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prob_06_03_07

# prob_06_03_07 - Problem 6.3-7 Given the integral below Find...

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Problem 6.3-7 Given: the integral below Find: Solutions using 1, 4 and 9 point quadrature. Solution: I exact 1 1 η 1 1 ξ 3 ξ 2 + 2 η 2 + d d 5.803 = := f ξ η , ( ) 3 ξ 2 + 2 η 2 + := One Point Rule: ξ 1 0. := η 1 0 := W 1 2. := I 1 W 1 W 1 f ξ 1 η 1 , ( ) 6 = := I 1 I exact I exact 100 3.398 = % error 4 Point Rule: ξ 1 1 3 := η 1 1 3 := W 1 1 := ξ 2 1 3 := η 2 1 3 := W 2 1 := I 2 W 1 W 1 f ξ 1 η 1 , ( ) W 2 W 1 f ξ 2 η 1 , ( ) + W 1 W 2 f ξ 1 η 2 , ( ) + W 2 W 2 f ξ 2 η 2 , ( ) + 5.714 = := I 2 I exact I exact 100 1.525 = % error 9 Point Rule: ξ 1 0.6 := η 1 0.6 := W 1 5 9 := ξ 2 0 := η 2 0 := W 2 8 9 := ξ 3 0.6 := η 3 0.6 := W 3 5 9 := I 3 W 1 W 1 f ξ 1 η 1 , ( ) W 2 W 1 f ξ 2 η 1 , ( ) + W 3 W 1 f ξ 3 η 1 , ( ) + W 1 W 2 f ξ 1 η 2 , ( ) W 2 W 2 f ξ 2 η 2 , ( ) + W 3 W 2 f ξ 3 η 2 , ( ) + + ... W 1 W 3 f ξ 1 η 3 , ( ) W 2 W 3 f ξ 2 η 3 , ( ) + W 3 W 3 f ξ 3 η
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