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Practice Midterm 15 with Solutions - ESAM Home Page

Practice Midterm 15 with Solutions - ESAM Home Page

Practice Midterm 15 with Solutions - ESAM Home

Math 234 Practice Midterm #1 Solutions Winter 2011 1. For the double integral ∫ 1 ∫ 1 0 y x 3 dx dy x 2 + y2 sketch the region of integration, then reverse the order of integration, and finally evaluate it. Solution: The boundaries of the region are x = y, x =1andy = 0. Noting that x runs from y to 1, this means that the shaded triangle in the figure is the region of integration. Changing the order of integration therefore gives ∫ 1 ∫ x 0 0 x 3 x 2 + y dy dx 2 = 0 ∫ 1 = ∫ 1 0 ∫ x x 3 0 ∫ 1 1 x 2 + y dy dx = 2 0 x 2 tan −1 (1) dx = 1 π 3 4 = π 12 . y 1 x 3 1 ( ) ∣ y y=x ∣∣∣ x tan−1 dx x y=0 2. Find the centroid of a solid homogenous (i.e., constant density) circular paraboloid of height h and radius a (at its widest). Solution: The equation of the paraboloid is z = h(x 2 + y 2 )/a 2 ,orz = hr 2 /a 2 ,sinceatr = a we want z = h. The constant density will cancel out at the end, and so we can simplify things by assuming it to be 1. The mass is then M = ∫ 2π ∫ a ∫ h 0 = 2πh 0 ( r 2 hr 2 /a 2 dzrdrdθ =2π 2 − r4 4a 2 )∣ ∣∣∣∣ a By symmetry, x = y = 0. In addition, z = 1 M ∫ 2π ∫ a ∫ h 0 = 2h ∫ ( ) a r − r5 a 2 0 a 4 0 0 ∫ ( a =2πh a2 4 = πha2 . 2 zdzrdrdθ = 4 ∫ a hr 2 /a 2 ha 2 0 = 2h a 2 ( r 2 2 − r6 6a 4 )∣ ∣∣∣∣ a 0 0 h − h r2 a 2 ) rdr =2πh ∣ 1 ∣∣∣∣ h 2 z2 ∫ ( a rdr = 2 ∫ ( a ha hr 2 /a 2 2 0 = 2h a 2 a 2 3 = 2h 3 . 0 r − r3 a 2 ) y=x h 2 − h 2 r4 a 4 ) rdr 3. Find the moment of inertia of a solid homogeneous sphere of radius a about a diameter, e.g., of the sphere x 2 + y 2 + z 2 ≤ a 2 about the z axis. Express the final answer as a constant times the mass of the sphere times a distance squared. Solution. This can be done in either cylindrical or spherical coordinates. Using the latter, we have ∫ 2π ∫ π ∫ a ∫ 2π ∫ π ∫ a I = ρr 2 R 2 sin ϕdRdϕdϑ = ρ R 2 sin 2 ϕR 2 sin ϕdRdϕdϑ 0 0 0 0 0 0 ∫ 2π ∫ π = ρ dϑ sin 3 ϕdϕ 0 0 ∫ a 0 R 4 dR = ρ2π 4 a 5 3 5 = 8π 15 ρa5 . 1 x

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