Multiple Integrals. Advanced Calculus. Lecture 1 Dr. Lahcen Laayouni. Department of Mathematics and Statistics McGill University.

Transcription

1 Lecture epartment of Mathematics and Statistics McGill University January 4, 27

2 ouble integrals Iteration of double integrals ouble integrals Consider a function f(x, y), defined over a rectangle = [a, b] [c, d]. Suppose, for simplicity, that f(x, y) for every (x, y) in. ouble integrals (continue) Let S be the solid bounded by the xy -plane (for a x b and c y d ), and the surface of the function f(x, y).the double integral of f(x, y) is the volume of the solid S x x y 2 y

3 ouble integrals Iteration of double integrals Partitioning of the domain We partition as fellows a = x < x < < x m = b c = y < y < < y n = d. The partition consists of mn rectangles R ij. ( i m, j n). y n =d y j y j y y =c x =a x x i x i x m =b R ij Riemann sum. Let (xij, y ij ) be any arbitrary point in R ij, then the Riemann sum is defined by R(f) = m n i= j= f(x ij, y ij ) A ij, A ij = x i y j = (x i x i )(y j y j ) x y 3 4

4 ouble integrals Iteration of double integrals The double integral over a rectangle The function f(x, y) is integrable over the rectangle and has double integral I = f(x, y)da. Iff the limit of the Riemann sum exists as n, m goes to, that is f(x, y)da = lim n,m m n i= j= f(x ij, y ij ) A ij. Remark The term da is called the area element and represents the limit of A = x y in the Riemann sum, written also dxdy or dydx.

5 ouble integrals Iteration of double integrals Example Find an approximate value of (x + y 2 )da, where = [, ] 2. Use a Riemann sum corresponding to the partition of into four smaller squares with points selected at the center of each. Solution /2 (/4,3/4) (3/4,3/4) (/4,/4) (3/4,/4) The partition of is formed by x = /2 and y = /2, which divide into four squares, each of area A = /4. The centers are (/4, /4), (/4, 3/4), (3/4, /4), and (3/4, 3/4), so that (x + y 2 )da ( ) 4 + ( ) 4 + ( ) 4 + ( ) 4 =.825 /2

6 ouble integrals Iteration of double integrals ouble integral over general domains If f(x, y) is defined and bounded on, R let ˆf be the extension of f { f(x, y) if (x, y) in ˆf(x, y) = otherwise ouble integral over general domains (continue) If is a bounded domain, then it is contained is some rectangle R with sides parallel to the coordinates axes. If ˆf is integrable over R, then f is integrable over and f(x, y)da = R ˆf(x, y)da.

7 ouble integrals Iteration of double integrals Properties of the double integral If f and g are integrable over and if a and b are constants, - f(x, y)da = if has zero area. 2- da = area of. 3- if f(x, y) then f(x, y)da = volume of S. 4- if f(x, y) then f(x, y)da = volume of S. 5- (af(x, y) + bg(x, y))da = a f(x, y)da + b g(x, y)da.

8 ouble integrals Iteration of double integrals Examples Let R be the rectangle a x b, c y d, then 2dA = 2 da = 2 area of R = 2(b a)(d c). R Let I = I = x 2 +y 2 x 2 +y 2 R (sin(y) + y 5 + )da. Using property (5), sin(y)da + x 2 +y 2 y 5 da + x 2 +y 2 da = I + I 2 + I 3. The domain is a symmetric disk of radius. Since sin(y) is an odd function of y, then sin(y)da = x 2 +y 2,y x 2 +y 2,y sin(y)da I =.

9 ouble integrals Iteration of double integrals Examples (continue) Similarly, y 5 is an odd function and is a symmetric domain about x -axis, thus I 2 =. Since I 3 = da = da = (area of ) = π. x 2 +y 2 x 2 +y 2 I = I + I 2 + I 3 = + + π = π. Additivity of domains If, 2,..., k are nonoverlapping domains on each of which f is integrable, then f is integrable on = 2 k and k f(x, y)da = f(x, y)da. j j=

10 ouble integrals Iteration of double integrals Example Let be the disk x 2 + y 2, then the integral I = x 2 y 2 da is the volume of a hemisphere of radius, thus I = Volume of the sphere 2 = 2π z...5 x..5. y

11 ouble integrals Iteration of double integrals y-simple domain The domain in the xy-plane is said to be y-simple if it is bounded by two vertical lines x = a and x = b, and two continuous graphs y = c(x) and y = d(x). x-simple domain The domain in the xy-plane is said to be x-simple if it is bounded by two horizontal lines y = c and y = d, and two continuous graphs x = a(y) and x = b(y).

12 ouble integrals Iteration of double integrals Iteration of double integrals If f(x, y) is continuous on the bounded y-simple domain given by a x b and c(x) y d(x), then f(x, y)da = b a dx d(x) c(x) f(x, y)dy. Similarly, if f(x, y) is continuous on the x-simple domain given by c y d and a(y) x b(y), then f(x, y)da = d c dy b(y) a(y) f(x, y)dx.

13 ouble integrals Iteration of double integrals Example Find V the volume of the solid lying above the square Q defined by x and y 2 and below the plane z = x + y. Solution Since the square is both y-simple and x-simple then we can iterate in either direction. For instance if we integrate first w.r.t x, then V = ( x + y)da = = Q 2 2 dy ( x + y)dx = 2 (/2 + y)dy = (y/2 + y 2 /2) 2 = 2 dy (x x 2 /2 + xy)

14 ouble integrals Iteration of double integrals Example Evaluate I = (, ), (, ), and (, ). T xy 2 da over the triangle T with vertices Solution Here the triangle is x-simple and y-simple. Integrating first w.r.t. y, we obtain I = dx x xy 2 dy = x 4 /3dx = x 5 /5 = /5 dx (xy 3 /3) x

15 ouble integrals Iteration of double integrals Solution (continue) Now integrating first w.r.t. x, we obtain I = dy xy 2 dx = dy (x 2 /2y 2 ) /2 y (y 2 y 4 )dy = /2 (y 3 /3 y 5 /5) = /5 y