Multiple Integrals. Introduction and Double Integrals Over Rectangular Regions. Philippe B. Laval KSU. Today

Transcription

1 Multiple Integrals Introduction and Double Integrals Over Rectangular Regions Philippe B. Laval KSU Today Philippe B. Laval (KSU) Double Integrals Today 1 / 21

2 Introduction In this section we define multiple integrals and learn how to compute them. We will see that like for integrals of functions of one variable (definite integrals), one diffi culty is finding an antiderivative. There is also an additional diffi culty for multiple integrals: the region of integration. Philippe B. Laval (KSU) Double Integrals Today 2 / 21

3 Review: Definite Integral The definite integral was defined when we were trying to solve the area problem. Given a function y = f (x) defined for a x b, we wanted to find the area between the graph of y = f (x), the x-axis, and the vertical lines x = a and x = b. We begin by dividing the interval [a, b] into n subintervals [x i 1, x i ], i = 1..n, of equal length. Let x denote the length of each subinterval. Clearly, x = b a n. In each subinterval [x i 1, x i ], we pick a point we denote x i. We form the Riemann sum definite integral was defined to be b a f (x) dx = lim n n i=1 f (x i ) x n i=1 f (x i ) xthe In the case that f (x) 0 on [a, b], the integral corresponds to the area below the graph (next slide). Philippe B. Laval (KSU) Double Integrals Today 3 / 21

4 Review: Definite Integral Figure: Approximating the area below a curve Philippe B. Laval (KSU) Double Integrals Today 4 / 21

5 Double Integrals and Volumes: Definition For the definite integral of a function of one variable, that is a function of the form f : R R, we integrated over over portions of its domain, most of the time over closed and bounded portions of its domain, that is over intervals. The functions we are about to learn how to integrate are functions of two variables or more, that is functions of the form f : R 2 R. The domain of such functions is a subset of R 2 that is the plane. We will be integrating over closed and bounded regions of the plane. The problem is that there are many possibilities for such regions ranging from a square or a rectangle to more complicated shapes. The region of integration is an added diffi culty when dealing with multiple integrals, a diffi culty we do not have for functions of one variable since we are always integrating over an interval. We will first consider integrals over a rectangular region. These are very simple. Then, we will look at more complicated regions. In spite of this added diffi culty, double integrals are defined in a manner similar to that of definite integrals. Philippe B. Laval (KSU) Double Integrals Today 5 / 21

6 Double Integrals and Volumes: Definition Suppose that we are given a function f (x, y) defined over a closed rectangle R = { (x, y) R 2 a x b, c y d } = [a, b] [c, d] As in the case of functions of one variable, we will first assume that f (x, y) 0 on R. Let S be the solid which lies above R and below the graph of f. In other words, S = { (x, y, z) R 3 0 z f (x, y), (x, y) R } We wish to find V, the volume of S (next slide) Philippe B. Laval (KSU) Double Integrals Today 6 / 21

7 Double Integrals and Volumes: Definition Figure: Solid above R below f (x, y) Philippe B. Laval (KSU) Double Integrals Today 7 / 21

8 Double Integrals and Volumes: Definition We begin by dividing R into subrectangles. For this, we divide the interval [a, b] into m subintervals [x i 1, x i ], i = 1..m, of equal length x = b a m. We also divide the interval [c, d] into n subintervals [y j 1, y j ], j = 1..n, of equal length y = d c n. This way, we obtain the subrectangles R ij = { (x, y) R 2 } x i 1 x x i, y j 1 y y j = [x i 1, x i ] [y j 1, y j ] The area of each R ij is ( A = ) x y. Next, in each subrectangle R ij, we pick a point denoted xij, y ij. We form the box with base R ij and height ( ) ( ) f xij, y ij. Its volume is f xij, y ij A Philippe B. Laval (KSU) Double Integrals Today 8 / 21

9 Double Integrals and Volumes: Definition Figure: Dividing a rectangle in subrectangles Philippe B. Laval (KSU) Double Integrals Today 9 / 21

10 Double Integrals and Volumes: Definition Figure: Approximating the volume of a solid above a rectangle, below f (x, y) Philippe B. Laval (KSU) Double Integrals Today 10 / 21

11 Double Integrals and Volumes: Definition Figure: Approximating a volume with boxes Philippe B. Laval (KSU) Double Integrals Today 11 / 21

12 Double Integrals and Volumes: Definition We can approximate the volume of S by adding the volume of each box obtained. In other words, we have m n V f ( xij, yij ) A i=1 j=1 Remark: The above sum is called a double Riemann sum. As in the case of functions of one variable, our approximation should get better the larger m and n are. We define the double integral of f over R to be Definition The double integral of f (x, y) over the rectangle R is: R f (x, y) da = providing the limit exists. lim m,n m n i=1 j=1 f ( x ij, y ij ) A Philippe B. Laval (KSU) Double Integrals Today 12 / 21

