Applications of integrals

Transcription

1 ApplicationsofIntegrals.nb Applications of integrals There are many applications of definite integrals and we cannot include all of them in a document intended to be a review. However there are some very important applications that all calculus students are expected to know and we will try to cover them here. Whatever application we use remember that the integral of a rate of change gives us an accumulated change. You can figure out many new applications just remembering that fact. In this section we will look at using definite integrals to find the area of a region,the average value of a function, the volume of a solid with known cross sectionsal areas, and solve differential equations. We will also revisit particle motion. Area of a Region We can use definite integrals to find the area between the graph of a function and the x-axis on a closed interval [a,b] as we saw in the section "Concepts of the Definite Integral." Combining that information with the First Fundamental Theorem of Calculus provides us with a way of finding exact areas of these regions. However, we have only looked at areas between curves and the x-axis when the graph of f was above the x-axis. What if the graph is below the x-axis or, even yet, part of the graph is above the x-axis and part is below? The answer to these questions is best answered by remembering that area is always positive (even though a definite integral can be negative). So if the graph of f is below the x-axis the definite integral represents the negative of the area (this is the concept of signed areas). Look at the example below: Example : Find the area of the region bounded by the curve f(x) = -x, the x-axis, and the lines x = 0 and x =. Solution: Look at the graph below. The shaded portion represents the area between the graph of f(x) = -x, the x-axis, and the lines x = 0 and x = fhxl=-x

2 ApplicationsofIntegrals.nb If we compute the definite integralÿ 0 -x x using the fundamental theorem we get the following: Ÿ 0 -x x = ÄÄÄÄÄÄÄÄÄÄ -x = ÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄ -0 = ÄÄÄÄÄÄÄÄÄÄ -6 The area of the shaded region is, therefore, ÄÄÄÄÄÄ 6. So even though the definite integral is negative, the area of the shaded region is considered to be positive. We can sum this concept up in this way; the area between the graph of a negative function and the x-axis on an interval is equal to the definite integral of the negative of the function (or, more simply, the absolute value of the integral). Remember that multiplying a function by negative one reflects the function in the x-axis. The area between the reflected graph and the original graph would be the same. What about the case where the graph of a function crosses the x-axis on the given integral? Look at the example below: Example : Find the area bounded by the graph of f(x) = x, the x-axis, and the lines x= - and x =. Solution: Look at the graph below. The shaded region is the area between the graph of f(x) = x and the x-axis on the interval [-,]. 8 6 fhxl=x

3 ApplicationsofIntegrals.nb Since the graph is below the x-axis on the interval [-,0] and above the x-axis on the interval [0,] we need to compute the area using two definite integrals. 0 Area =Ÿ - -x x + Ÿ 0 x x = ÄÄÄÄÄÄÄÄÄÄ -x ÄÄÄÄÄÄ x 0 = (0 + ) + ( + 0) = 8 Note that if we just evaluated Ÿ - x x we would get 0 which would not, clearly, represent the area of the region described. So for functions that cross the x-axis on the interval in question you need to separate the region into areas below and above the x-axis and integrate the opposite of the function whenever it is below the x-axis (or take the absolute value of the integral). Then just add up all the individual areas. This is summarized below: Finding Total Area Analytically To find the area between the graph of y = f(x) and the x-axis over the interval [a, b] analytically,. partition [a, b] with the zeros of f,. integrate f over each subinterval,. add the absolute values of the integrals. Now, what about the area between two curves? It can be shown using basic properties that it does not matter whether the curves in question are above or below the x-axis to compute the area between them. Look at the example below:

4 ApplicationsofIntegrals.nb Example : Find the area bounded by the graphs of f(x) = x and g(x) = x +. Solution: Look at the graph below. The shaded region is the area enclosed by the graphs of f(x) = x and g(x) = x+. 6 ghxl=x+ fhxl=x - - Since the two graphs intersect at x = - and x = (you can find these points of intersection algebraically by setting the two functions equal and solving for x) these are our limits of integration. Geometrically, to find the area between the two curves we simply find the area under each curve on the given integral and subtract the area under the curve on the bottom from the area under the curve on the top. See the graphs below. 6 ghxl=x+ fhxl=x - - This graph shows the area under the curve g(x) = x + on the interval [-, ]. The area under this curve can be found using the definite integral Ÿ - Hx + L x

