in the shape of an inverted cone has a height of 15 meters and a base radius of 4 meters and is filled with water to a depth of 12 meters. Determine the amount of work needed to pump all of the water to the top of the tank. Assume that the density of the water is 1000 kg/m3. Solution Okay, in this case we cannot just determine a force function, ( )F x that will work for us. So, we

are going to need to approach this from a different standpoint. Let’s first set 0x = to be the lower end of the tank/cone and 15x = to be the top of the tank/cone. With this definition of our x’s we can now see that the water in the tank will correspond to the interval [ ]0,12 .

So, let’s start off by dividing [ ]0,12 into n subintervals each of width x∆ and let’s also let *

ix be

any point from the ith subinterval where 1,2,i n= K . Now, for each subinterval we will

approximate the water in the tank corresponding to that interval as a cylinder of radius ir and

height x∆ . Here is a quick sketch of the tank. Note that the sketch really isn’t to scale and we are looking at the tank from directly in front so we can see all the various quantities that we need to work with.

The red strip in the sketch represents the “cylinder” of water in the ith subinterval. A quick application of similar triangles will allow us to relate ir to *

ix (which we’ll need in a bit) as follows.

**

4 415 15

ii i

i

r r xx

= ⇒ =

Okay, the mass, im , of the volume of water, iV , for the ith subinterval is simply,

density i im V= × We know the density of the water (it was given in the problem statement) and because we are approximating the water in the ith subinterval as a cylinder we can easily approximate the volume,

( ) ( )2radius heightiV π≈ , and hence the mass of the water in the ith subinterval.

The mass for the ith subinterval is approximately,

( ) ( ) ( )2 22 * *640415 91000 1000i i i im r x x x x xπ π π ≈ ∆ = ∆ = ∆

To raise this volume of water we need to overcome the force of gravity that is acting on the

volume and that is, iF m g= , where 29.8 msg = is the gravitational acceleration. The force to

raise the volume of water in the ith subinterval is then approximately,

( ) ( )2*64099.8i i iF m g x xπ= ≈ ∆

Next, in order to reach to the top of the tank the water in the ith subinterval will need to travel approximately *15 ix− to reach the top of the tank. Because the volume of the water in the ith subinterval is constant the force needed to raise the water through any distance is also a constant force. Therefore the work to move the volume of water in the ith subinterval to the top of the tank, i.e. raise it a distance of *15 ix− , is then approximately,

( ) ( ) ( ) ( )2* * *640915 9.8 15i i i i iW F x x x xπ≈ − = − ∆

The total amount of work required to raise all the water to the top of the tank is then approximately the sum of each of the iW for 1,2,i n= K . Or,

( ) ( ) ( )2* *6409

19.8 15

n

i ii

W x x xπ=

≈ − ∆∑

To get the actual amount of work we simply need to take n → ∞ . I.e. compute the following limit,

( ) ( ) ( )2* *6409

1lim 9.8 15

n

i in iW x x xπ

→∞=

= − ∆∑

This limit of a summation should look somewhat familiar to you. It’s probably been some time, but recalling the definition of the definite integral we can see that this is nothing more than the following definite integral,

( ) ( ) ( )

( ) ( )

12 122 2 3640 6409 90 0

123 4640 19 4 0

9.8 15 9.8 15

9.8 5 7,566,362.543J

W x x dx x x dx

x x

π π

π

= − = −

= − =

∫ ∫

As we’ve seen in the previous example we sometimes need to compute “incremental” work and then use that to determine the actual integral we need to compute. This idea does arise on occasion and we shouldn’t forget it!

