Fall 2003 Final Solutions

Fall 2003 Final Solutions - Fall, 2003 - Final Exam...

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Unformatted text preview: Fall, 2003 - Final Exam Solutions 1a. Evaluating the double integral we get: Z 1 Z y 1 / 3 y 2 xy 2 dxdy = Z 1 • 1 2 x 2 y 2 ‚ y 1 / 3 y 2 dy = 1 2 Z 1 ( y 8 / 3- y 6 ) dy, = • 3 11 y 11 / 3- 1 7 y 7 ‚ 1 = 1 2 µ 3 11- 1 7 ¶ = 5 77 . 1b. Below is a plot of the region of integration, R . 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x y x = y 2 x = y 1/3 R 1c. Switching the order of integration, we get Z 1 Z y 1 / 3 y 2 xy 2 dxdy = Z 1 Z √ x x 3 xy 2 dydx 2. We must use the order of integration, drdzdθ . Therefore, we must use two triple integrals to compute the volume - one for the cylindrical part and one for the spherical part (see the Figure below). The equation for the cylinder is r = 1 and the equation for the sphere is r 2 + z 2 = 4. The curve of intersection of these surfaces is a circle of radius 1 at the height, z = √ 3 (obtained by plugging the equation for the cylinder into the equation for the sphere). The volume is then given by V = Z 2 π Z √ 3 Z 1 rdrdzdθ + Z 2 π Z 2 √ 3 Z √ 4- z 2 rdrdzdθ.-1-0.5 0.5 1-1-0.5 0.5 1 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x y z 1 3a. Using spherical coordinates and the order of integration, dρdφdθ , the integral is given by Z 2 π Z π/ 2 π/ 3 Z 2 ρ 2 sin φ cos 2 φdρdφdθ. 3b. Evaluating the integral, we get Z 2 π Z π/ 2 π/ 3 Z 2 ρ 2 sin φ cos 2 φdρdφdθ = 2 π Z π/ 2 π/ 3 • 1 3 ρ 3 sin φ cos 2 φ ‚ 2 dφ, = 16 π 3 Z π/ 2 π/ 3 sin φ cos 2 φdφ = 16 π 3 •- 1 3 cos 3 φ ‚ π/ 2 π/ 3 , = 16 π 3 µ 1 24 ¶ = 2 π 9 ....
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This note was uploaded on 09/23/2007 for the course MATH 1920 taught by Professor Pantano during the Fall '06 term at Cornell.

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Fall 2003 Final Solutions - Fall, 2003 - Final Exam...

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