F Chapter
12
d (x) = f (x), dx
Applications of Integration
Recall that if f and F are functions such that
then we write f (x) dx = F (x), or, more generally, The constant c
Remember that including the constant of integration c gives the most general antiderivative.
f (x) dx = F (x) + c, and we call F (x) or F (x)+c an indenite integral or antiderivative of f . Also, the denite integral of f from a to b is denoted by
b
f (x) dx,
a
and has the value
b a
f (x) dx = F (b) F (a).
We have studied these ideas in detail earlier, focusing especially on techniques for nding antiderivatives, and familiarity with them is essential for the work of this chapter. If you are not thoroughly familiar with this work, now is the time to go back and revise it!
Copyright c 2004 by the School of Mathematics and Applied Statistics, University of Wollongong
121
122
CHAPTER 12. APPLICATIONS OF INTEGRATION
12.1 The Areas of Regions in the Plane
12.1.1 The area under a curve
Consider the region of the plane lying below the curve y = f (x) and above the x-axis, and between the vertical lines x = a and x = b, as in the diagram:
y
y=f ( x )
A 0 a b x
Also, assume for the moment that f (x) 0, at least in the range a x b, again as in the diagram. (We will consider later the case when this is not so.) How can we calculate the area A of this region? Consider the point x, f (x) on the curve, and another nearby point x + x, f (x + x) , as in the diagram below, and consider the narrow strip of area which lies under the arc of the curve between these two points. (Remember that we typically use the symbol in front of another symbol to indicate a small change in the quantity which followsx is a small change in x, for example.)
12.1. THE AREAS OF REGIONS IN THE PLANE
123
y
y=f ( x )
f(x+Dx) f(x) DA 0 a x x+ D x b x
Then the area A of the narrow strip is given approximately by It is clear from the diagram that this formula cannot be expected to give the exact value for the area A, but only an approximation: we are approximating A by the area of the rectangle shown in the diagram, and this approximation is in error by the area of the small wedge-shaped region lying above the rectangle and below the curve. If we now divide the whole region between x = a and x = b into a number of similar narrow strips, then the total area A will be approximately equal to the sum of the areas of all the strips, each of which has the form f (x)x, for appropriate values of x. (Again, refer to the diagram.) If the region is divided into n intervals of equal length x, and if we let the corresponding x-values be x1 , x2 , x3 , . . . , xn1 , xn , then we have A = sum of areas of n rectangles
n
A height base = f (x)x.
=
i=1
f (xi )x.
If we make the strips narrower, there will be more of them to cover the whole gure, and the diagram makes it plausible that the total error in the approximation will be less. If we continue the process, simultaneously letting n and x 0, then we obtain in the limit the actual area A of the region:
n
A =
n b
lim
f (xi )x
i=1
=
a
f (x) dx.
124
CHAPTER 12. APPLICATIONS OF INTEGRATION
The argument
For completeness, it should be noted that our argument above was not mathematically rigorous. A completely rigorous argument is outside the scope of this course.
This is the outline of the argument leading to the following fundamental fact about areas in the plane:
The Area Under a Curve
If f is integrable on [a, b] and if f (x) 0 for all x [a, b], then the area A of the region bounded by the curve, the x-axis, and the lines x = a and x = b is given by
b
A=
a
f (x) dx.
There is some terminology associated with the argument we have just outlined which is fairly standard and which we will use again several times in these notes. When we divide an interval [a, b] into narrow sub-intervals of length x, as we did above, then it is usual to call these sub-intervals, or the lengths of these sub-intervals, elements of length. Also, in the argument above, each small element of length x dened a corresponding small sub-area, denoted by A, of the total area A under the curve, and these sub-areas are similarly referred to as elements of area. Later, we will discuss elements of volume and elements of arc length. Lengths vs. widths vs. heights Sometimes, also, it is convenient to refer to the elements of length as elements of width. If we are thinking of them simply as sub-intervals (or lengths of sub-intervals), then its appropriate to speak of elements of length, but when we are thinking of them as the bases of the corresponding rectangles, the phrase elements of width is more suitable. Later, when we take our elements of length on the y-axis instead of on the x-axis, it is sometimes better to speak of elements of height. How do we calculate area in the case when the graph of f (x) lies below the x-axis between a and b, instead of above it?
12.1. THE AREAS OF REGIONS IN THE PLANE
125
If f (x) 0 for all x [a, b], then all we need do to nd the area bounded by it and the x-axis is, intuitively, to turn the whole picture upside-down, and then use the idea we have just developed. (Remember that an area cannot be negative!) Mathematically, this means that we perform the same integration as before, but reverse the sign of the answer:
The Area Above a Curve
If f is integrable on [a, b] and if f (x) 0 for all x [a, b], then the area A of the region bounded by the curve, the x-axis, and the lines x = a and x = b is given by
b
A=
f (x) dx.
a
An alternative argument Instead of just adapting the formula already developed for the area under a curve lying above the axis, we could construct an argument from scratch to deal with the case of a curve lying below the axis. That is, we would divide the axis into small elements, add up the corresponding elements of area, and convert the sum into an integral. We would then obtain exactly the expression above. You should try this as an exercise; but in any case, you will see this type of argument used repeatedly in this chapter of the notes. And what if the graph of f (x) actually crosses the x-axis, perhaps several times, between a and b? The answer is simple: split the region up into as many separate regions as necessary, in such a way that f (x) is either above the axis or below the axis in each of them, use whichever of the above two methods is appropriate in each region separately, and nally
126
CHAPTER 12. APPLICATIONS OF INTEGRATION add the answers.
Also, an essential aid in getting this process right is to sketch the graph of the given function rst. It is worth emphasising again that areas cannot be negative, so it is a good idea to check that each of the numbers going into the nal sum is non-negative. This provides one way of guarding against error. Saving on notation Note that we could combine all our results above about areas into a single statement by saying that the area A enclosed by the curve y = f (x) and the x-axis between the lines x = a and x = b is given by
b
A=
a
f (x) dx.