5 ApplicationsofIntegrals.nb 5 Now look at the second graph. 6 ghxl=x+ fhxl=x - - This graphs shows the area under the curve f(x) = x on the interval [-, ]. The area under this curve can be found using the definite integral Ÿ - x x So the area between the two graphs is the difference of the two areas. Analytically, this would be the difference between the two definite integrals Area between curves = Ÿ - Hx + L x- Ÿ - x x This, of course, can be written as a single integral Area between curves = Ÿ + L - x D x = ÄÄÄÄÄÄ x + x - ÄÄÄÄÄÄ x =.5 - = H ÄÄÄÄ L - H ÄÄÄÄ - + ÄÄÄÄ Notice that if we were to shift both graphs equally upward or downward the area between the two curves would not change. That is why we do not need to concern ourselves with whether the graphs are above or below the x-axis, only which of the two graphs dominates (or is "on top of" the other). The following theorem summarizes the procedure: L

6 ApplicationsofIntegrals.nb 6 Area between Curves If f and g are continuous functions and f(x) g(x) over the interval [a,b], then the area bounded by y = f(x) and y = g(x) for a x b is given exactly by Area = Ÿ a f HxL - ghxld x Sometimes the two graphs will change dominance on an interval. Don't let this confuse you. We just need to partition the interval into subintervals (by finding where the graphs intersect) and setting up a sum of definite integrals. Look at the example below. Example : Find the area of the region bounded by f(x) = - x and g(x) = -x on the interval [-, ]. Solution: Look at the graph below The shaded region is the area between the two curves on the interval [-,]. Notice that the two functions change dominance at x = - (one of the two points of intersection). On the interval [-, -] the function g(x) = - x is dominant and on the interval [-, ] the function f(x) = - x is dominant. So we need to write two separate integrals to find the area. - Area between curves = Ÿ - H - x LD x + Ÿ - x L - H- xld x = 6 ÄÄÄÄ

7 ApplicationsofIntegrals.nb 7 The general procedure for analytically finding the area between two curves where the functions change dominance over the interval is similar to finding the area between a curve and the x-axis where the graph crosses the x-axis over the interval. The procedure is outilined below. Finding the Area between Curves when the Functions Change Dominance To find the area between the graphs of y = f(x) and y = g(x) on the interval [a, b] where f and g change dominace over the interval.. partition [a, b] with the intersections of f and g,. write integrals which represent the area between the curves on the partitions and evaluate,. add the values of the integrals Average Value of a Function We can also use definite integrals to find the average value of a function because it has a very easy to understand geometric explanation. We know that if we want to find the average of a list of numbers we simply add up all the numbers and divide by the total number of numbers. What if we wanted to average all the y values of a function on an interval? At first this would seem impossible since even on a very small interval a function has an infinite number of values. How can we add up an infinite number of values? We do this every time we evaluate a definite integral! Look at the definition again. A definite integral is defined to be the limit as n Ø of the product of y values of a function times really small changes along the x-axis. All of these really small changes along the x-axis add up to be the entire interval from x = a to x = b. So if we evaluate the definite integral of a function on an interval and divide it by the difference b - a then we will get the average y value of that function on that interval. This is summarized below:

8 ApplicationsofIntegrals.nb 8 Average value of f = ÄÄÄÄÄÄÄÄÄÄ b-a Ÿ b a f HxL x from a to b Look at the example below and the included graph to see the geometric interpretation. Example 5: Find the average value of the function f(x) = x on [0,]. Solution: From the formula A.V. = ÄÄÄÄÄÄÄÄÄÄ - Ÿ 0 x x = ÄÄÄÄ ÄÄÄÄÄÄ x 0 = ÄÄÄÄ ( ÄÄÄÄ 8-0) = ÄÄÄÄ. The graph below illustrates, geometrically, that the average value of the function f(x) = x on the interval [0, ] is the height of a rectangle with the same area on the given interval. y fhxl=x y= ÅÅÅÅÅ x Volumes of Solids with Known Cross Sectional Areas Just as we found the area between a curve and the x-axis and the area between two curves using definite integrals we can also find the volumes of solids. The key to this is seeing the analogy. In finding the area between a curve and the x-axis in effect we were finding the area of small pieces (rectangles whose height was determined by the function and whose width was approaching zero) and then adding these small areas up to get the whole area by writing a Riemann sum and taking the limit of the sum as the number of terms in the sum tended to infinity (giving us a definite integral). The procedure is pretty much the same in finding volumes except that rather than finding areas of small pieces we will be finding volumes of small pieces and adding up all of their contributions. The process is summarized below:

9 ApplicationsofIntegrals.nb 9 To Compute a Volume using an Integral. Divide the solid into small pieces whose volume we can easily approximate. Add the contributions of all the pieces, obtaining a Riemann sum that approximates the total volume. Take the limit as the number of terms in the sum tends to infinity, giving a definite integral for the total volume. Sometimes we are asked to find the volume of a solid of revolution. This is where we take a region in the x-y plane and revolve it about either a vertical or horizontal line. What we need to do is "slice" the solid into thin discs (or washers if there is a hole in the region) and write a formula for the volume of the disc (or washer). Remember that a disc is just a short cylinder whose volume is given by pr h. So we just need to write this formula in terms of our function and set up a definite integral. Let's look at the example below: Example 6: Find the volume of the solid obtained when the region enclosed by the graph of f(x) = x, the x-axis, and the vertical lines x = 0 and x = is revolved about the x-axis. Solution: Look at the graph below. y fhxl=x x

10 ApplicationsofIntegrals.nb 0 When this region is revolved about the x-axis it will sweep out a region in space that looks like a volcano on its side (or maybe the end of a trumpet). Can you see if we slice this solid vertically (perpendicular to the axis of revolution) we will obtain discs standing on end? Picture the rectangular element drawn on the graph revolving about the x-axis. The solid obtained will look like a coin. fhxl=x x Dx The width of these discs is a change along the x-axis so we designate it Dx and the radius of these discs is the distance from our axis of revolution (the x-axis) and the graph of the function, f(x) = x. So the volume of one of the discs is given by V = phx L Dx. for all of the discs we get V º ph f HxLL Dx = phx L Dx. As the thickness of each slice tends to zero (this is when the number of slices increases without bound), we get

11 ApplicationsofIntegrals.nb Ÿ 0 phx L x = Ÿ 0 px x = pi ÄÄÄÄÄÄ x5 5 M 0 = p[ ÄÄÄÄÄÄ 5-0] = p ÄÄÄÄÄÄÄÄÄ 5 If we took the same region and revolved it about the y-axis we will get a solid with a "hole" in it. Basically we will just be computing the volume without the hole and then subtracting the volume of the hole from it. A vertical cross section is shown below so that you can visualize the hole. You can see that the solid we get is a bowl shape. We are trying to find the volume of the material in the bowl itself; not the volume the bowl can hold. Think of finding the volume of the solid cylinder of material necessary to make the bowl and subtract the volume of material the bowl would hold. Look at the example below. Example 7: Find the volume of the solid obtained when the region enclosed by the graph of f(x) = x, the x-axis, and the vertical lines x = 0 and x = is revolved about the y-axis. Solution: Look at the graph below. The shaded region is a cross-sectional view of the solid generated with a rectangular element drawn in. y Dy - - x Our slices will be perpendicular to the axis of revolution (the y-axis). Can you see that we will get a washer shape? è!!! y Dy The volume of a washer is just the volume of a larger disc minus the volume of a

12 ApplicationsofIntegrals.nb smaller disc. So we will be identifying "outside" and "inside" radii in this problem. The outside radius is the distance from the axis of revolution to the outside of the region; in this case, (the distance from the axis of revolution to the "far end" of the rectangular element). The inside radius is the distance from the axis of revolution to the graph of y = x (the "near end" of the rectangular element), namely x. However, notice that the height of each of our washers is a change along the y-axis or Dy so we will need to have all of our quantities in terms of y. Solving y = x for x gives us x = è!!! y. So the volume of each of our washers is as follows: Volume = volume of larger disc minus volume of smaller disc = phl Dy - pi è!!! y M Dy The volume of our solid can then be estimated as V º ph - I è!!! y M )Dy This will give us the definite integral Ÿ 0 ph - yl y. Notice that the limits of integration are also in terms of y; when x = 0, y = 0 and when x =, y =. Evaluating the definite integral gives p[y - ÄÄÄÄÄÄ y ]» = p[(6-8) -(0-0)] = 8p 0 Sometime we are asked to find the volumes of solids that are not formed by revolving about an axis. The procedure remains the same; determine how to slice the solid to get cross sections of known area, multiply these areas by the width of the slice, and then write a Riemann sum and definite integral that represents the total volume. Look at the example below. Example 8: Find the volume of the solid whose base is the region in the xy-plane bounded by the curve y = è!!! x, the x-axis, and the line x = and whose cross-sections perpendicular to the x-axis are squares with one side in the xy-plane.