Calculus I

© 2007 Paul Dawkins 9 http://tutorial.math.lamar.edu/terms.aspx

http://tutorial.math.lamar.edu/terms.aspx

Calculus II

© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

Preface Here are my online notes for my Calculus II course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus II or needing a refresher in some of the topics from the class. These notes do assume that the reader has a good working knowledge of Calculus I topics including limits, derivatives and basic integration and integration by substitution. Calculus II tends to be a very difficult course for many students. There are many reasons for this. The first reason is that this course does require that you have a very good working knowledge of Calculus I. The Calculus I portion of many of the problems tends to be skipped and left to the student to verify or fill in the details. If you don’t have good Calculus I skills, and you are constantly getting stuck on the Calculus I portion of the problem, you will find this course very difficult to complete. The second, and probably larger, reason many students have difficulty with Calculus II is that you will be asked to truly think in this class. That is not meant to insult anyone; it is simply an acknowledgment that you can’t just memorize a bunch of formulas and expect to pass the course as you can do in many math classes. There are formulas in this class that you will need to know, but they tend to be fairly general. You will need to understand them, how they work, and more importantly whether they can be used or not. As an example, the first topic we will look at is Integration by Parts. The integration by parts formula is very easy to remember. However, just because you’ve got it memorized doesn’t mean that you can use it. You’ll need to be able to look at an integral and realize that integration by parts can be used (which isn’t always obvious) and then decide which portions of the integral correspond to the parts in the formula (again, not always obvious). Finally, many of the problems in this course will have multiple solution techniques and so you’ll need to be able to identify all the possible techniques and then decide which will be the easiest technique to use. So, with all that out of the way let me also get a couple of warnings out of the way to my students who may be here to get a copy of what happened on a day that you missed.


2. In general I try to work problems in class that are different from my notes. However, with Calculus II many of the problems are difficult to make up on the spur of the moment and so in this class my class work will follow these notes fairly close as far as worked problems go. With that being said I will, on occasion, work problems off the top of my head when I can to provide more examples than just those in my notes. Also, I often


Calculus II

© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

don’t have time in class to work all of the problems in the notes and so you will find that some sections contain problems that weren’t worked in class due to time restrictions.

3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible in writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are.



Calculus II


Arc Length In this section we are going to look at computing the arc length of a function. Because it’s easy enough to derive the formulas that we’ll use in this section we will derive one of them and leave the other to you to derive. We want to determine the length of the continuous function ( )y f x= on the interval [ ],a b . Initially we’ll need to estimate the length of the curve. We’ll do this by dividing the interval up into n equal subintervals each of width x∆ and we’ll denote the point on the curve at each point by Pi. We can then approximate the curve by a series of straight lines connecting the points. Here is a sketch of this situation for 9n = .

Now denote the length of each of these line segments by 1i iP P− and the length of the curve will then be approximately,

11

n

i ii

L P P−=

≈ ∑

and we can get the exact length by taking n larger and larger. In other words, the exact length will be,

11

limn

i in iL P P−→∞

=

= ∑

Now, let’s get a better grasp on the length of each of these line segments. First, on each segment let’s define ( ) ( )1 1i i i i iy y y f x f x− −∆ = − = − . We can then compute directly the length of the line segments as follows.

( ) ( )2 2 2 21 1 1i i i i i i iP P x x y y x y− − −= − + − = ∆ + ∆

By the Mean Value Theorem we know that on the interval [ ]1,i ix x− there is a point *

ix so that,

( ) ( ) ( )( )

( )

*1 1

*

i i i i i

i i

f x f x f x x x

y f x x

− −′− = −

′∆ = ∆


Calculus II


Therefore, the length can now be written as,

( ) ( )

( )

( )

2 21 1 1

22 * 2

2*1

i i i i i i

i

i

P P x x y y

x f x x

f x x

− − −= − + −

′= ∆ + ∆

′= + ∆

The exact length of the curve is then,

( )

11

2*

1

lim

lim 1

n

i in in

in i

L P P

f x x

−→∞=

→∞=

=

′= + ∆

∑

∑

However, using the definition of the definite integral, this is nothing more than,

( ) 21

b

aL f x dx′= + ⌠

⌡

A slightly more convenient notation (in my opinion anyway) is the following.