This formula applies regardless of any changes of sign that f might have between a and b, so it seems at rst sight an improvement over our earlier approach, where we needed to use dierent formulae to handle f positive and f negative, and to combine them when f changed sign. But in practice, the new formula gives no improvement at all: if we want to calculate an integral of the form
b
f (x) dx,
a
we will have work out where f is positive and where it is negative, so that we can replace it by f in one case and f in the other, because otherwise we will not be able to perform the integration. Thus, the new formula saves slightly on notation, but doesnt actually save us any work!
Example 12.1
Find the area A of the region bounded by the parabola y = x2 and the x-axis between x = 0 and x = 5. We should always sketch the given region rst. Here, we have:
Solution
12.1. THE AREAS OF REGIONS IN THE PLANE
127
y
y=x
2
0
Therefore, the area is
5
5
x
A =
0
x2 dx x3 3 53 . 3
5
= = Example 12.2 Solution
0
Find the area A of the region bounded by the curve y = x2 7x + 10 and the x-axis from x = 3 to x = 4. Our graph here is:
y
y=x -7x+10
2
3 0 2
4 5 x
From the graph we can see that f (x) 0 for x [3, 4]. Therefore,
4
A =
3
(x2 7x + 10) dx
(note the minus sign!)
128
CHAPTER 12. APPLICATIONS OF INTEGRATION x3 7 2 x + 10x = 3 2 = = 13 . 6
4
3
43 7 2 4 + 10 4 + 3 2
33 7 2 3 + 10 3 3 2
Example 12.3 Solution
Find the area bounded by the curve y = tan x and the x-axis from x = /4 to x = /3. Our graph is:
Calculator?
All the calculations here except for the very last one can be done easily enough without a calculator, but you are free to use a calculator for the other steps if you need to.
Splitting the region up appropriately into two sub-regionsrefer to the graphwe have
0 /3
A =
tan x dx +
/4 0
tan x dx
(note the combination of signs here!)
0
= ln sec x
+ ln sec x
/4
/3 0
= ln sec 0 + ln sec(/4) + ln |sec /3| ln sec 0 3 ln 2 2 1.0397. =
12.1. THE AREAS OF REGIONS IN THE PLANE
129
12.1.2 The area between two curves
Suppose we are now given two curves, y = f (x) and y = g(x), with the graph of f lying above the graph of g, as shown:
y y = f(x) y y = g(x) A1 A2 a 0 b x a 0 b x
and that we wish to calculate not their separate areas, but the area A that lies between them:
y A
y = f(x)
y = g(x) a 0 b x
If we carry out another argument just like the one we used earlier to calculate the area under a single curve, we nd the following:
The Area Between Two Curves
If f and g are integrable on [a, b], and f (x) g(x) for x [a, b], then the area A of the region bounded by the curves y = f (x) and y = g(x) between the lines x = a and x = b is given by
b
A=
a
f (x) g(x) dx.
1210 Exercise
CHAPTER 12. APPLICATIONS OF INTEGRATION
Try to outline this argument as an exercise. The key observation is that the small rectangles constructed in this case will yield elements of area A of the form A height base = f (x) g(x) x, and this leads to the integral above. An important point needs to be emphasised here. Our graphs of f and g above showed the curves lying entirely above the x-axis between a and b. Situating the curves like this makes it a little easier to see the argument which leads to the integral. But in fact, the argument is completely independent of the positions of the curves relative to the x-axis: all that matters are their positions relative to each other. Correspondingly, no assumption is needed in the result above about the positions of the two curves f and g in relation to the x-axiswe did not need to specify that f (x) 0 or g(x) 0, for example. A question Using our formula for the area enclosed by a curve and the x-axis, we can see that the area A enclosed by the graph of the function y = f (x) g(x) and the x-axis between the lines x = a and x = b is given by the integral
b
A=
a
f (x) g(x) dx
(note too that f (x) g(x) 0 in this region). This integral is also, as we have just seen, the expression for the area enclosed between the graphs of f and g. Can you see why this should be so?
Example 12.4
Calculate the area of the region bounded by the line f (x) = 2x and the parabola g(x) = x2 for x between 0 and 2. Our sketch of the region is:
Solution
12.1. THE AREAS OF REGIONS IN THE PLANE
1211
y 4
g(x)=x
2
0 f(x)=2x
Therefore, the area is
2
2
x
A =
0
(2x x2 ) dx x3 3 8 3
2
= = = Example 12.5 Solution
x2 4 4 . 3
0
0
Find the area between the curves f (x) = 1 + sin x and g(x) = cos x for x between 0 and . Our sketch is:
1212
CHAPTER 12. APPLICATIONS OF INTEGRATION
and the area is therefore
A =
0
(1 + sin x) cos x dx
0
=
x cos x sin x
= cos sin + cos 0 = (1) + 1 = + 2. Example 12.6 Solution Find the area between the parabola y = x2 + 1 and the line y = 2x + 1. Our sketch is:
y 5
y=x +1 1 y=2x+1
Therefore, the area is
2
2
0
2
x
A =
0 2
(2x + 1) (x2 + 1) dx 2x x2 dx x3 3
2 0
=
0
=
x2 8 3
= 4 = 4 . 3
12.1. THE AREAS OF REGIONS IN THE PLANE
1213
Note that this is the same answer as for Example 12.4. This is not a coincidence! Compare the two examples to see why, and note the connection with the comments made just before Example 12.4. Example 12.7 Solution Findthe area of the region bounded by the x-axis and the curves y = x and y = 6 x.
