Lecture 15 Multiple Integration (Relevant section from Stewart, Section 15.1) We now turn to the integration of scalar-valued functions f : Rn R, i.e., f(x ,x , , c ), over → 1 2 · · · n regions in Rn. The need to perform such integrations is common in Physics. For example, we may wish to find: 1. The total charge Q in a region R R3 that is enclosed by a closed surface S, e.g., a spherical ⊂ surface of radius R centered at a point p R3, ∈ 2. The probability of finding an electron in the 1s state of a hydrogen atom within the distance α/a from the nucleus, where α 0 and a is the so-called Bohr radius. 0 ≥ 0 In all cases, the idea of integrating scalar-valued functions of several variables is a natural generalization of the integration of real-valued functions of a single variable, i.e., f : R R over the real line R. → We simply have to keep in mind the “Spirit of Calculus,” where we subdivide the region of interest and perform a summation over the subregions. Double Integrals (Integration in R2) (Relevant sections from Stewart, Section 15.1-2) We begin with the integration of functions of two variables, f(x,y), over regions in the plane R2. Suppose that a function f : R R2 is defined → over a region R R2. ⊂ Example: f(x,y) is the (2D or “areal”) density of a thin plate at (x,y), a measure of the amount of mass per unit area around the point (x,y). 110 The amount of mass ∆m in a small rectangle ∆x, ∆y centered at (x,y) is ∆m ∼= f(x,y)∆x, ∆y Question: What is the total mass of the plate? Answer: M = f(x,y) dA ( where dA = dx dy or dy dx) Z ZR Suppose region R is contained in the rectangular region a x b ≤ ≤ c y d ≤ ≤ In the “Spirit of Calculus,” we perform the following procedures: b a 1. Divide [a, b] into n subintervals Ii = [xi 1,xi], i = 1, 2, ,n, of length ∆x = − . − · · · n d c 2. Divide [c, d] into m subintervals Jj = [xj 1,xj], j = 1, 2, ,m, of length ∆y = − . − · · · m This produces n m tiny rectangles of area ∆A =∆x∆y. × 3. From each rectangle, ∆A , pick a sample point (x ,y ), where x I ,y J . ij i∗ j∗ i∗ ∈ i j∗ ∈ j 4. Evaluate f(xi∗,yj∗) at the sample point. 5. Now form the sum n m Snm ′ ′f(xi∗,yj∗)∆A, Xi=1 jX=1 where the primes indicate that we sum only over those rectangles ∆Aij for which ∆Aij contains points in R. 111 Claim: In the limits n,m , so that ∆A =∆A 0, the Riemann sums S , i.e., →∞ ij → mn Snm “ f(x,y) dA ” → Z ZR where the expression in quotes denotes the “ double integral of the function f over the region R. ” Particular applications: 1. If f(x,y) is the mass density function ρ, then R ρ dA is the total mass of the plate. R R 2. If f(x,y) is the charge density function ρ, then R ρ dA is the total charge on the plate. R R 3. For the special case where f(x,y) = 1, then R dA = the area of R, since we are simply summing over the areas ∆A of rectangles thatR contain R points of R. 112 Practical Calculation: We consider the summations involved in the Riemann sums as the result of a “nested ‘for’ loop”: 1. Column-wise Scan of R sum := 0 for i from 1 to n (a x b) ≤ ≤ for j from j (i) to j (i) (y (x) y y (x)) lower upper lower ≤ ≤ upper sum := sum + f( )∆x∆y end for loop end for loop 2. Row-wise Scan of R sum:= 0 for j from 1 to m (c x d) ≤ ≤ for i from i (j) to i (j) (x (y) x x (y)) lower upper lower ≤ ≤ upper sum := sum + f( )∆x∆y end for loop end for loop 113 The simplest case is when the region R is a rectangular region, i.e., a x b, c y d, which ≤ ≤ ≤ ≤ we often write as the “tensor product”, [a, b] [c, d]. In this case, we don’t have to worry about what × the “upper” or “lower” indices of integration are: We always integrate x from a to b and y from c to d. We illustrate with an example. Example: Evaluate the double integral (x2 + 5y) dA, (1) Z ZR where R is the rectangular region, 0 x 10, 0 y 3. ≤ ≤ ≤ ≤ Method No. 1: Columnwise scan We scan the rectangular region with vertical slices of thickness dx centered at x. For each x ∈ [0, 10], we scan the vertical slice from y = 0 to y = 3. As a result, Outer integration: 0 x 10. • ≤ ≤ Inner integration: 0 y 3. • ≤ ≤ This is a nested integration - a continuous version of the columnwise nested summation that we presented earlier. y Region R dx 3 x 0 x 10 Columnwise integration over region R We consider the inner integration first: For each x [0, 10], we have the integral ∈ 3 (x2 + 5y) dy. (2) Z0 114 We “nest” this integral inside the outer integral over x. The result is 10 3 (x2 + 5y) dy dx. (3) Z0 Z0 This is known as an iterated integral (because we’ll have to perform two integrations). To evaluate this integral, we consider the inner integral in (2) first. Recall that x is considered to be fixed: For each fixed x, we integrate over y. We expect that the result will probably be dependent on x. We use the Fundamental Theorem of Calculus to evaluate the inner integral. Since it involves an integration over y, we’ll use the partial antiderivative of the integrand with respect to y, i.e., 3 1 y=3 (x2 + 5y) dy = x2y + y2 . (4) 5 Z0 y=0 Note that I have written “y = 0” and “y = 3” for the limits, instead of simply “0” and “10”. The latter is certainly acceptable, but when dealing with several variables, it is often helpful to indicate exactly which variable is going to be replaced with a value. In this case, it would be easy to replace x with 0 and 3. Let us now go ahead and evaluate the above antiderivative: 3 1 y=3 (x2 + 5y) dy = x2y + y2 5 Z0 y=0 45 = 3x2 + . (5) 2 We now insert this result into the outer integral, i.e., 10 45 45 10 3x2 + dx = x3 + x 2 x Z0 0 450 = 1000 + 2 = 1225. (6) Therefore, the value of the integral over the rectangular region, as obtained by Method No. 1, is (x2 + 5y) dA = 1225. (7) Z ZR Method No. 2: Row-wise scan We scan the rectangular region with horizontal slices of thickness dy centered at y. For each y [0, 3], we scan the vertical slice from x = 0 to x = 10. As a result, ∈ 115 Outer integration: 0 y 3. • ≤ ≤ Inner integration: 0 x 10. • ≤ ≤ This is a continuous version of the second nested summation that we presented earlier. y Region R 3 y dy x 0 10 Rowwise integration over region R We consider the inner integration first: For each y [0, 3], we have the integral ∈ 10 (x2 + 5y) dx. (8) Z0 We “nest” this integral inside the outer integral over y. The result is 3 10 (x2 + 5y) dx dy. (9) Z0 Z0 Again, we consider the inner integral in (8) first. Recall that y is considered to be fixed: For each fixed y, we integrate over x. We expect that the result will probably be dependent on y. Once again, we use the Fundamental Theorem of Calculus to evaluate the inner integral. Since it involves an integration over x, we’ll use the partial antiderivative of the integrand with respect to x, i.e., 10 1 x=10 (x2 + 5y) dx = x3 + 5xy . (10) 3 Z0 x=0 Now evaluate the above antiderivative: 10 1 x=10 (x2 + 5y) dx = x3 + 5xy 3 Z0 x=0 1000 = + 50y. (11) 3 116 We now insert this result into the outer integral, i.e., 3 1000 1000 3 + 50y dy = y + 25y2 3 3 Z0 0 = 1000 + 225 = 1225. (12) Therefore, using Method No. 2, the value of our integral over the rectangular region is (x2 + 5y) dA = 1225, (13) Z ZR in agreement with that of Method No. 1. This example illustrates that, as probably expected, the value of the integral does not depend on the method used to evaluate it. In fact, for sufficiently “nice” functions and boundary curves, the value of the double integral, f(x,y) dA, (14) Z ZD should be independent of the method of integration used. What does “nice” mean? Well, for example, 1. f(x,y) is continuous over the region D and 2. the boundary ∂D of D is piecewise differentiable, that is, it is the union of a finite number of curves Ci, each of which have well-defined tangent vectors at all points, except possibly at the endpoints. That being said, we may relax the requirement in 1) to read f(x,y) is piecewise continuous over the region D. The formal mathematical result is known as Fubini’s Theorem. You can find a short discussion in Stewart’s textbook, p. 961. The “meaning” of this double integral Think of the integrand f(x,y) = x2 + 5y as a weighting function over the region of integration R – it could represent the mass density of a rectangular plate.