2220pracprelim2sol - Math 222 Practice Prelim II Solutions...

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Unformatted text preview: Math 222 Practice Prelim II Solutions Spring 2010 Calculators, books, and notes are not permitted. Show all your work. Formulas that might be useful: cos2 θ = 1 (1 + cos 2θ ) 2 sin2 θ = 1 (1 − cos 2θ ) 2 1. Sketch the region of integration for the integral 2 −2 4 y2 0 4−x dz dx dy and find the limits of integration when the order of integration is changed to dy dz dx . Do not evaluate the integral. Solution. We can read the answer off from the figure: 4 0 0 4−x √ x √ −x dy dz dx 2. Evaluate the integral e−x R 2 −y 2 −z 2 x2 + y 2 dx dy dz where R is the ball x2 + y 2 + z 2 ≤ a2 of radius a (an arbitrary constant). Solution. Use spherical coordinates. We have x2 + y 2 = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ2 sin2 φ , so the integral becomes 2π 0 0 π 0 a e−ρ ρ2 sin φ dρ dφ dθ = ρ sin φ 2 2π π a dθ 0 0 dφ 0 ρe−ρ dρ = π 2 −e−a + 1 2 2 = 2π π e −2 −ρ2 a 0 3. Using polar coordinates, find the area of the region inside the circle (x − 1)2 + y 2 = 1 and outside the circle x2 + y 2 = 1 . Solution. In polar coordinates the circle (x − 1)2 + y 2 = 1 becomes (r cos θ − 1)2 + r 2 sin2 θ = 1 , which 1 the points ( 2 , ± √ 3 ), 2 simplifies to r = 2 cos θ . The two circles intersect at where θ = ± π . By symmetry we 3 can compute the area of the upper half of the region and double the answer. Thus the area is Math 222 Practice Prelim II Solutions Spring 2010 π /3 2 cos θ π /3 2 0 1 r dr dθ = 0 r2 = 2 cos θ 1 π /3 0 π /3 dθ = 0 4 cos2 θ − 1 dθ π /3 0 2 + 2 cos 2θ − 1 dθ = θ + sin 2θ √ π 3 =+ 3 2 z ≥ 0 and compute the volume of this region. 4. Sketch the region in R3 defined by the inequalities y 2 + z 2 ≤ 1 , x ≤ y , x ≥ 0 , and Solution. The region is shown at the right. There are several choices on the order of integration that could be used. One that works fairly nicely is dz dx dy . √ 0 1 0 0 y 1−y 2 1 y 0 1 dz dx dy = 0 1− y2 dx dy = 0 y 1− y 2 dy (1 − y 2 )3/2 = −3 1 0 = 1 3 The problem could also be done using cylindrical coordinates with the roles of x and z switched, so y = r sin θ , z = r cos θ , and x = x . The integral is then π /2 0 0 1 0 r sin θ π /2 1 r dx dr dθ = 0 0 r 2 sin θ dr dθ = r3 3 1 0 − cos θ π /2 0 = 1 3 5. Use an appropriate change of variables to evaluate 1 R 4 x2 + 9 y 2 dA where R is the region inside the ellipse 4x2 + 9y 2 = 1 and above the line 3y = −2x . Solution. This is similar to a homework problem. We first simplify by setting u = 2x and v = 3y , so x = u/2 and y = v/3 , hence dx dy = 1 6 du dv . The ellipse 4x2 + 9y 2 = 1 becomes the circle u2 + v 2 = 1 and the line 3y = −2x becomes v = −u , so the given region R in the xy plane becomes the region D in the uv -plane inside u2 + v 2 = 1 and by 0 ≤ r ≤ 1 and − π ≤ θ ≤ 4 √ 3π 4 above v = −u . Switching to polar coordinates in the uv -plane, the region D is defined . Then we have 3π/4 −π/4 0 1 D 1 1 du dv = 2 + v2 6 u 11 1 θ r dr dθ = 6r 6 3π/4 −π/4 1 r 0 = π 6 Math 222 Practice Prelim II Solutions Spring 2010 constant density δ = 1 . 6. Find the center of mass of the upper hemisphere x2 + y 2 + z 2 ≤ a2 , z ≥ 0 , with Solution. By symmetry the center of mass must lie on the z axis. We use spherical coordinates to compute 2π 0 0 π /2 0 a R z dx dy dz : 2π π /2 a ρ cos φ ρ2 sin φ dρ dφ dθ = 0 dθ 0 sin φ cos φ dφ 0 π /2 0 ρ3 dρ a4 4 πa4 = 4 sin2 φ = 2π 2 ρ4 4 a 0 1 = 2π 2 Now we divide this by the volume of the hemisphere, 2 πa3 , to get the z -coordinate of the 3 πa4 /4 = 3a/8 . Thus the center of mass is (0, 0, 3a/8) . center of mass: 2πa3 /3 ...
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