This is a slightly more complicated example than the previous ones, so we will be a little more systematic in the way we write out the solution. Step 1. As always, the rst step is to draw a graph. We can make a rough sketch easily enough without doing much calculation, because we have a fair idea what the two curves look like individually. But a reasonably accurate sketch will be needed herewe will need to know where the two curves intersect, in particularso we will do the algebra required for this rst. For possible intersections, we must have y = x = 6 x,
At x = 9?
What happened at x = 9? The answer is that we introduced a superuous solution to the equation when we squared. If there is any doubt, check that apparent solutions really are solutions, by substituting in.
Therefore, any intersections must occur at x = 4 and x = 9. However, x = 9 does not satisfy 6 x = x, while x = 4 does. Therefore, the sole point of intersection is at x = 4. (In fact, x = 9 lies outside the range 0 x 6 in any case, but we could easily nd an example where a spurious solution did lie in the range of interest.) Step 2. We now have enough information to draw a reasonably detailed diagram:
and if we square both sides and regroup terms, we get x2 13x+36 = 0, and, factorising, (x 4)(x 9) = 0.
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CHAPTER 12. APPLICATIONS OF INTEGRATION
Step 3. Now we proceed to nd the area. We will have to split the interval [0, 6] into the two intervals [0, 4] and [4, 6] because (as we see from the sketch) the region is bounded above by the curve y = 6 x from x = 0 to x = 4, and by the curve y = x from x = 4 to x = 6. Thus we have
4
A =
0
6
x dx +
4 4
(6 x) dx x2 2
6
= =
2 3/2 x 3
0
+ 6x
4
16 +2 3 22 . = 3
12.1.3 Areas bounded by the y-axis
How do we nd areas, as in the diagram below, which are bounded by a curve and the y-axis (rather than by a curve and the x-axis, as in our applications so far)?
The answer, intuitively at least, is simple: Turn the picture on its side, since this interchanges the two axes, and then use the previous methods! What does this entail mathematically? It means that we proceed as in our earlier applications, but that we interchange the roles of the variables x and y everywhere. Thus, we will need to express x as a function of y, and then
12.1. THE AREAS OF REGIONS IN THE PLANE integrate with respect to y, between limits which are y-values. In summary, we have the following.
1215
The Area Between a Curve and the y-Axis
Suppose that the graph of x = g(y) lies to the right of the y-axis between y = c and y = d. Then the area A of the region bounded by x = g(y) and the yaxis between the horizontal lines y = c and y = d is given by
d
A=
c
g(y) dy.
Example 12.8
Find the area bounded by the curve y = ln x and the y-axis from y = 1 to y = 1. The graph reveals that the curve lies to the right of the y-axis:
Solution
y 1
y =ln x
x -1
Rewriting x as a function of y, we have x = ey . Then the area is
d
A =
c 1
g(y) dy ey dy
1
=
1216
CHAPTER 12. APPLICATIONS OF INTEGRATION = ey
1 1
= e e1 . Example 12.9 Find the area enclosed by the line y = x1 and the parabola y 2 = 2x+6 by (a) (b) Solution integrating with respect to x, and integrating with respect to y.
Again, we will be a little more systematic in our approach. Step 1. We nd the intersection(s) of the curves: we have y 2 = 2x + 6 = (x 1)2 , and expanding, regrouping and factorising gives (x 5)(x + 1) = 0. Hence, respectively. (Substitution conrms that both pairs of x, y-values are genuine solutionswhich we should expect in this case, since we did not square to nd them.) Step 2. We draw the graph: x = 5, 1 and y = 4, 2,
y=x-1 y 4 (5,4)
2x + 6
-3 5 x
- 2x + 6
(-2,-1) y =2x+6
2
12.1. THE AREAS OF REGIONS IN THE PLANE Step 3. (a) Now we can nd the area in the two ways required.
1217
First, we will integrate with respect to x. Here we are required to split the interval [3, 5] into the two sub-intervals [3, 1] and [1, 5], since the graph shows that the curves cross x = 1. The area over [3, 1], doubly-shaded in the diagram, is
1 3
2x + 6 2x + 6
dx = 2
1 3
2x + 6 dx,
and therefore we have A = 2 = 2 =
1 3
5
2x + 6 dx +
1 1 3
2x + 6 (x 1) dx
5
1 2 (2x + 6)3/2 2 3
+
1 2 x2 (2x + 6)3/2 +x 2 3 2
1
25 1 1 2 3/2 1 (4) + (16)3/2 + 5 (4)3/2 + + 1 3 3 2 3 2 = 18.
(b)
Second, we will integrate with respect to y. Expressing x as a function of y in both cases, we nd that y = x 1 implies x = y + 1 and that Therefore,
4
1 y 2 = 2x + 6 implies x = (y 2 6). 2
A =
2
1 (y + 1) (y 2 6) dy 2
4
= =
y2 1 + y y 3 + 3y 2 6
2
64 4 8 16 +4 + 12 + 2 + 6 2 6 2 6 = 18,
as before.
1218
CHAPTER 12. APPLICATIONS OF INTEGRATION
Which method? Having solved the above problem in two ways, we can see that the second solution was distinctly easier than the rst: we didnt have to split the area into two parts, and the integration itself was simpler. With practice, you will get to be able to judge in advance with some accuracy which of several methods is best to apply to a given problem. Try to see what it was about this problem which made the second method better than the rst.
12.1.4 Areas in polar coordinates
Consider the following graph of a polar curve r = f ():
q=p/2
C r=f(q)
q=b
DA Dq B a 0 b
q=a
q=0
We wish to nd the area A of the region bounded by the curve r = f () between the rays OB (corresponding to = ) and OC (corresponding to = ). Consider the wedge-shaped element of area A of our region dened by some element of angle . If is small, then Area of sector
Remember: The area of a sector of angle of a circle 1 of radius r is 2 r 2 .