2

1b

a

dyL dxdx

= +

⌠⌡

In a similar fashion we can also derive a formula for ( )x h y= on [ ],c d . This formula is,

( )2

21 1

dd

cc

dxL h y dy dydy

′= + = +

⌠⌠ ⌡

⌡

Again, the second form is probably a little more convenient. Note the difference in the derivative under the square root! Don’t get too confused. With one we differentiate with respect to x and with the other we differentiate with respect to y. One way to keep the two straight is to notice that the differential in the “denominator” of the derivative will match up with the differential in the integral. This is one of the reasons why the second form is a little more convenient. Before we work any examples we need to make a small change in notation. Instead of having two formulas for the arc length of a function we are going to reduce it, in part, to a single formula. From this point on we are going to use the following formula for the length of the curve. Arc Length Formula(s)


Calculus II


L ds= ∫ where,

( )

( )

2

2

1 if ,

1 if ,

dyds dx y f x a x bdx

dxds dy x h y c y ddy

= + = ≤ ≤

= + = ≤ ≤

Note that no limits were put on the integral as the limits will depend upon the ds that we’re using. Using the first ds will require x limits of integration and using the second ds will require y limits of integration. Thinking of the arc length formula as a single integral with different ways to define ds will be convenient when we run across arc lengths in future sections. Also, this ds notation will be a nice notation for the next section as well. Now that we’ve derived the arc length formula let’s work some examples.

Example 1 Determine the length of ( )ln secy x= between 04

x π≤ ≤ .

Solution In this case we’ll need to use the first ds since the function is in the form ( )y f x= . So, let’s get the derivative out of the way.

2

2sec tan tan tansec

dy x x dyx xdx x dx

= = =

Let’s also get the root out of the way since there is often simplification that can be done and there’s no reason to do that inside the integral.

2

2 21 1 tan sec sec secdy x x x xdx

+ = + = = =

Note that we could drop the absolute value bars here since secant is positive in the range given. The arc length is then,

( )

40

40

sec

ln sec tan

ln 2 1

L x dx

x x

π

π

=

= +

= +

∫

Example 2 Determine the length of ( )32

2 13

x y= − between 1 4y≤ ≤ .

Solution


Calculus II


There is a very common mistake that students make in problems of this type. Many students see that the function is in the form ( )x h y= and they immediately decide that it will be too difficult to work with it

in that form so they solve for y to get the function into the form ( )y f x= . While that can be done here it will lead to a messier integral for us to deal with. Sometimes it’s just easier to work with functions in the form ( )x h y= . In fact, if you can work with

functions in the form ( )y f x= then you can work with functions in the form ( )x h y= . There really

isn’t a difference between the two so don’t get excited about functions in the form ( )x h y= . Let’s compute the derivative and the root.

( )21

21 1 1 1dx dxy y ydy dy

= − ⇒ + = + − =

As you can see keeping the function in the form ( )x h y= is going to lead to a very easy integral. To see

what would happen if we tried to work with the function in the form ( )y f x= see the next example. Let’s get the length.

4

1

432

1

23

143

L y dy

y

=

=

=

∫

As noted in the last example we really do have a choice as to which ds we use. Provided we can get the function in the form required for a particular ds we can use it. However, as also noted above, there will often be a significant difference in difficulty in the resulting integrals. Let’s take a quick look at what would happen in the previous example if we did put the function into the form ( )y f x= . Example 3 Redo the previous example using the function in the form ( )y f x= instead. Solution In this case the function and its derivative would be,

2 13 33 31

2 2x dy xy

dx

− = + =

The root in the arc length formula would then be.


Calculus II


( )

( )( )

( )

( )

222 33 33 22

2 2 13 3 33 3 32 2 2

1111 1xx

x x x

dydx

++ + = + = =

All the simplification work above was just to put the root into a form that will allow us to do the integral. Now, before we write down the integral we’ll also need to determine the limits. This particular ds requires x limits of integration and we’ve got y limits. They are easy enough to get however. Since we know x as a function of y all we need to do is plug in the original y limits of integration and get the x limits of integration. Doing this gives,

( )32

20 33

x≤ ≤

Not easy limits to deal with, but there they are. Let’s now write down the integral that will give the length.

( )

( )

( )32

2 3 233 32

13 320

1x

xL dx

+=

⌠⌡

That’s a really unpleasant looking integral. It can be evaluated however using the following substitution.

