midterm_solution

midterm_solution - Problem 1 a Volume of the given shape...

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Unformatted text preview: Problem 1 a) Volume of the given shape can be found as V = integraldisplay 1 integraldisplay 1 − 1 h ( x, y ) dxdy (1) Note that “ h ( x, y ) dxdy ” is the differential volume of a rectangular cube of di- mensions dx , dy and h ( x, y ). b) Total mass the given shape can be found as M = integraldisplay 1 integraldisplay 1 − 1 integraldisplay h ( x,y ) ρ ( x, y, z ) dzdxdy. (2) Note that in equation (2), ρ ( x, y, z ) dzdxdy is the differential mass. Also, one should be always careful about the ranges of the integral and the order of dx , dy and dz appearing in the integral. Problem 2 a) Analytical integral for the given function f ( x ) =- x 2 + 2 x is found as integraldisplay 2 (- x 2 + 2 x ) dx =- 1 3 x 3 + x 2 vextendsingle vextendsingle vextendsingle vextendsingle 2 = 4 3 . (3) b) Using simple Trapezoidal rule, the given integral is found as integraldisplay 2 (- x 2 + 2 x ) dx ∼ = 2- 2 [ f (0) + f (2)] = 0 . (4) c) Using simple Simpson’s rule with n = 2, the given integral is found as integraldisplay 2 (- x 2 + 2 x ) dx = 2- 6 [ f (0) + 4 f (1) +...
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midterm_solution - Problem 1 a Volume of the given shape...

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