Applications of the Double Integral Mass Density of a Laminate The double integral has many interpretations other than volume. In this section, we examine several of those di erent interpretations. Many of these intepretations will involve a lamina of a region in the xy-plane, which is a solid whose base is the region and whose height is 1. To begin with, let us suppose that the lamina has a mass-density of (x; y) ; measured in units of mass per unit area ( is the Greek letter mu, pronounced mew ). In particular, suppose the x and y-axes are both partitioned into h- ne partitions. Then the box containing the point (x; y) has a small mass m and a small base with area A; and the mass density function is then de ned (x; y) = lim h! m A That is, we de ne the mass density function so that m (x; y) A As a result, the mass M of the lamina is approximately the sum of the masses m jk of the boxes in the partition M X X m jk = X X (x j ; y k ) A jk j j k 1 k
The total mass is then the limit of the approximations as h approaches : X X M = lim (x j ; y k ) A jk = (x; y) da h! j k That is, the total mass is the double integral over the region of the mass density. EXAMPLE 1 What is the mass of the lamina of the unit square with a height of 1 and a mass density of (x; y) = (x + 2y) kg m 2 Solution: The mass M of the lamina satis es M = (x + 2y) da where is the unit square. Thus, M = = = Z 1 Z 1 Z 1 Z 1 = 1:5 kg (x + 2y) dydx xy + y 2 1 dx (x + 1) dx Notice that if (x; y) = 1 for all xy in a region, then the mass is simply the product of the height 1 and the area of the base : As a result, the area A of a region is given by A = da EXAMPLE 2 Find the area of the region bounded by x = ; x = 1; y = and y = x: Solution: The area of the region is given by A = da = Z 1 Z x dydx 2
Evaluating the double integral leads to Z 1 A = xdx = x2 1 2 = 1 2 Check your reading: What type of region is given in example 2? Centers of Mass In elementary physics courses, it is shown hat if n objects with masses m 1 ; m 2 ; : : : ; m n have positions (x 1 ; y 1 ) ; (x 2 ; y 2 ) ; : : : ; (x n ; y n ) ; respectively, then their center of mass is the point in the plane with coordinates x = m 1x 1 + : : : + m n x n m 1 + : : : + m n ; y = m 1y 1 + : : : + m n y n m 1 + : : : + m n If a lamina with mass-density (x; y) is partitioned into n boxes with masses m 1 ; m 2 ; : : : ; m n and positions (x 1 ; y 1 ) ; (x 2 ; y 2 ) ; : : : ; (x n ; y n ) ; respectively, then the center of mass of the lamina is approximately the same as the center of mass of the boxes: approximately given by x m 1x 1 + : : : + m n x n m 1 + : : : + m n ; y m 1y 1 + : : : + m n y n m 1 + : : : + m n However, since m 1 + : : : + m n M; where M is the mass of the lamina, and since m j (x j ; y j ) A j ; where A j is the area of the base of the j th box, the center of mass of the boxes can be approximated by x x 1 (x 1 ; y 1 ) A 1 + : : : + x n (x n ; y n ) A n M y y 1 (x 1 ; y 1 ) A 1 + : : : + y n (x n ; y n ) A n M Since the numerators are approximately the same as double integrals, we are led to de ne the center of mass of the lamina of a region to be the point in the xy-plane with coordinates x = 1 x (x; y) da; y = 1 y (x; y) da M M where M is the mass of the lamina. EXAMPLE 3 Find the center of mass of the lamina of the unit square with mass density (x; y) = (x + 2y) kg m 2 3
Solution: In example 1, we saw that the mass of the lamina is M = 1:5 kg. Thus, x = 1 1:5 x (x; y) da = 1 1:5 Z 1 Z 1 Z 1 = 1 1:5 = :5556 x (x + 2y) dydx x 2 + x dx Likewise, the y-coordinate of the center of mass is y = 1 y (x; y) da 1:5 = 1 1:5 = :6111 Z 1 Z 1 y (x + 2y) dydx When the mass density of the lamina is (x; y) = 1 for all (x; y) in the region ; then the center of mass is de ned only by the region itself and is thus called the centroid of : Indeed, the mass M reduces to the area A of the region and thus, x = 1 xda; y = 1 yda A A Moreover, if the region is symmetric about a line l; then as we will show in the exercises, the centroid of the region must lie on the line l: EXAMPLE 4 Find the centroid of the region bounded by x = ; x = 2; y = x; :y = x Solution: First, we compute the area of the region A = da = Z 2 Z x x dydx = 4 4