A area of sector of circle 1 2 r . = 2 We now divide the whole region between OB and OC into many such small wedge-shaped regions. The sum of the elements of area for these regions will then give an approximation to the total area of the region BOC. Specically, if we divide into n regions, each with angle , and evaluate the function at the point i in the ith region, then we will
12.1. THE AREAS OF REGIONS IN THE PLANE have A
1219
n
i=1
1 f (i )2 . 2
As 0 and as n , we have (much as we found earlier for the area under a curve):
The Area in Polar Coordinates
The area A of the region bounded by the curve r = f () between the rays OB (corresponding to = ) and OC (corresponding to = ) is given by A = = 1 2 1 2
r 2 d
f () d.
2
Example 12.10 Solution
Find the area of the region bounded by the cardioid r = 1 cos . Step 1. Our graph here is the following (revise your notes on polar coordinates if this is unfamiliar):
q=p/2
q=0 r =1 - cosq
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CHAPTER 12. APPLICATIONS OF INTEGRATION
Note that in order to draw this curve, we must consider all values of such that 0 2. Step 2. We need to determine and . From the graph (or the process used in nding the graph), we nd that the interval of integration is from = 0 to = 2. Step 3. Techniques?
Revise your techniques of integration if need be. In this chapter, we wont normally stop to detail the integration methods we are using.
The area is therefore A = = 1 2 1 2
2 0 2 0
(1 cos )2 d 1 2 cos + cos2 d
2
1 1 1 2 sin + sin cos + = 2 2 2 = 3 . 2
0
Example 12.11 Solution
Find the area in the rst quadrant within the cardioid r = 1 cos . From the sketch in the previous example, we see that the values of which give rise to points in the rst quadrant are those in the range 0 /2. Hence our range of integration will be from = 0 to = /2. Therefore, 1 A = 2
/2 0
(1 cos )2 d
/2 0
1 1 1 = 2 sin + sin cos + 2 2 2 = Example 12.12 3 1. 8
Find the area inside one leaf of the eight-leafed rose r = 6 sin 4.
12.1. THE AREAS OF REGIONS IN THE PLANE Solution
1221
To sketch a single leaf of the rose, all we need do is nd two consecutive values of for which r = 0. These can be found by the following method: 6 sin 4 = 0 implies sin 4 = 0, which implies 4 = n,
for any n N. Therefore, 6 sin 4 = 0 when = n , 4 for n N.
Hence, one leaf of the rose is dened by 0 /4, and our graph is:
Therefore, the area is A = = 1 2 1 2
/4 0 /4
[6 sin 4]2 d 36 sin2 4 d
0 /4
= 8
0
sin2 4 d
= Example 12.13
9 . 4
Find the area of the region that lies within the polar curve r = 1+2 cos and outside the circle r = 2. The geometry is relatively complicated here, and an accurate diagram is vital. We nd:
Solution
1222
CHAPTER 12. APPLICATIONS OF INTEGRATION
The points of intersection of the two curves occur at = and 3 = , as shown on the graph (check the calculations for this). 3 Also, the area we want, as shaded, will be found as the dierence between the area enclosed by the curve r = 1 + 2 cos and the area enclosed by the curve r = 2, for 3 3 in each case. Thus if we change notation slightly, writing r1 = 1 + 2 cos our area A will be given by A = 1 2
2 2 (r1 r2 ) d
and r2 = 2,
1 = 2 = =
0
/3 /3
(1 + 2 cos )2 22 d 1 + 4 cos + 4 cos2 4 d
1 2 2
/3
/3 0
4 cos + 4 cos2 3 d
=
5 3 . 2 3
12.2. VOLUMES OF REVOLUTION
Exercise 12A
1223
1. Find the area bounded by (a) the x-axis and the parabola given by y = 2x x2 ,
4. (a) Let y = f (x) = sin x, x [0, 2]. Sketch the graph of f , and nd the area bounded by f on [0, 2]. (b) Let y = f (x) = |sin x| for x [0, 2]. Sketch the graph of f , and nd the area bounded by f on [0, 2]. (c) Decide whether the following statement is true or false: If f is integrable on [a, b], then the area bounded by f
b
(b) the graph of the equation y 2 = x and the line x = 4, (c) the function y = x2 and the line y = x, (d) the functions y = sin x and y = cos x and y = 0 for x between 0 and , 2
(e) the line y = x + 2 and the parabola y = x2 , (f) the lines y = x, y = 1 bola y = x ,
x 4
from a to b is
a
f (x) dx.
and the hyper-
5. After making suitable sketches, nd the areas of the regions: (a) bounded by the cardioid r = 77 sin ; (b) bounded by the cardioid r = a+a sin , for a > 0; (c) inside the large loop of the limaon c r = 2 4 sin ; (d) inside the lemniscate r 2 = 5 cos 2; (e) inside one leaf of the three-leafed rose r = 4 cos 3; (f) outside the circle r = 2 and inside the lemniscate r 2 = 8 cos 2; (g) in the rst quadrant which is inside the cardioid r = 3+3 cos and outside the cardioid r = 3 + 3 sin .
(g) the lines y = 2, y = x and the parabola y 2 = x, and which lies above y = 1, (h) the parabola y 2 = x + 1 and the line y = x + 1. 2. The area bounded by the parabola y = the x-axis, and the line x = 8 3 2 is divided into two equal parts by the line x = c. Find c. 3. Find the area above the x-axis bounded by x2 y 2 the graph of the ellipse 2 + 2 = 1. a b x2 ,
12.2 Volumes of Revolution
In this section we will discuss two methods for nding the volume of a solid of revolution, that is, the 3-dimensional solid obtained when a region of the plane is rotated about some axis in the plane. The methods are called the disc method and the shell method.