13 ApplicationsofIntegrals.nb Solution: The region is shown below: y x For this problem you need to imagine that the squares are sitting in the region perpendicular to the x-axis and are rising from the page. As we proceed from left to right the squares get bigger because the length of the side of the square is the height of the function y = è!!! x. The area of the square cross-sections is given by ( è!!! x M and the thickness of each slice is a change along the x-axis and hence Dx. So the volume of one slice is given by ( è!!! x M Dx (the volume of a prism is area of the base times the height). So the process looks like this: V º I è!!! x M Dx = I è!!! x M x = ÄÄÄÄÄÄ x 0 = - 0 = 0 This leads us to the following definition. Definition: Volume of a Solid The volume of a solid of know integrable cross section area A(x) from x = a to x = b is given by V = Ÿ a b AHxL x. To apply this definition, follow the procedure outlined below.

14 ApplicationsofIntegrals.nb Finding the Volume of a Solid by Slicing. Sketch the solid and a typical cross section.. Find a formula for A(x), the cross-sectional area.. Find the limits of integration.. Integrate A(x) to find the volume. Finding the volumes of solids of revolution or "built-up" geometrical figures can be daunting at first and the above examples are very simple for the purpose of illustration. You will need to practice this process to make yourself comfortable with it. Look at your homework and worksheets for more practice. Particle Motion Revisited Now that we understand areas between a curve and the x-axis especially we can look at finding the total distance a particle travels along a straight line. If a particle has a positive velocity (remember that velocity is a vector with both magnitude and direction) then the graph of this velocity will be entirely above the x-axis. Under these conditions the definite integral of the velocity function on [a,b] will represent the total distance traveled. Suppose, however, that the particle changes direction. This will be indicated graphically by the velocity function being below the x-axis. Now the definite integral will not represent the total distance traveled by the particle but the displacement of the particle. Displacement is the distance a particle is from where it started as opposed to the total distance it has traveled. Think of it this way; if you drive to Memphis and back (from Nashville) you have traveled approximately 0 miles so your total distance is 0. But since your are back where you started your displacement is 0! In rectilinear problems on the AP exam never assume that the displacement and distance traveled are the same until you investigate it. To find the total distance traveled always take the velocity function, set it equal to zero and then set up a sign chart to see if the particle is actually changing direction (sometimes a particle will stop and then proceed in the same direction. This is the case when a factor of the velocity function is a square.) Then you can simply find the position of the particle at each of the pertinent points and add up all of the individual distances traveled to get the total distance traveled. Look at the example below.

15 ApplicationsofIntegrals.nb 5 Example 9: A particle, initially at rest, moves along the x-axis so that its acceleration at any time t 0 is given by a(t) = t -. The position of the particle when t = is x(t) =. Find the total distance traveled by the particle from t = 0 to t =. (This in an actual AP question with other parts but we will just concentrate on finding the total distance traveled) Solution: We first need to find the velocity function which will require us to integrate the acceleration function. We will also need an initial condition in order to find the constant of integration. This one is given verbally, and sometimes students do not see it. The problem says that the particle is initially at rest. This translates into v(0) = 0. So v(t) = Ÿ ahtl t= Ÿ H t - L t = t - t + C. Since v(0) = 0, this implies that C=0. So v(t) = t - t. Now, let's investigate the velocity function by first looking at the graph. v 0 5 vhtl t We can see that part of the velocity graph is below the t-axis (indicating motion to the left) and part is above the t-axis (indicating motion to the right). So we need to find when this change occurs. Factoring out t and setting the velocity equal to zero gives us tht - ) = 0 which implies that t = 0, ±. Since the problem indicates that we are only concerned with the motion of the particle for t 0 we discard the -. Setting up a sign chart (not shown here) results in the following; the particle is moving to the left on (0,) and moving to the right on (, ) so we only have one change in direction. We can now calculate the total distance traveled. Since the particle is moving to the left on (0,) the definite integral of v(t) on (0,) will be negative so we will integrate the opposite of v(t) on this interval. However, since the particle is moving to the