As a result, we have x = 1 4 = 1 4 = 1 4 xda Z 2 Z x Z 2 x 2x 2 dx xdydx = 4 3 and since the region is symmetric about the line y = ; we must have y = Check your eading:what type of region is given in example 4? Joint Probability Suppose that (X; Y ) is the outcome of a certain experiment and suppose that it is known that (X; Y ) must occur in some region S in the plane. Then S is called the sample space for the experiment and X and Y are called random variables. Moreover, a function p (x; y) is called a joint probability density of the experiment if the small probability P that (X; Y ) is in a small region in S containing (x; y) is given by P = p (x; y) A where A is the area of the small region. Like a mass density, a joint probability density is essentially the probability per unit area of the experiment. 5
Thus, if is a region inside of S and if the x and y-axes are both partitioned into h- ne partitions, then the probability P ((X; Y ) in ) that (X; Y ) is in is approximated by the iemann sum P ((X; Y ) in ) X X P jk = X X p (x j ; y k ) A jk j j k The actual probability of (X; Y ) being in is the limit of the approximations as h approaches : X X P ((X; Y ) in ) = lim p (x j ; y k ) A jk = p (x; y) da h! j k As a result, the density p (x; y) must be non-negative and since there is a 1% chance of (X; Y ) being in the sample space S; we require also that p (x; y) da = 1 (1) S k EXAMPLE 5 A point (X; Y ) is chosen at random from the unit square with probability density 1 if x 1 and y 1 p (x; y) = otherwise What is the probability that the point is inside the rectangle given by [:1; :6] [:3; :8]? Solution: Since the outcome (X; Y ) must be in the unit square, the unit square is the sample space S for our experiment. Moreover, notice that Z 1 Z 1 Z 1 p (x; y) da = 1dydx = dx = 1 S thus showing that p (x; y) does satisfy (1) and is a probability density. Indeed, in this experiment, the probability density can be considered the probability of choosing the point (x; y) at random from the sample space S: Moreover, the probability that a random selection of (X; Y ) from S will be in P [(X; Y ) in ] = p (x; y) da = Z :6 Z :8 :1 = :25 :3 1dydx 6
Thus, there is a 25% chance that (X; Y ) will be in ; which is about a 1 in 4 chance. Suppose that the random variable X of a given event has a probability density of p 1 (x) and suppose that the random variable Y of an additional event has a probability density of p 2 (y). Then X and Y are said to be independent random variables if their joint density function is p (x; y) = p 1 (x) p 2 (y) That is, two events are independent if their joint density function is the product of the density functions of the random variables of the individual events. EXAMPLE 6 At a certain restaurant, customers must wait an average of 1 minutes for a table. From the time they are seated until they have nished their meal requires an additional 3 minutes, on average. What is the probability that a customer will spend less than an hour at the restaurant, assuming that waiting for a table and completing the meal are independent events? Solution: Waiting times are often modeled by exponential probability densities. Indeed, if X and Y are the random variables for waiting for a table and completing the meal, respectively, then their probability density functions are respectively p 1 (x) = if x < 1 1 e x=1 if x p 2 (y) = if y < 1 3 e y=3 if y Since the events are independent, the joint probability for the two events is if x < or y < p (x; y) = p 1 (x) p 2 (y) = 1 3 e x=1 e y=3 if x; y Our goal is to determine the probability that the combined time X +Y is under 6 minutes. That is, we want to know the probability associated with the region in the 1st quadrant with upper bound x + y = 6: 7