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CHAPTER 12. APPLICATIONS OF INTEGRATION
12.2.1 The disc method
Suppose that we are given a function y = f (x), for x between a and b, we know that f (x) 0 in that range, and we wish to rotate the graph of f about the x-axis and compute the volume V of the solid of revolution so obtained. The following graph represents the given information:
Our approach to nding an expression for the volume will be as follows. We rotate the area shown about the x-axis, producing the required solid of revolution. We consider, as shown on the graph, a typical small segment or element of the x-axis between a and b, of width or length x. We estimate the element of volume V contributed to the total volume of revolution V by rotation of this segment about the x-axis, as in the diagram below. Finally, we produce an integral expression for the total volume V from the expression for V .
12.2. VOLUMES OF REVOLUTION
1225
Since the element of volume produced by the rotation has the shape of a thin disc, the method we are using is called the disc method. In more detail, consider the thin strip or element of area in the plane which lies below the curve and above the x-axis and has base equal to the chosen segment, as shown in the diagram above. We approximate the strip by a rectangle of height f (x), for some point x in the interval. Since the width of the rectangle is x, the length of the segment, the corresponding disc has radius f (x) (the height of the rectangle) and thickness x (the width of the rectangle), as in the following diagram:
1226
CHAPTER 12. APPLICATIONS OF INTEGRATION
Therefore, the element of volume V is given approximately by V area of circle of radius f (x) thickness = f (x) x. In our terminology from earlier, we have expressed the element of volume V in terms of the element of length x. We now follow the procedure we used in a number of earlier arguments. We add together large numbers of such elements of volume, we take a limit as their number goes to innity and their size goes to zero, and we obtain in the limit an integral expression for the total volume V . Specically, we obtain
b 2
V
=
a b
f (x) y 2 dx.
a
2
dx
=
In summary, then, we have the following.
The Disc Method
Suppose that y = f (x) 0 for a x b. Then the volume of revolution V obtained by rotating the graph of f between x = a and x = b about the x-axis is given by
b
V
=
a b
f (x) dx y 2 dx.
a
2
=
For emphasis, let us note here that in the disc method, our element of areathe narrow stripis taken perpendicular to the axis of rotation. (When we look at the shell method shortly, we will nd that our element of area is parallel to the axis of rotation.)
12.2. VOLUMES OF REVOLUTION Example 12.14
1227
Find the volume of the solid generated by rotating the area bounded by y = x and the x-axis, between the points x = 1 and x = 4, about the x-axis. Directly from the disc method formula, we have
4
Solution A diagram?
This example is simple enough for us to proceed without drawing a diagram rst, but normally that should be our rst step.
V
=
1 4
( x)2 dx x dx
=
1
= =
x2 2
4
1
15 . 2
But to reinforce the ideas we have usedwe will be using them frequently!we will also show how to solve the problem directly, bypassing the general formula for the volume. We start by sketching the situation. First, for the curve and the resulting solid, we have:
Then, drawing the small disc (or element of volume) on its own, but turned on its face, we have:
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CHAPTER 12. APPLICATIONS OF INTEGRATION
The volume of the small disc, our element of volume, is given by V radius2 thickness = ( x)2 x = xx, and when we sum large numbers of these elements and take a limit, we obtain an integral expression for the entire volume V , namely,
4
V =
1
x dx =
15 , 2
as we found by applying the formula directly.
12.2.2 The disc method in non-standard cases
The disc method applies most naturally to a rotation about the x-axis, but the same idea can be used for rotation about other lines in the plane. We will examine rst how to use it for a rotation about the y-axis, and we will then investigate how to use it in some less standard applications. Once you understand these applications, you should be in a position to adapt the method to t any given situationrotation about a line parallel to an axis, for example. We start by considering a rotation about the y-axis. The idea here is basically quite simple: we ip our diagram over so that the x-axis becomes the y-axis, and vice versa. Mathematically, this ip corresponds to interchanging the roles of the variables x and y everywhere. In fact, we encountered exactly this idea earlier, in subsection 12.1.3, where we used our knowledge of how to calculate the area between a curve and the x-axis to calculate the area between a curve and the yaxis. Repeating the procedure from subsection 12.1.3, we simply need to express x as a function of y, and then integrate with respect to y, between limits which are y-values.
12.2. VOLUMES OF REVOLUTION
1229
We argue as follows. Our disc method formula, in its standard form, says that the volume V for a rotation about the x-axis is given by
b
V
=
a b
f (x) dx y 2 dx.
a
2
= x = g(y)?
If f is one-to-one on [a, b], then f will have an inverse f 1 , and we will have x = g(y) = f 1 (y) and c = f (a) and d = f (b).
If we instead want the volume V for a rotation about the y-axis, then our formula will be
d
V
=
c d
g(y) dy x2 dy,
c
2
=
where x = g(y) expresses x as a function of y, given that y = f (x), and where c and d are the appropriate limits on the y-axis. Let us see an example of the formula in use. Find the volume of the solid generated when the region bounded by y = x, y = 2 and x = 0 is revolved about the y-axis. The original equation y = x becomes x = y 2 , while the boundaries specied by y = 2 and x = 0 give us limits of integration c = 0 and d = 2 with respect to y. Therefore, applying the formula we just developed, the volume is
2
Example 12.15 Solution
V
=
0 2
(y 2 )2 dy y 4 dy
0 2
= y5 = 5 = 32 . 5
0
Once again, let us solve the problem directly, bypassing the new formula. First, as always in such a case, we sketch the situation. The region to be rotated is as follows:
1230
CHAPTER 12. APPLICATIONS OF INTEGRATION
y
y= x
2 Dy or 2 (x=y )
0
4
x
the volume of revolution itself is:
y 2 Dy r 0 4 x x=y
2
and the small disc, or element of volume, obtained by slicing through the large body is:
r=y
2
Dy
Now the volume of the small disc, our element of volume, is given approximately by V radius2 thickness = (y 2)2 y = y 4y, and when we sum large numbers of these elements and take a limit, we obtain an integral expression for the entire volume V , namely,
2
V =
0
y 4 dy =
32 , 5
as before.