2 13 33 31

2 2x xu du dx

− = + =

( )

32

0 12 3 43

x u

x u

= ⇒ =

= ⇒ =

Using this substitution the integral becomes,

4

1

432

1

23

143

L u du

u

=

=

=

∫

So, we got the same answer as in the previous example. Although that shouldn’t really be all that surprising since we were dealing with the same curve. From a technical standpoint the integral in the previous example was not that difficult. It was just a Calculus I substitution. However, from a practical standpoint the integral was significantly more difficult than the integral we evaluated in Example 2. So, the moral of the story here is that we can use either formula (provided we can get the function in the correct form of course) however one will often be significantly easier to actually evaluate. Okay, let’s work one more example.


Calculus II


Example 4 Determine the length of 212

x y= for 102

x≤ ≤ . Assume that y is positive.

Solution We’ll use the second ds for this one as the function is already in the correct form for that one. Also, the other ds would again lead to a particularly difficult integral. The derivative and root will then be,

2

21 1dx dxy ydy dy

= ⇒ + = +

Before writing down the length notice that we were given x limits and we will need y limits for this ds. With the assumption that y is positive these are easy enough to get. All we need to do is plug x into our equation and solve for y. Doing this gives, 0 1y≤ ≤ The integral for the arc length is then,

1 2

01L y dy= +∫

This integral will require the following trig substitution. 2tan secy dy dθ θ θ= =

0 0 tan 0

1 1 tan4

y

y

θ θπ

θ θ

= ⇒ = ⇒ =

= ⇒ = ⇒ =

2 2 21 1 tan sec sec secy θ θ θ θ+ = + = = = The length is then,

( )

( )( )

340

4

0

sec

1 sec tan ln sec tan21 2 ln 1 22

L dπ

π

θ θ

θ θ θ θ

=

= + +

= + +

∫

The first couple of examples ended up being fairly simple Calculus I substitutions. However, as this last example had shown we can end up with trig substitutions as well for these integrals.


Calculus II

Preface

Here are my online notes for my Calculus II course that I teach here at Lamar University. Despite the fact that these are my “class notes” they should be accessible to anyone wanting to learn Calculus II or needing a refresher in some of the topics from the class. These notes do assume that the reader has a good working knowledge of Calculus I topics including limits, derivatives and basic integration and integration by substitution. Calculus II tends to be a very difficult course for many students. There are many reasons for this. The first reason is that this course does require that you have a very good working knowledge of Calculus I. The Calculus I portion of many of the problems tends to be skipped and left to the student to verify or fill in the details. If you don’t have good Calculus I skills and you are constantly getting stuck on the Calculus I portion of the problem you will find this course very difficult to complete. The second, and probably larger, reason many students have difficulty with Calculus II is that you will be asked to truly think in this class. That is not meant to insult anyone it is simply an acknowledgement that you can’t just memorize a bunch of formulas and expect to pass the course as you can do in many math classes. There are formulas in this class that you will need to know, but they tend to be fairly general and you will need to understand them, how they work, and more importantly whether they can be used or not. As an example, the first topic we will look at is Integration by Parts. The integration by parts formula is very easy to remember. However, just because you’ve got it memorized doesn’t mean that you can use it. You’ll need to be able to look at an integral and realize that integration by parts can be used (which isn’t always obvious) and then decide which portions of the integral correspond to the parts in the formula (again, not always obvious). Finally, many of the problems in this course will have multiple solution techniques and so you’ll need to be able to identify all the possible techniques and then decide which will be the easiest technique to use. So, with all that out of the way let me also get a couple of warnings out of the way to my students who may be here to get a copy of what happened on a day that you missed.




Calculus II





Calculus II

Center of Mass In this section we are going to find the center of mass or centroid of a thin plate with uniform density ρ. The center of mass or centroid of a region is the point in which the region will be perfectly balanced horizontally if suspended from that point. So, let’s suppose that the plate is the region bounded by the two curves ( )f x and on the interval [a,b]. So, we want to find the center of mass of the region below.