As a result, the probability that (X; Y ) is in is P (X + Y 6) = P [(X; Y ) in ] = p (x; y) da = 1 3 e x=1 e y=3 da Since x + y = 6 is the same as y = 6 x; we have a type I integral: P (X + Y 6) = 1 3 Z 6 Z 6 x e x=1 e y=3 dydx = :7982 Thus, there is a 79.8% chance that a customer will spend less than an hour at the restaurant. Check your eading: Can we evaluate the integral in example 6 as a type II integral? Expected Value Since a probability density is similar in concept to a mass density, we de ne the expected values of the random variables X and Y to be the coordinates of the center of mass of a lamina of the sample space S with density p (x; y) : However, since p (x; y) da = 1 this implies that the expected values of X and Y are given by X = xp (x; y) da; Y = yp (x; y) da S S Often we denote the expected values as E (X) and E (Y ) ; respectively, and we assume they represent the most likely outcome of the experiment. S EXAMPLE 7 What is the expected time for waiting for a table and completing the meal in example 6. Solution: Since the sample space is the 1st quadrant, the expected 8
time for waiting on a table is E (X) = x 1 3 e x=1 e y=3 da = 1 3 = 1 1 S Z 1 Z 1 Z 1 xe x=1 dx = 1 1 = 1 1 xe x=1 e y=3 dydx Likewise, E (Y ) = 3: That is, the expected values of the 2 events are the average waiting time and the average dining time, respectively. Exercises: Find the mass of the lamina with the given mass density 1. x = ; x = 1; y = ; y = 2 2. x = ; x = y; y = ; y = 1 (x; y) = 2 kg per square meter (x; y) = 2 kg per square meter 3. x = ; x = ; y = ; y = sin(x) 4. y = ; y = 1; x = y; x = y 2 (x; y) = 2x kg per square meter (x; y) = 2x kg per square meter 5. y = 1 x 2 ; y = x 2 1 6. y = 1; y = cosh (x) (x; y) = x 2 + y 2 kg per square meter (x; y) = y kg per square meter Find the center of mass of the lamina of the following regions with the given mass density. (Use 1-6 above) 7. x = ; x = 1; y = ; y = 2 8. x = ; x = y; y = ; y = 1 (x; y) = 2 kg per square meter (x; y) = 2 kg per square meter 9. x = ; x = ; y = ; y = sin(x) 1. y = ; y = 1; x = y; x = y 2 (x; y) = 2x kg per square meter (x; y) = 2x kg per square meter 11. y = 1 x 2 ; y = x 2 1 12. y = 1; y = cosh (x) (x; y) = x 2 + y 2 kg per square meter (x; y) = y kg per square meter Find the centroid of the following regions. 13. x = ; x = 1; y = ; y = x 14. x = ; x = 1; y = ; y = x 2 15. x = 1; x = 1; y = ; y = 1 jxj 16. y = ; y = 1; x = ; x = sin (y) 17. y = ; y = 1; x = y 2 ; y = x 2 18. y = ; y = 1; x = y 2 ; y = x Show the following are joint probability density functions over the given sample spaces. Then nd the expected valeus of the random variables X and Y: 9
19. Sample Space: [; 1] [; 5] 2. Sample Space: [; 1] [; 4] p (x; y) = :2 p (x; y) = :25 21. Sample Space: [; 1] [; 1] 22. Sample Space: [; 1] [; 1] p (x; y) = x + y p (x; y) = 4xy 23. Sample Space: 1st Quadrant 24. Sample Space: 1st Quadrant p (x; y) = :2e x y=5 p (x; y) = 2 e x 1+y 2 25. Sample Space: Entire Plane 26. Sample Space: [1; 1] [; 1] p (x; y) = 1 4 e jxj jyj p (x; y) = 1 x e xy 27. A bank operates both a drive-up window and an indoor teller window. On a randomly selected day, let X = the percentage of the work-day that the drive-up window is in use and let Y = percentage of the work-day that the indoor teller window is in use (Note: both X and Y are between and 1 ). Observation over a period of time leads them to the following joint density for X and Y : 6 p (x; y) = 5 x + 6 5 y2 if x; y 1 otherwise 1. (a) Show that p (x; y) is a joint probability density? (b) What is the probability that two windows combined are in use less than half of the time? (i.e., what is Pr (X + Y :5)?) (c) What is the probability of the drive-up window being used less than the indoor teller window? (d) What is the expected value of X? What is the expected value of Y? Interpret your result. 28. What is the joint probability density for the coordinates of a point (X; Y ) chosen uniformly randomly from the triangle with vertices (; ) ; (1; ) ; and (; 1)? What is the probability that X < Y? What is the probability that 2X < Y? 29. Suppose we are given radioactive substances A and B with decay rates > and > ; respectively. Then the probability that A will decay in the time interval [s; s + ds] and B will decay in the time interval [t; t + dt] is given by e s t dsdt for s; t : 1. (a) What is the probability density on the rst quadrant? (b) What is the probability that A decays before B (in terms of and )? (c) What is the probability that both will have decayed by time T: 1