12.2. VOLUMES OF REVOLUTION
1231
We now move to some rather dierent, non-standard kinds of applications of the disc method. Here, merely remembering the disc method formula is of limited use. Instead, we must rst analyse and understand exactly what the problem is that we have to solve. Only then will we be able to tell which of our formulae, if any, is appropriate to the problem, or, if none is appropriate, how to proceed to a solution from rst principles. Example 12.16 Find the volume of the solid generated by rotating the area bounded by y = 0, x = 1 and y = x3 1. Solution 1. about the line x = 1, 2. about the y-axis. The that region we are required to rotate is the shaded region in the rst diagram below. Note carefully that we are not rotating the region about either of the coordinate axes, but about the line x = 1. Because of this, neither of the two disc method formulae that we have derived can be applied, since they were both for the case of rotation about one of the coordinate axes. The best way of proceeding in a non-standard case like this, where none of our formulae applies directly to the problem, is to work out the required integral expression for the volume from rst principlesthat is, to divide the volume into small elements, and then to convert that expression into an integral, just as we did when developing the original formula for the disc method. (In Example 12.15, we did this as an alternative to direct application of the formula, as well as by applying the formula directly. In the present case, working from rst principles is by far the most reliable procedure.) Thus the diagram of the situation is as follows:
y x=1 1 Dy x x 1 1-x y=x
3
1232
CHAPTER 12. APPLICATIONS OF INTEGRATION
We wish to break our solid of revolution up into small elements of volume, based on a division of the region into small elements of area, and we rst need to be clear in which direction the elements of area runhorizontally or vertically. The answer is given by the observation we made earlier: in the disc method, elements of area are taken perpendicular to the axis of rotation. Since our axis of rotation here is vertical (though its not one of the coordinate axes), our elements of area must therefore run horizontally, as already indicated in the diagram. Hence, arguing broadly as we have done several times already, we take an element of height y perpendicular to the line x = 1, form the long thin horizontal strip or element of area of height y and width 1 x (the distance from the curve to the line x = 1), calculate the volume V of the disc dened by the rotation of this strip about the line x = 1 to be approximately V radius2 thickness = (1 x)2 y, and nally convert this to an integral expression for the total volume:
1
V =
0
(1 x)2 dy.
Two diagrams representing this process are as follow:
y 1 y=x
3
Dy r x=1 x
12.2. VOLUMES OF REVOLUTION
1233
Now we calculate the integral. Since we are integrating with respect to y, we need to express x as a function of y in the integral namely, x = y 1/3 and so we have
1
V
=
0 1
(1 x)2 dy (1 y 1/3 )2 dy
1
=
0
=
0
(1 2y 1/3 + y 2/3 ) dy
1
3 3 = y y 4/3 + y 5/3 2 5 = 2. . 10
0
Once again, this case is not one where we can directly apply one of our formulae, so it is best to proceed from rst principles. It is important to be clear here what the problem is asking it is rather easy to misinterpret it. We are asked to rotate the shaded region in the diagram below about the y-axis. At rst, it sounds as though this will call for nothing more than a routine application of the variation of the disc method that we used in Example 12.15 above. But if we look more carefully, we see that this is an incorrect interpretation of the problem: we cannot use that method, because it applies to the region enclosed between a curve and the y-axis, while the region in the present problem is enclosed between a curve and the x-axis.
y 1
y=x
3
x=y
1/3
1/3
r=y
{}
inside
outside
Dy x
r=1
1234
CHAPTER 12. APPLICATIONS OF INTEGRATION
Referring now to the next diagram, which shows the rotated region, we see that the volume which we have to compute is not the volume of the bowl-shaped region, but the volume which is outside of the bowl-shaped region and enclosed by the cylinder of radius 1 about the y-axis. Thus, the volume is best computed by taking the dierence of two volumes: V = volume of outside region volume of inside region. We will compute each of these volumes by applying a suitable form of the disc method.
y 1 y=x
3
Dy r x= -1 0 x=1 x
Refer again to the diagrams, now to the details rather than just to the broad features. We see that by taking an element of height y along the y-axis, forming the thin horizontal strip or element of area of dened by y, perpendicular to the axis or rotation, as always with the disc method, and calculating the elements of the outside and the inside volumes, we have V 1 y x2 y, giving the integral expression for the total volume: V = volume of outside region volume of inside region
1 1
=
0 1
1 dy 1 dy
1
x2 dy
0 1
=
0
(y 1/3 )2 dy
0 1
=
0
1 dy
y 2/3 dy
0
12.2. VOLUMES OF REVOLUTION 3 = y y 5/3 5 0
1 1
1235
0
=
2 . 5
We will also outline a similar but slightly dierent way of approaching the problem. Here, rather than treating the required volume as the dierence of two volumes, an outer volume and an inner volume, we compute a volume element directly for the volume we are interested in. The following diagram shows precisely the volume element that we have to estimate:
Using the appropriate expressions for the radii, obtained from the formulae for the curves involved in the problem, we get V
2 2 r1 r2 y
2 2 = r1 r2 y
= 1 (y 1/3 )2 y
1
(since r1 = 1 and r2 = x = y 1/3 )
=
0
(1 y 2/3 ) dy,
and this is essentially the same integral as obtained previously. Understand the problem rst Several of the examples above were ones in which we were not able to apply a ready-made formula to arrive at the answer, and in which we instead had to proceed from rst principles, deriving the required integral expression by subdivision into elements of volume. You should be prepared to confront this possibility whenever you encounter a problem involving a volume of revolution. The best step to take rst with all such problems is to produce an accurate sketch of the situation; only then will it become easy to determine what approach is best, and, in
1236
CHAPTER 12. APPLICATIONS OF INTEGRATION
particular, whether or not one of the ready-made formulae is applicable. (Similar comments apply to the shell method, which we introduce next.) An illustration of this is given by the preceding example, which we solved using two dierent approaches. There was in fact very little dierence between the two approaches: this was indicated by the fact that we obtained not just the same numerical answerwhich must happen whatever our method!but also essentially the same integral expression. The two approaches simply result from two slightly dierent ways of viewing the same situation, but it gives you maximum exibility in dealing with problems involving solids of revolution if you can get to feel comfortable with both of them.
