Quiz 8 - Math 53: Multivariable Calculus October 31, 2007...

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Unformatted text preview: Math 53: Multivariable Calculus October 31, 2007 Quiz 8 Lecturer: Prof. Michael Hutchings GSI: Gary Sivek Name: 1. (7 pts) Find the center of mass of the part of the disk x2 + y 2 = 4 which lies in the first quadrant and has constant density. Let the density be some real number k . Then the mass, since density is constant, 1 is just k times the area, which is k · 4 · 4π = πk . Now if we let R be the region contained in this quarter-circle, then the x-coordinate of the center of mass is: x= 1 M 0 1 = π R π /2 x · k dA = cos θ dθ · 1 kπ 2 0 2 π /2 0 0 2 r cos θ · kr drdθ 2 r3 1 r dr = · sin θ|π/2 · 0 π 3 π /2 0 = 0 8 . 3π A similar computation gives y, and since 88 , x = y . Then the center of mass is 3π 3π cos θ dθ = 1 = . π /2 0 sin θ dθ, we have 1 1 x2 0 2. (6 pts) Rewrite −1 1−y f (x, y, z ) dz dy dx as an iterated integral in the order dx dy dz . (Don’t forget to draw a picture!) It should be clear from the picture and the limits given that z ranges from 0 to 1. Next, z goes from 0 to 1 − y , which means y is bounded by 1 − z . Last, for the √ √ bounds in x, we use the bound y = x2, from which − y ≤ x ≤ y . So this is: 1 0 0 1−z √ y f (x, y, z ) dx dy dz. √ −y 3. (6 pts) Find the volume of the “spherical cap” which is the part of the solid inside x2 + y 2 + z 2 = 2 lying above z = 1. Remember that a plane not through the origin usually means spherical is not the best choice of a coordinate system – it can be done, though!. But cylindrical will be easier. The surfaces intersect when x2 + y 2 + 12 = 2, or x2 + y 2 = 1, so: 2π 1 0 2π 1 21 V = 0 √ 2−r 2 2π 1 0 r dz dr dθ = 0 √ (r 2 − r2 − r) dz dr dθ = 0 2π r −1 (2 − r2 )3/2 − 3 2 dθ 0 = 0 2π = 0 −1 3 / 2 −1 1 − − ·2 dθ 3 2 3 √ √ 4 2π − 5π 22 5 dθ = − . 3 6 3 4. (6 pts) Given E = {(x, y, z ) | 0 ≤ x ≤ 3, 0 ≤ y ≤ x, 0 ≤ z ≤ x + y }, find This presentation of E implies a straightforward setup: 3 0 0 x 0 x+y 3 x E xy dV . xy dz dy dx = 0 3 (x2y + xy 2) dy dx xy 3 xy dx + 2 30 x4 x4 + dx 3 2 5x4 x5 dx = 6 6 3 0 22 x = 0 3 = 0 3 = 0 = 0 81 243 = . 6 2 ...
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