3. What are the expected values for the decays of A and B, respectively, in exercise 29 (as functions of and )? Explain. 31. If X is uniformly randomly distributed in [; 1] ; then it has a probability density of :1 if x 1 p (x) = otherwise What is the probability that two numbers chosen uniformly randomly from [; 1] have a sum greater than 1? 32. What is the probability that two numbers chosen uniformly randomly from [; 1] have a product greater than 1? (see exercise 31). 33. The probability density for a point (x; y) chosen uniformly randomly in the unit square is 1 if x; y 1 p (x; y) = if otherwise What is the probability that y will be less than x 2 in the unit square? 34. A can of mixed nuts contains 1 kg of cashews, peanuts, and almonds. For a randomly selected box, let X and Y represent the weights of the cashews and almonds, respectively, and suppose that the joint density function of these variables is 24xy if x 1; y 1; x + y 1 p (x; y) = if x < or y < Show that p(x; y) is a probability density, and then compute the expected values of X and Y. What is the probability that half of the weight of the contents of the can will be cashews? 35. Your arrive at a restaurant where the time T between two customers being seated has a density of p (t) = e t for t : What is the probability that you have less than 1 minutes to wait before being seated if there is one person ahead of you if the person two ahead of you is seated just as you arrive? (Hint: Let S be the time between you and the customer ahead of you, and let T be the time between the customer ahead of you and the customer just seated. Assume S and T are independent with densities p (s) = e s for s and q (t) = e t for t ; respectively). 36. Suppose that on a certain internet server, the time T between the arrival of two successive packets is exponentially distributed with density p (t) = :1e :1t for t What is the probability of 3 packets being received in less than 5 milliseconds? (Hint: Let T be the waiting time between rst and second arrival, and let S be the time between the second and third arrival. Assume S and T are independent). Exercises 37 and 38 explore population densities. 11
37. Bacteria is growing on a slide which corresponds to [; 2][; 5], where distances are in centimeters. If the density of the bacteria is determined to be (x; y) = 1xy (2 x) (5 y) bacteria cm 2 then about how many bacteria are on the slide? 38. A certain town is overlaid by the unit square with sides of length 1 mile. The population density is then measured to be (x; y) = 1xy people per square mile What is the approximate population of that town? 39. Suppose that a region [ S has mass density (x; y) Show that if the center of mass of a laminate of is (x 1 ; y 1 ) and if the center of mass of a laminate of S is (x 2 ; y 2 ) ; then the center of mass of a laminate of [ S has coordinates x = x 1M 1 + x 2 M 2 M 1 + M 2 ; y = y 1M 1 + y 2 M 2 M 1 + M 2 where M 1 is the mass of and M 2 is the mass of S: 4. Show that if p 1 (x) and p 2 (y) are probability densities for x and y, respectively, then p (x; y) = p 1 (x) p 2 (y) is a probability density on 2 : 41. Write to Learn: Torque is the tendency of a force to cause an object to rotate about a given axis. In particular, if the force is gravity, then the torque 12
of a laminate of a region with mass density (x; y) about a line x = is given by torque = (x ) (x; y) da Write a short essay which shows that the torque is zero if and only if = x 42. Write to Learn: Suppose x is a random variable with probability density p 1 (x) ; and suppose that y is a random variable with density p 2 (y) : If their joint probability density p (x; y) can be written as p (x; y) = p 1 (x) p 2 (y) then x and y are said to be independent random variables. Write a short essay which shows that if x and y are independent, then P (a x b and c y d) = P (a x b) P (c y d) for all [a; b] [c; d] : 43. Write to Learn: The convolution for x of two functions f (t) and g (t) is given by (f g) (t) = Z t f (t u) g (u) du Show that if X and Y are non-negative random variables with probability densities f and g; respectively, then f g is the probability density for their sum, X + Y: 44. What is the probability density for X + Y in exercise 43 if we remove the restriction that X and Y be non-negative? 13