12.2.3 The shell method
Suppose that we are given a function y = f (x), for x between a and b, we know that f (x) 0 in that range, and we wish to rotate the region between the graph of f and the x-axis about the y-axis and to compute the volume of the region enclosed. The following graph represents the situation:
We consider, as shown on the graph, a typical element of length x of the x-axis between a and b, we form the strip or element of area
12.2. VOLUMES OF REVOLUTION
1237
lying above it and extending up to the curve, and we estimate the element V of volume produced by rotation of the strip about the y-axis, as in the following diagram:
When we rotate the strip, we generate a shell or hollow cylinder, which gives the method its name, the shell method. Now the shell has the following dimensions: its wall has thickness x, its height is approximately f (x), for some value of x in the element of the x-axis, and its radius is approximately x, and its circumference therefore approximately 2x To estimate the volume V of the shell in the simplest way, let us imagine cutting the shell down one side and unrolling it, so that we obtain a rectangular solid with the dimensions shown in the diagram below:
1238
CHAPTER 12. APPLICATIONS OF INTEGRATION
The volume element V is approximated by the volume of this rectangular solid, so we have V length height depth = 2xf (x)x. Then converting this expression into an integral in the usual way yields the following integral expression for the total volume V :
The Shell Method
Suppose that y = f (x) 0 for a x b. Then the volume of revolution V obtained by rotating the region between the graph of f and the x-axis about the y-axis is given by
b
V
= 2
a b
xf (x) dx xy dx.
a
= 2
For emphasis, let us note here that in the shell method our element of area is parallel to the axis of rotation, in contrast to the disc method, where the element is taken perpendicular to the axis of rotation. Example 12.17 Find the volume of the solid generated by rotating the area bounded by y = 0, y = x3 and x = 1 about the y-axis. A direct application of the shell method yields the answer
1
Solution Caution
Note that y is a function of x here, and that we must substitute in the expression in x which denes y before we integrate.
V
= 2
0 1
xy dx x x3 dx
1
= 2
0
= 2
0
x4 dx
12.2. VOLUMES OF REVOLUTION x5 = 2 5 = 2 . 5
1
1239
0
Let us also approach the problem from rst principles, not because it is necessary in this case, but because we will soon encounter cases where it is. The relevant diagrams are as follows:
y 1 y=x
3
x 0 Dx 1 x
y
y=f(x)
-1
0 r=x
1
x Dx
Unrolling the shell shown in the second diagram, we obtain a strip with dimensions as follows:
1240
CHAPTER 12. APPLICATIONS OF INTEGRATION
This gives us the approximation V length height depth = 2x x3 x, which then leads to the same integral expression
1
V = 2
0
x4 dx
as before.
12.2.4 The shell method in non-standard cases
As we did earlier in the case of the disc method, we will now examine some applications of the shell method where the shell method formula itself is not directly applicable (or is not easily applicable), and where we therefore have to argue from rst principles. Example 12.18 Solution Find the volume of the solid that is generated when the region that is enclosed by y = x3 , y = 1 and x = 0 is revolved about the line y = 1. Study the statement of the problem carefully here, referring to the diagram below: make sure you are clear that this is a case where we cannot directly apply our shell method formula. Thus, we proceed from rst principles. The diagram representing the problem is as follows:
y 1
y=x
3
x=y =g(y)
1/3
1-y Dy y x 1
12.2. VOLUMES OF REVOLUTION
1241
Note that our axis of rotation here is horizontal, and that our elements of area must therefore be horizontal too, since the elements always run parallel to the axis of rotation in the shell method. Now, taking an element of height y on the y-axis and the long thin horizontal strip it denes, we form a shell by rotating the strip about the line y = 1, as shown:
y
y=x
3
x=y =g(y)
1/3
1
Dy 0 1 x
The radius r of the shell is given by r = 1 y. The shell is lying on its side in the diagram, so in referring to its height we are referring to how far it extends in the x-direction, so the height is given by the expression for x as a function of y, namely, x = y 1/3 = g(y), say. Unrolling the shell, we obtain a rectangular solid with the dimensions shown:
This then yields the approximation V length height depth = 2r g(y) y = 2(1 y)y 1/3y.
1242
CHAPTER 12. APPLICATIONS OF INTEGRATION
Therefore, producing an integral in the usual way, we nally obtain
1
V
= 2
0 1
(1 y)y 1/3 dy y 1/3 y 4/3 dy
1
= 2
0
3 3 = 2 y 4/3 y 7/3 4 7 = 2 = Example 12.19 Solution 9 . 14 3 3 4 7
0
Find the volume of the solid generated by revolving the region bounded by x + y = 4, x = 0 and y = 0 about the y-axis. We will solve the problem in two ways, rst by the shell method, and second by the disc method. First, we can apply the standard shell method formula, after drawing a diagram to check its applicability (you should do this as an exercise), and we nd
4
V
= 2
0 4
x(4 x) dx (4x x2 ) dx
4 2
= 2
0
x3 = 2 2x 3 = 2 32 = 64 . 3 64 3
0
Of course, after such a comparatively straightforward solution, there is no need to look for a second and probably more complicated method of solution! We wish, however, to show an example of a problem to
12.2. VOLUMES OF REVOLUTION
1243
which both of our methods apply, so we will try the disc method as well. Our diagram is the rst one shown below, and we see that we cannot directly apply the disc method formula, so we argue from rst principles. Consider an element of height y on the y-axis and the thin horizontal strip it denes, as in the diagram:
Now consider the disc with volume V obtained by rotating the strip about the y-axis, as shown:
Next, we isolate the disc to see its dimensions more clearly:
1244
CHAPTER 12. APPLICATIONS OF INTEGRATION
Reading the dimensions from the diagram and forming an expression for V , as in many previous problems, we nd V (4 y)2 y. Therefore, converting to an integral, we nally have
4
V
=
0 4
(4 y)2 dy 16 8y + y 2 dy y3 3
4
=
0
= 16y 4y 2 + = 64 , 3
0
as already found by the shell method. Note that the two integrals which the two methods gave here were quite dierent, though of course they still yielded the same numerical values. Check the details Quite a few details were omitted in the application of the disc method above. You should ll these in to consolidate your understanding of the procedure.