( )g x

We’ll first need the mass of this plate. The mass is,

( )

( ) ( )

Area of plateb

a

M

f x g x dx

ρ

ρ

=

= −∫

Next we’ll need the moments of the region. There are two moments, denoted by Mx and My. The moments measure the tendency of the region to rotate about the x and y-axis respectively. The moments are given by, Equations of Moments

( ) ( )( )( ) ( )( )

2 212

b

xa

b

y a

M f x g x dx

M x f x g x dx

ρ

ρ

= −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

= −

⌠⎮⌡

∫

The coordinates of the center of mass, ( , )x y , are then,


Calculus II

Center of Mass Coordinates

( ) ( )( )( ) ( )

( ) ( )( )

( ) ( )( )( ) ( )

( ) ( )( )2 2

2 2

1

11 12

2

b

bayb a

a

b

bax

ba

a

x f x g x dxMx x f x g x

M Af x g x dx

f x g x dxM

dx

y f x g x dxM Af x g x dx

−= = = −

−

−⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = = −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

−

⌠⎮⌡ ⌠⎮

⌡

∫∫

∫

∫

where,

( ) ( )b

aA f x g x d= −∫ x

Note that the density, ρ, of the plate cancels out and so isn’t really needed. Let’s work a couple of examples. Example 1 Determine the center of mass for the region bounded by ( )2sin 2y = x 0y =, on the

interval 0,2π⎡ ⎤

⎢ ⎥⎣ ⎦.

Solution Here is a sketch of the region with the center of mass denoted with a dot.

Let’s first get the area of the region.

( )

( )

20

20

2sin 2

cos 2

2

A x

x

π

π

=

= −

dx

=

∫


Calculus II


Now, the moments (without density since it will just drop out) are,

( )

( )

( )

( )

( ) ( )

( ) ( )

22 20 0

2 2200 0

4 220

00

2sin 2 2 sin 2 integrating by parts...

1 cos 4 cos 2 cos 2

1 1sin 4 cos 2 sin 24 2

22

x yM x dx M x x dx

x dx x x x dx

x x x x x

π π

π ππ

π ππ

ππ

= =

= − = − +

⎛ ⎞= − = − +⎜ ⎟⎝ ⎠

==

∫ ∫

∫ ∫

The coordinates of the center of mass are then,

22 4

22 4

x

y

π π

π π

= =

= =

Again, note that we didn’t put in the density since it will cancel out.

So, the center of mass for this region is ,4 4π π⎛ ⎞⎜ ⎟⎝ ⎠

.

Example 2 Determine the center of mass for the region bounded by 3y x= and y x= . Solution The two curves intersect at and 0x = 1x = and here is a sketch of the region with the center of mass marked with a box.

We’ll first get the area of the region.

Calculus II


1 3

0

1342

0

2 13 4

512

A x x dx

x x

= −

⎛ ⎞= −⎜ ⎟⎝ ⎠

=

∫

Now the moments, again without density, are

( ) ( )1 1 360

031 41 2

2 7 0

10 552

0

12

1 1 12 2 7

2 15 5 528

15

yxM x x x dM x x dx

x x dxx x

x x

= −= −

= −⎛ ⎞= −⎜ ⎟⎝ ⎠ ⎛ ⎞

= −⎜ ⎟= ⎝ ⎠

=

⌠⎮⌡

∫

∫

x

The coordinates of the center of mass is then,

1 5 125 12 255 28 35 12 7

x

y

= =

= =

The coordinates of the center of mass are then,12 3,25 7

⎛ ⎞⎜ ⎟⎝ ⎠

.