16 ApplicationsofIntegrals.nb 6 right on the interval (,) the definite integral will represent the distance traveled on this interval. So Total distance traveled = Ÿ 0 -vhtl t+ Ÿ vhtl t = Ÿ 0 -H t - tl t + Ÿ H t - tl t t - t D» 0 - t D» = - + (8 - (-)) = 0 So the total distance traveled by the particle is 0. Note that the displacement of the particle is given by Ÿ 0 H t - tl t= 8 So the particle traveled a total of 0 units but ends up only 8 units to the right of where it started. We could also have found the total distance traveled using the position function which we didn't find in this solution but we would have in the original problem since it was asked for. The position function, x(t), is found by integrating the velocity function and using the initial condition x() = to solve for the constant of integration. This gives us x(t) = t - t + Now we find where the particle is t = 0 (the beginning of the interval), at t = (when it changes direction, and at t = (the end of the interval). x(0) = x() = x() = So from t = 0 to t = the particle traveled - = unit. From t = to t = the particle traveled - = 9 units. So the total distance traveled is +9 = 0 units!

17 ApplicationsofIntegrals.nb 7 Solving Differential Equations We have seen how to sketch solution curves of a differential equation using a slope field (see BC addendum entitled "Differential Equations:Slope Fields") and how to approximate numerical solutions (see BC addendum entitled "Differential Equations: Euler's Method"). Now we will look at how to find the equation of a solution curve for certain differential equations. Remember, not all differential equations can be solved ananlytically. That's why we looked at slope fields and Euler's method. This section provides a method for solving those differential equations that can be solved analytically. First a definition. Differential Equation An equation of the form dy ÄÄÄÄÄÄÄ dx = f HxL is a differential equation. Every time we antidifferntiate to find Ÿ f HxL x, we are actually solving a differential equation. Remember that we are actually finding a family of functions whose derivative is the differential equation. If we are given an initial value (for instance some y-value paired with some x-value) then we can solve for a particular solution. We have done this many times with differential equations that are soley functions of x. The difference in this section is that the right-hand side of the differential equation may now be a function of y, or both x and y, instead of just x. In summary, the problem of finding a function y of x when we are given its derivative and its value at a particular point is called an initial value problem. The value of f for one value of x is the initial condition of the problem. When we find all the functions y that satisfy the differential equation we have solved the differential equation. When we then find the particular solution that fulfills the initial condition, we have solved the intial value problem. The method that we will use to solve differntial equations that are functions of y or of both x and y is called separation of variables. Look at the definition below.

18 ApplicationsofIntegrals.nb 8 Separable Differential Equations A differential equation y' = f(x, y) is separable if f can be expressed as a product of a function of x and a function of y. The differential equation then has the form dy ÄÄÄÄÄÄÄ dx = ghxl hhyl If h(y) 0, we can separate the variables by dividing both sides of the equation by h(y), obtaining ÄÄÄÄÄÄÄÄÄ hhyl Ÿ Ÿ ÄÄÄÄÄÄÄÄÄ hhyl dy ÄÄÄÄÄÄÄ dx = ghxl dy ÄÄÄÄÄÄÄ dx x = Ÿ ghxl x ÄÄÄÄÄÄÄÄÄ hhyl y = Ÿ ghxl x With x and y now separated, the integrated equation provides the solution we seek by expressing y either explicitly or implicitly as a function of x, up to an arbitrary constant. Let's look at an example. Example 0: Given ÄÄÄÄÄÄÄ dy dx = ÄÄÄÄÄÄÄÄ -x, with the initial condition that y() =, find the particular solution to the differntial y equation. Solution: First we separate the variables. y dy ÄÄÄÄÄÄÄ dx = -x