13 Double Integrals and Volumes: Definition Remark: It can be proven that the limit exists if f is continuous on R. Remark: To understand the notation, it helps to draw a parallel between the definite integral and double integrals. In the case of the definite integral, the width of each subinterval was called x. As the number of subintervals went to, the width of each subinterval went to 0. We called it dx. Similarly, for functions of two variables, the area of each subrectangle is A = x y. As the number of subrectangles approaches, the area of each subrectangle approaches 0. We call it da. You can think of da as being da = dxdy, the area of each subrectangle as well as the length of their sides are approaching 0. Remark: The definition of the integral does not require f (x, y) to be positive on R. When f (x, y) 0 on R, the integral is exactly the volume of the solid S. Remark: We can approximate double integrals by using double Riemann sums. In the next slides, we show a way to find the exact value of double integrals. Philippe B. Laval (KSU) Double Integrals Today 13 / 21

14 Iterated Integrals Over Rectangular Regions Let f (x, y) be a function which is continuous on R = [a, b] [c, d]. When we write d c f (x, y) dy, we mean that x is held as a constant and we integrate with respect to y. Similarly, when we write b a f (x, y) dx, we mean that y is held as a constant and we integrate with respect to x. Example Find 2 1 x 2 ydy Example Find π 0 sin x sin ydy Example Find π 0 sin x sin ydx Philippe B. Laval (KSU) Double Integrals Today 14 / 21

15 Iterated Integrals Over Rectangular Regions Definition An integral of the form [ b ] d a c f (x, y) dy dx is called an iterated integral. Usually, we omit the brackets and write b d a c f (x, y) dydx. Remark: To compute it, we integrate in two steps. First, we evaluate the integral inside, the one with respect to y, holding x as a constant. This will give us a function of x. We then integrate that function with respect to x. Remark: There is another iterated integral, obtained by switching the order of x and y: [ d ] b c a f (x, y) dx dy = d b c a f (x, y) dxdy. In this case, we first evaluate the inside integral, the one with respect to x, holding y as a constant. It will give us a function of y which we then integrate with respect to y. Philippe B. Laval (KSU) Double Integrals Today 15 / 21

16 Iterated Integrals Over Rectangular Regions: Examples Example Evaluate π 2 0 Example Evaluate π 0 π 0 sin x sin ydydx π 2 0 sin x sin ydxdy Philippe B. Laval (KSU) Double Integrals Today 16 / 21

17 Relationship Between Iterated Integrals and Double Integrals Over Rectangular Regions: Fubini s Theorem Iterated integrals are important in evaluating double integrals, as the next theorem shows us. The theorem is stated without proof. Theorem Suppose that f (x, y) is continuous on the rectangle R = [a, b] [c, d]. Then b d d b f (x, y) da = f (x, y) dydx = f (x, y) dxdy R a c Remark: Note that the order of integration does not matter. We can first integrate with respect to y, then to x. We can also do the reverse. Make sure you use the correct limits of integration. For the dx integral, the limits must be for the x variable. For the dy integral, the limits must be for the y variable. Philippe B. Laval (KSU) Double Integrals Today 17 / 21 c a

18 Relationship Between Iterated Integrals and Double Integrals Over Rectangular Regions: Fubini s Theorem Example Compute R ( x 3y 2 ) da where R = { (x, y) R 2 0 x 2, 1 y 2 }. Example Compute R y sin (xy) da where R = [1, 2] [0, π]. Example Find the volume of the solid bounded by z = f (x, y) = x 2 2y , the planes x = 2, y = 2 and the three coordinate planes. Philippe B. Laval (KSU) Double Integrals Today 18 / 21

19 Relationship Between Iterated Integrals and Double Integrals Over Rectangular Regions: Fubini s Theorem In some cases, the integral can be a little bit easier to evaluate. This happens when f can be written as the product of two functions, one a function of x, the other one a function of y. Theorem If f (x, y) = g (x) h (y), in other words if f (x, y) can be written as the product of two functions, one a function of x, the other one a function of y, then b d f (x, y) da = g (x) dx h (y) dy Example Evaluate π 2 0 π 0 R sin x sin ydydx a c Philippe B. Laval (KSU) Double Integrals Today 19 / 21

20 Double Integrals Over Rectangular Regions: Summary There are two diffi culties with computing multiple integrals: finding antiderivative and the region of integration. We learned to compute iterated integrals, that is integrals of the form b d a c f (x, y) dydx or d b c a f (x, y) dxdy. We learned to compute double integrals over a rectangular region, that is integrals of the form f (x, y) da where R = [a, b] [c, d]. R A double integral is evaluated in terms of iterated integrals. More precisely, if R = [a, b] [c, d], then Fubini s theorem tells us that f (x, y) da = b d a c f (x, y) dydx = d b c a f (x, y) dxdy R If the graph of z = f (x, y) is above the xy-plane then f (x, y) da is the volume of the solid with cross section R, between the xy-plane and z = f (x, y). Philippe B. Laval (KSU) Double Integrals Today 20 / 21 R

21 Exercises See the problems at the end of my notes on double integrals over rectangular regions. Philippe B. Laval (KSU) Double Integrals Today 21 / 21