Calculus II


Preface Here are my online notes for my Calculus II course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus II or needing a refresher in some of the topics from the class. These notes do assume that the reader has a good working knowledge of Calculus I topics including limits, derivatives and basic integration and integration by substitution. Calculus II tends to be a very difficult course for many students. There are many reasons for this. The first reason is that this course does require that you have a very good working knowledge of Calculus I. The Calculus I portion of many of the problems tends to be skipped and left to the student to verify or fill in the details. If you don’t have good Calculus I skills, and you are constantly getting stuck on the Calculus I portion of the problem, you will find this course very difficult to complete. The second, and probably larger, reason many students have difficulty with Calculus II is that you will be asked to truly think in this class. That is not meant to insult anyone; it is simply an acknowledgment that you can’t just memorize a bunch of formulas and expect to pass the course as you can do in many math classes. There are formulas in this class that you will need to know, but they tend to be fairly general. You will need to understand them, how they work, and more importantly whether they can be used or not. As an example, the first topic we will look at is Integration by Parts. The integration by parts formula is very easy to remember. However, just because you’ve got it memorized doesn’t mean that you can use it. You’ll need to be able to look at an integral and realize that integration by parts can be used (which isn’t always obvious) and then decide which portions of the integral correspond to the parts in the formula (again, not always obvious). Finally, many of the problems in this course will have multiple solution techniques and so you’ll need to be able to identify all the possible techniques and then decide which will be the easiest technique to use. So, with all that out of the way let me also get a couple of warnings out of the way to my students who may be here to get a copy of what happened on a day that you missed.




Calculus II






Calculus II


Surface Area In this section we are going to look once again at solids of revolution. We first looked at them back in Calculus I when we found the volume of the solid of revolution. In this section we want to find the surface area of this region. So, for the purposes of the derivation of the formula, let’s look at rotating the continuous function

( )y f x= in the interval [ ],a b about the x-axis. Below is a sketch of a function and the solid of revolution we get by rotating the function about the x-axis.

We can derive a formula for the surface area much as we derived the formula for arc length. We’ll start by dividing the integral into n equal subintervals of width x∆ . On each subinterval we will approximate the function with a straight line that agrees with the function at the endpoints of the each interval. Here is a sketch of that for our representative function using 4n = .

Now, rotate the approximations about the x-axis and we get the following solid.


Calculus II


The approximation on each interval gives a distinct portion of the solid and to make this clear each portion is colored differently. Each of these portions are called frustums and we know how to find the surface area of frustums. The surface area of a frustum is given by, 2A rlπ= where,

( )1 2 1

2

1 radius of right end2

radius of left end

r r r r

r

= + =

=

and l is the length of the slant of the frustum. For the frustum on the interval [ ]1,i ix x− we have,

( )( )

( )

1

2 1

1 1length of the line segment connecting and

i

i

i i i i

r f x

r f x

l P P P P−

− −

=

=

=

and we know from the previous section that,

( ) [ ]2* *

1 11 where is some point in ,i i i i i iP P f x x x x x− − ′= + ∆

Before writing down the formula for the surface area we are going to assume that x∆ is “small” and since ( )f x is continuous we can then assume that,

( ) ( ) ( ) ( )* *1andi i i if x f x f x f x−≈ ≈

So, the surface area of the frustum on the interval [ ]1,i ix x− is approximately,


Calculus II


( ) ( )

( ) ( )

11

2* *

22

2 1

i ii i i

i i

f x f xA P P

f x f x x

π

π

−−

+ =

′≈ + ∆

The surface area of the whole solid is then approximately,

( ) ( ) 2* *

12 1

n

i ii

S f x f x xπ=

′≈ + ∆ ∑

and we can get the exact surface area by taking the limit as n goes to infinity.

( ) ( )

( ) ( )

2* *

1

2

lim 2 1

2 1

n

i in ib

a

S f x f x x

f x f x dx

π

π

→∞=

′= + ∆

′= + ⌠⌡

∑

If we wanted to we could also derive a similar formula for rotating ( )x h y= on [ ],c d about the y-axis. This would give the following formula.

( ) ( ) 22 1

d

cS h y h y dyπ ′= + ⌠

⌡

These are not the “standard” formulas however. Notice that the roots in both of these formulas are nothing more than the two ds’s we used in the previous section. Also, we will replace ( )f x with y and

( )h y with x. Doing this gives the following two formulas for the surface area. Surface Area Formulas

2 rotation about axis

2 rotation about axis

S y ds x

S x ds y

π

π

= −

= −

∫∫

where,

( )

( )

2

2

1 if ,

1 if ,

dyds dx y f x a x bdx

dxds dy x h y c y ddy

= + = ≤ ≤

= + = ≤ ≤

There are a couple of things to note about these formulas. First, notice that the variable in the integral itself is always the opposite variable from the one we’re rotating about. Second, we are allowed to use either ds in either formula. This means that there are, in some way, four formulas here. We will choose the ds based upon which is the most convenient for a given function and problem. Now let’s work a couple of examples.