19 ApplicationsofIntegrals.nb 9 Now we integrate both sides of the equation. Ÿ y dy ÄÄÄÄÄÄÄ dx x = Ÿ -x x Ÿ y y = Ÿ -x x Integrating we get y ÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄ -x the equation but = C Note: We get a constant of intergation on both sides of we combine the constants on one side. Multiplying both sides of the equation by results in Note: We actually get C but this is just another constant. y + x = C We recognize this equation as that of a circle. We can solve for C by using the intitial condition that y = when x =. + = C ï C = 5 So our particular solution is x + y = 5. Many times we are given the differential equation in words rather than in an outright eqaution. That's why we have spent so much time translating words into equations and vice-versa. Look at the following example. Example : The rate of growth of a population of bacteria is directly proportional to the amount of bacteria at any time t, given in hours. If the initial population of the bacteria culture is 00 and hour later the population is 50, find the population of the bacteria when t =. Solution: The problem says that the rate of growth is directly proportional to the number of bacteria present. This gives us the differential equation dp ÄÄÄÄÄÄÄ dt = kp where k is the constant of proportionality

20 ApplicationsofIntegrals.nb 0 Separating the variables gives us ÄÄÄÄÄ P dp ÄÄÄÄÄÄÄ dt = k Integrating both sides gives us or Ÿ ÄÄÄÄÄ P dp ÄÄÄÄÄÄÄ dt t = Ÿ k t Ÿ ÄÄÄÄÄ P P = Ÿ k t Integrating both sides gives us ln P = kt + C Now we solve for P. e ln» p» = e kt + C P = e kt e C = Ce kt since e C is just another constant. Now we use the initial condition P(0) = 00 (the popultion was intitially 00) to solve for C. 00 = C e k 0 î C = 00 so P = 00 e kt and the other condition that at t = the population was 50 to solve for k. 50 = 00 e k.5 = e k ln.5 = k ï kº 0.055

21 ApplicationsofIntegrals.nb So our mathematical model (the particular solution to the given differntial equation) is P(t) = 00 e t So now we can use our model to predict the population of the bacteria when t =. P() = 00 e HL ª 506 bacteria. Let's look at another common differential equation known as Newton's Law of Heating/- Cooling. Newton's Law of Heating and Cooling As we learned in class the rate at which an object's temperature is changing at any given time is proportional to the difference between its temperature and the temperature of the surrounding medium. So, if T is the temperature of the object at time t, and T s is the surrounding temperature, then dt ÄÄÄÄÄÄÄÄ dt = -k HT - T s L. Example : A hard boiled egg at 98 C is left on a table at room temperature (0 C). After 5 minutes, the egg's temperature is found to be 8 C. How much longer will it take the egg to reach 0 C? Solution: We have ÄÄÄÄÄÄÄÄ dt dt temperature of the egg. = -kht - T s L with T s = 0 C and T(0) = 98 (the initial Separating variables gives us ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ HT-0L dt ÄÄÄÄÄÄÄÄ dt = -k Integrating both sides gives us Ÿ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ HT-0L dt ÄÄÄÄÄÄÄÄ dt t = Ÿ -k t or

22 ApplicationsofIntegrals.nb Ÿ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ HT-0L T = Ÿ -k t Simplifying gives us ln T - 0 = -kt + C T - 0 = Ce -kt or solving for T gives us T = 0 + Ce -kt Since our egg is initially at 98 C we can solve for C. 98 = 0 + Ce -kh0l î C = 78 so T = e -kt Now we use the fact that at t = 5 the temperature of the egg is 8 C to solve for k. 8 = e -kh5l 8 = 78 e -5 k ln 8/78 = -5k ï k º 0.09 so our mathematical model is T = e t Now we use the model to predict when the temperature of the egg reaches 0 C. 0 = e t 0 = 78 e t ln 0/78 = -0.09t ï t º 0.05 minutes. Since the question was how much longer will it take the egg to reach 0 C we subtract the original 5 minute period from 0.05 minutes and we get 5.05 minutes. That concludes our review on applications of integration.

23 ApplicationsofIntegrals.nb Worksheet a: Area of a Region Worksheet b: Area between Two Curves Worksheet c: Average Value of a Function Worksheet d: Volumes of Solids of Revolution Worksheet e: Revolving about a Line»» to the x-axis Worksheet f: Revolving about the y-axis Worksheet g: Revolving about a line»» to the y-axis Worksheet h: Volumes of Solids between Two Curves Worksheet i: Volumes of Solids of Known Cross-sectional Areas Worksheet j: Particle Motion Revisited