Example 12.20
Find the volume of the solid generated by revolving the region bounded by y = x3 and y = x2 about the y-axis. Once again, we will solve this problem by applying both our methods, rst the shell method and then the disc method. The shell method. Our initial diagram is:
Solution
12.2. VOLUMES OF REVOLUTION
1245
y y=x 1 x -x x 0 Dx
And the diagram showing an element, or shell, of volume formed in the usual way from an element of length x is:
3 3 2 3
y=x
2
1
x
y 1
Dx
f(x)=x -x 0 x
2
3
Now the points of intersection of the two curves are given by the solutions of the equation x3 = x2 , namely, x = 0 and x = 1. Hence, when we unroll our strip, we obtain a diagram with the dimensions shown:
1246
CHAPTER 12. APPLICATIONS OF INTEGRATION
Here, the height is f (x) = x2 x3 . Therefore, we have V 2x(x2 x3 )x, and nally
1
V
= 2
0 1
x(x2 x3 ) dx (x3 x4 ) dx
1
= 2
0
x4 x5 = 2 4 5 = 2 = The disc method. . 10 1 20
0
Now our initial diagram is:
y x=y 1 r1 r2 0 1 Dy x
1/3
x=y
1/2
The diagram showing an element, or disc, of volume formed in the usual way from an element of length y is:
y 1 r1 Dy r2 0 x
12.2. VOLUMES OF REVOLUTION And the diagram showing the disc in isolation is:
1247
r1 Dy r2
Using the dimensions shown, we nd V =
2 2 r1 r2 y
(y 1/3 )2 (y 1/2 )2 y
= (y 2/3 y)y, giving nally
1
V
=
0
y 2/3 y dy
1
y2 3 = y 5/3 5 2 = = as already found. , 10 3 1 5 2
0
12.2.5 A comparison of the methods
Probably the most important point to note is the similarity, rather than the dierence, between the disc and shell methods. The similarity lies in the common idea which underlies both methods: that of estimating a typical element of volume, and then of converting this estimate into an integral expression for the total volume. This idea leads to the formulae for the two methods, and allows us to nd volumes in situations where neither of our formulae applies, and it is also fundamental in many other applications of integration, both in this chapter and more widely. The idea is in fact the most important idea of this entire chapter, and once you understand it well, you will nd that everything else in the chapter ows naturally from it.
1248
CHAPTER 12. APPLICATIONS OF INTEGRATION
One specic point of comparison between the disc and shell methods, noted earlier, is that in the disc method, the element of area is perpendicular to the axis of rotation, while in the shell method, the element of area is parallel to the axis of rotation. We have seen that there are some applications where both methods can be used. They must, of course, lead to the same answer, but often one will yield an easier integral than the other. Thus, trying both methods may be justied if the rst method used leads to a dicult integral.
Exercise 12B 1. Find the volume generated when the given areas are revolved about the given axes: (a) the area bounded by x + y = 2, y = 0 and x = 0, about the x-axis; about the y-axis, (b) the area bounded by y = sin x, x = 0, y = 0 and x = about the x-axis; about the y-axis, (c) the area bounded by y = sin x for 0 x , by x = 0 and by y = 1 2 about the x-axis; about the y-axis, (d) the area in the rst quadrant bounded by y = x3 and y = x about the x-axis; about the y-axis. 2. Find the volume generated when the area bounded by y = x2 , x = 2, and y = 0 is revolved about (a) the x-axis, (b) the y-axis, (c) the line x = 2, (d) the line y = 4.
3. Find the volume generated when the area in the rst quadrant bounded by y = x2 , y = 4, and x = 0 is revolved about (a) the x-axis, (c) the line x = 2, (b) the y-axis, (d) the line y = 4.
4. Find the volume generated if the region above the x-axis and between the x-axis y2 x2 and the graph of the ellipse 2 + 2 = 1 a b is rotated about the x-axis.
12.3 The Arc Length of a Curve
When we consider a curve drawn in the plane, we can reasonably ask what is the length of some segment of the curve. The mathematical term for this length is the arc length of the segment, and the computation of arc length is the topic of this section. If you have a clear understanding of the work of this chapter so far, you should nd the arc work quite straightforward, because it simply
12.3. THE ARC LENGTH OF A CURVE
1249
uses yet again the fundamental idea that is the theme of this chapter: calculating a quantity by nding an expression for an element of the quantity, and then converting that expression into an integral expression for the whole quantity. We have done this so far for area (in a number of forms) and for volumes of revolution (in two ways), and we now apply the idea to arc lengths. First, we note an important but technical point. It turns out that some curves are too irregular to have a properly dened arc length at all. While we dont meet such curves very often in applications, they certainly exist, and any properly stated formula for arc length must exclude them from its scope. To indicate the kinds of considerations that are involved, it can be shown, for example, that for a curve y = f (x), the condition that the derivative f (x) be a continuous function of x is enough ...