Example 1 Determine the surface area of the solid obtained by rotating 29y x= − , 2 2x− ≤ ≤


Calculus II


about the x-axis. Solution The formula that we’ll be using here is, 2S y dsπ= ∫ since we are rotating about the x-axis and we’ll use the first ds in this case because our function is in the correct form for that ds and we won’t gain anything by solving it for x. Let’s first get the derivative and the root taken care of.

( ) ( )( )

12 2

12 2

1 9 22 9

dy xx xdx x

−= − − = −

−

2 2

2 2 2

9 31 19 9 9

dy xdx x x x

+ = + = = − − −

Here’s the integral for the surface area,

2

22

329

S y dxx

π−

=−

⌠⌡

There is a problem however. The dx means that we shouldn’t have any y’s in the integral. So, before evaluating the integral we’ll need to substitute in for y as well. The surface area is then,

22

22

2

2

32 99

6

24

S x dxx

dx

π

π

π

−

−

= −−

=

=

⌠⌡

∫

Previously we made the comment that we could use either ds in the surface area formulas. Let’s work an example in which using either ds won’t create integrals that are too difficult to evaluate and so we can check both ds’s. Example 2 Determine the surface area of the solid obtained by rotating 3y x= , 1 2y≤ ≤ about the y-axis. Use both ds’s to compute the surface area. Solution Note that we’ve been given the function set up for the first ds and limits that work for the second ds. Solution 1 This solution will use the first ds listed above. We’ll start with the derivative and root.

231

3dy xdx

−=


Calculus II


4 42 3 3

4 4 23 3 3

1 9 1 9 11 19 9 3

dy x xdx x x x

+ + + = + = =

We’ll also need to get new limits. That isn’t too bad however. All we need to do is plug in the given y’s into our equation and solve to get that the range of x’s is 1 8x≤ ≤ . The integral for the surface area is then,

8 43

23

1

1 483 3

1

9 123

2 9 13

xS x dxx

x x dx

π

π

+=

= +

⌠⌡

⌠⌡

Note that this time we didn’t need to substitute in for the x as we did in the previous example. In this case we picked up a dx from the ds and so we don’t need to do a substitution for the x. In fact if we had substituted for x we would have put y’s into the integral which would have caused problems. Using the substitution

4 13 39 1 12u x du x dx= + =

the integral becomes,

145

10

14532

10

3 32 2

18

27

145 10 199.4827

S u du

u

π

π

π

=

=

= − =

∫

Solution 2 This time we’ll use the second ds. So, we’ll first need to solve the equation for x. We’ll also go ahead and get the derivative and root while we’re at it.

3 23dxx y ydy

= =

2

41 1 9dx ydy

+ = +

The surface area is then,

2 4

12 1 9S x y dyπ= +∫

We used the original y limits this time because we picked up a dy from the ds. Also note that the presence of the dy means that this time, unlike the first solution, we’ll need to substitute in for the x. Doing that


Calculus II


gives,

2 3 4 4

1

145

10

3 32 2

2 1 9 1 9

18

145 10 199.4827

S y y dy u y

u du

π

π

π

= + = +

=

= − =

∫

∫

Note that after the substitution the integral was identical to the first solution and so the work was skipped. As this example has shown we can use either ds to get the surface area. It is important to point out as well that with one ds we had to do a substitution for the x and with the other we didn’t. This will always work out that way. Note as well that in the case of the last example it was just as easy to use either ds. That often won’t be the case. In many examples only one of the ds will be convenient to work with so we’ll always need to determine which ds is liable to be the easiest to work with before starting the problem.


cálculo i, aplicación de la integral

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