Multiple Integrals. Chapter 4. Section 7. Department of Mathematics, Kookmin Univerisity. Numerical Methods.

Transcription

1 4.7.1 Multiple Integrals Chapter 4 Section 7

2 4.7.2 Double Integral R f ( x, y) da

3 4.7.3 Double Integral Apply Simpson s rule twice R [ a, b] [ c, d] a x, x,..., x b, c y, y,..., y d 0 1 n 0 1 h ( b a) / n, k ( d c) / m m R f ( x, y) da R b f ( x, y) da f ( x, y) dy dx a c d

4 4.7.4 Apply Simpson s rule for y y c jk, j 0,1,..., m c j d ( m / 2) 1 k f ( x, y) dy f ( x, y0) 2 f ( x, y2 j ) 3 j 1 m /2 4 f ( x, y ) (, ) j ( d c) k f ( x, ) 2 j1 f x ym 4 k f ( x, y) dydx f ( x, y ) dx 2 f ( x, y ) dx ( m / 2) 1 b d b b a c 0 2 j 3 a a j 1 m /2 b b 4 f ( x, y2 j 1) dx f ( x, ym) dx a a j 1 ( d c) k b f ( x, ) 180 a y 4 dx R y f ( x, y) da

5 4.7.5 for i 0,1,..., n let x a ih then for j 0,1,..., m Apply Simpson s rule for x b a i ( n / 2) 1 h f ( x, y j ) dx f ( x0, y j) 2 f ( x2i, y j) 3 i1 n /2 ( b a) h f 4 f ( x2i 1, y j) f ( xn, y j) (, ) 4 j y j i1 180 x Substitute this into the Simpson's rule for y then R f ( x, y) da

6 4.7.6 Composite Simpson s Rule R f ( x, y) da b a c d ( n / 2) 1 n /2 hk f ( x, y) dydx ( 0, 0) 2 ( 2, 0) 4 ( 2 1, 0) (, 0) 33 f x y f x i y f x i y f xn y i1 i1 ( m/ 2) 1 ( n/ 2) 1 n /2 2 f ( x0, y2 j) 2 f ( x2i, y2 j) 4 f ( x2i 1, y2 j) f ( xn, y2 j) j 1 i1 i1 m/ 2 ( n / 2) 1 n/ 2 4 f ( x0, y2 j 1) 2 f ( x2i, y2 j 1) 4 f ( x2i 1, y2 j 1) f ( xn, y2 j 1) j 1 i1 i1 ( n / 2) 1 n /2 f ( x0, ym) 2 f ( x2i, ym) 4 f ( x2i 1, ym) f ( xn, ym) i1 i1 + Error term

7 4.7.7 Error term Composite Simpson s Rule ( m / 2) 1 4 k( b a) h f ( 0, y0) f( 2 j, y2 j) E x j 1 x ( m / 2) f( 2 j1, y2 j1) f (, ) ( ) b m y m d c k f ( x, ) 4 dx a j x x y f f If and are continuous on then E can be written as x y ( d c)( b a) 4 f 4 f ( ˆ, ˆ) E h (, ) k 180 x y

8 4.7.8 h0.15, k 0.25 Example n 4, m 2, ln( x 2 y) dydx

9 4.7.9 h0.15, k We have Example n 4, m 2, ln( x 2 y) dydx i0 j0 f f (0.15)(0.25) ln( x 2 y) dydx wi, j ln( xi 2 y j ) ( x, y), ( x, y) x ( x 2 y) y ( x 2 y) (0.5)(0.6) E (0.15) max (0.25) max (, ) x y R ( x 2 y) ( x, y) R ( x 2 y) Actual value is (accurate to within ) 4

10 Example 2 Transform R {( x, y) 1.4 x 2.0, 1.0 y 1.5} into Rˆ {( u, v) 1 u 1, 1 v 1} using the linear transformations ln( x 2 y) dydx u (2x ), and v (2y ) then 1 1 ln( x 2 y) dy dx ln(0.3u 0.5v 4.2) dv du 1 1

11 , , ,3 3,2 3,1 3, Example Gaussian Quadrature formula for n 3 in both u and v u v r 0, u v r u v r The associated weights are c 0.88, c c ln( x 2 y) dy dx c i c i1 j ln( x 2 y) dydx ln(0.3r 0.5r 4.2) 3, 3, j 3, i 3, j

12 Not a Rectalgular region b a d( x) c( x) Apply Simpson's rule with h, k( x) 2 2 b d ( x) b kx ( ) (, ) (, ( )) 4 (, ( ) ( )) (, ( ))] a f x y dy dx c( x) f x c x f x c x k x f x d x dx a 3 h k( a) f ( a, c( a)) 4 f ( a, c( a) k( a)) f ( a, d( a)) k( a h) f ( a h, c( a h)) 4 f ( a h, c( a h) 3 k( a h)) f ( a h, d( a h)) kb ( ) f ( b, c( b)) 4 f ( b, c( b) k( b)) f ( b, d( b)) 3 b a d ( x) c( x) f ( x, y) dy dx

13 Not a Rectalgular region Graphically

14 Gaussian Quadrature For each a x b, transform [ c( x), d( x)] into [ 1, 1] ( d( x) c( x)) t ( dx c( x) f ( x, y) f x, 2 and d( x) c( x) dy dt 2 Then for each x in [ a, b] apply Gaussian Quadrature to the integral ( d( x) c( x)) t ( dx c( x) f ( x, y) dy f x, dt 2 d( x) 1 c( x) 1

15 Gaussian Quadrature Result: b a d ( x) c( x) f ( x, y) dydx b d( x) c( x) ( d( x) c( x)) r d( x) c( x) a 2 2 n cn, jf x, n, j dx j 1 After this transform [a, b] into [-1, 1] and then apply Gaussian Quadrature to approximate the integral on the right side of this equation. Accurate with small computations Widely used

16 Example 3

17 Example 4 The center of mass of a solid region D with density function σ occurs at where (= mass)

18 Example z x y z, 2 ( x, y, z) x y Center of mass is (0,0,1.6) Apply 2 2 Triple Integral Algorithm with 5 Gaussian nodes in each dimension n m p 5

19 Example 4 Approximate center of mass is x, y, z (0,0, )