Lecture 32: Double Integrals PDF

1 Lecture 32 : Double integrals In one variable calculus we had seen that the integral of a nonnegative function is the area under the graph. The double integral of a nonnegative function f (x, y) defined on a region in the plane is associated with the vo...

1 Lecture 32 : Double integrals In one variable calculus we had seen that the integral of a nonnegative function is the area under the graph. The double integral of a nonnegative function f (x, y) defined on a region in the plane is associated with the volume of the region under the graph of f (x, y). The definition of double integral is similar to the definition of Riemannn integral of a single variable function. Let Q = [a, b] × [c, d] and f : Q → R be bounded. Let P1 and P2 be partitions of [a, b] and [c, d] respectively. Suppose P1 = {x0 , x1 ,..., xn } and P2 = {y0 , y1 ,..., yn }. Note that the partition P = P1 × P2 decomposes Q into Pn mn Pmsub-rectangles. Define mij = inf{f (x, y) : (x, y) ∈ [xi−1 , xi ] × [yj−1 , yj ]} and L(P, f ) = i=1 j=1 mij 4yj 4xi. Similarly we can define U (P, f ). Define lower integral and upper integral as we do in the single variable case. We say that f (x, y) is integrable if both lower and upper integral of f (x, y) are equal. If the function f (x, y) is integrable on Q then the double integral is denoted by RR RR f (x, y)dxdy or f (x, y)dA. Q Q The proof of the following theorem is similar to the single variable case. Theorem: If a function f (x, y) is continuous on a rectangle Q = [a, b] × [c, d] then f is integrable on Q. Fubini’s Theorem: In one variable case, we use the second FTC for calculating integrals. The fol- lowing result, called Fubini’s theorem, provides a method for calculating double integrals. Basically, it converts a double integral into two successive one dimensional integrations. Theorem 32.1: Let f : Q = [a, b] × [c, d] → à R be continuous. ! Then à ! RR R R d b Rb Rd f (x, y)dxdy = f (x, y)dx dy = f (x, y)dy dx. Q c a a c We will not present the proof of the previous theorem, instead we present a geometric interpre- tation of it. Geometric interpretation: Let f (x, y) > 0 for every (x, y) ∈ Q and f be continuous. Consider the solid S enclosed by Q, the planes x = a, x = b, y = c, y = d and the RR surface z = f (x, y). From the way we have defined the double integral, we can consider the value f (x, y)dxdy as the volume of Q S. We will now use the method of slicing and calculate the volume of S. Rb For every y ∈ [c, d], A(y) = f (x, y)dx is the area of the cross section of the solid S cut by a a plane parallel to the xz-plane. à Therefore, it!follows from the method of slicing that Rd Rb Rd f (x, y)dx dy = A(y)dy c a c is the volume of the solid S. The other two successive single integrals compute the volume of S by integrating the area of the cross section cut by the planes parallel to the yz-plane. Double integral over general bounded regions: We defined the double integral of a function which is defined over a rectangle. We will now extend the concept to more general bounded regions. Let f x, y) be a bounded function defined on a bounded region D in the plane. Let Q be a rectangle such that D ⊆ Q. Define a new function fe(x, y) on Q as follows: fe(x, y) = f (x, y) if (x, y) ∈ D and fe(x, y) = 0 if (x, y) ∈ Q \ D.  2 Basically we have extended the definition of f to Q by making the function value equal to 0 outside D. e RR If f (x, y) is integrable RR over Q, then we say that f (x, y) is integrable over D and we define f (x, y)dxdy = fe(x, y)dxdy. We find that defining the concept of double integral over a more D Q RR general region D is a trivial one, but the important question is how to evaluate f (x, y). D If D is a general bounded domain, then there is no general method to evaluate the double integral. However, if the domain is in a simpler form (as given in the following result) then there is a result to convert the double integral in to two successive single integrals. Fubini’s theorem (stronger form) : Let f (x, y) be a bounded function over a region D. 1. If D = {(x, y) : a ≤ x ≤ b and f1 (x) à ≤ y ≤ f2 (x)} ! for some continuous functions f1 , f2 : RR Rb f2R(x) [a, b] → R, then f (x, y)dxdy = f (x, y)dy dx. D a f1 (x) 2. If D = {(x, y) : c ≤ y ≤ d and g1 à (y) ≤ x ≤ g2 (y)} ! for some continuous functions g1 , g2 : RR Rd g2R(y) [c, d] → R, then f (x, y)dxdy = f (x, y)dx dy. D c g1 (y) If we use the method of slicing, as we did earlier, we can get the geometric interpretation of the previous theorem. Let us illustrate the method given in the previous theorem with some examples. RR Example 1: Let us evaluate the integral (x + y)2 dxdy where D is the region bounded by the D lines joining the points (0, 0), (0, 1) and (2, 2). Note that the domain D (see Figure 1) is the form given in the first part of the previous theorem with a = 0, b = 2, f1 (x) = x and f2 (x) = x2 + 1. x RR 2 R2 2R+1 Therefore, by the previous theorem (x + y) dxdy = ( (x + y)2 dy) dx. D 0 x R2 R1 x2 Example 2: Let us evaluate ( e dx) dy. Note that we are given two consecutive single integrals. 0 y 2 First we have to integrate w.r.to x and then w.r.to y. If we directly integrate then the calculation becomes complicated. So we will use Fubini’s theorem and change the order of integration (i.e., dxdy to dydx). Note that when we change the order of integration the limits will change. We will first use Fubini’s theorem and convert the consecutive single integrals in to a double integral over a domain D. Note that the integrals are of the form given in the R 2second R 1 part of the 2 previous theorem. By the previous theorem (going from right to left) we have 0 ( y/2 ex dx) dy = RR y D2 f (x, y)dxdy where D2 = {(x, y) : 0 ≤ y ≤ 2 and 2 ≤ x ≤ 1} (see Figure 2). Now we will use the first part of the previous theorem and convert this double integral into two consecutive single RR R1 R2x 2 integrals. By the first part of the previous theorem, f (x, y)dxdy = ( ex dy) dx = e − 1. D2 0 0