Practice Exam 4 Solution

# Let q be the tetrahedron bounded by the coordinate

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Unformatted text preview: dz dr dθ = m= 3 2π = 0 0 2π = 0 0 4 12 z r + r2 z dz dr dθ = 2 3 8r + 4r2 dr dθ 0 0 0 3 0 0 2π 0 0 0 2π 2π 4 4r2 + r3 dθ = 3 = 144π 36 + 36 dθ = 72θ 0 0 0 10. Let Q be the tetrahedron bounded by the coordinate planes and the plane 2x + 5y + z = 10. Find the mass and center of mass of Q is the density at a point P (x, y, z ) is directly proportional to the distance from P to the xz -plane. First, notice that the distance of a point from the xz plane is given by |y |. Since the volume Q is a subset of the ﬁrst octant y is always positive, thus δ (x, y, z ) = ky for some constant of proportionality k . Next, notice that if we decompose Q with respect to z , then the top surface is given by z = 10 − 2x − 5y and the bottom surface is given by z = 0. The region R in the plane that deterines the outer pair of limits of integration in our triple integrals is a triangle in the ﬁrst quadrant of the xy -plane. This triangle is bounded above by y = 1 (10 − 2x) [set z = 0 in the equation for the 5 plane and then solve for y ] and below by y = 0. Finally, we see that 0 ≤ x ≤ 5 [notice tha x-intercept of the plane is at x = 5.] From this, we can set up triple integrals for the mass and for each of the three moments: 1 5 (10−2x) 5 m= 0 0 10− 2 x − 5 y ky dz dy dx = 0 0 1 5 (10−2x) 5 =k 0 0 5 0 5 10 − 2x 5 =k 0 −x 2 10 − 2x 5 5−x dx = k 3 2 − 5 0 5 3 0 dy dx = kyz 0 5 5y 2 − xy 2 − y 3 3 10 − 2x 5 1 5 (10−2x) 5 0 10y − 2xy − 5y 2 dy dx = k 2 10− 2 x − 5 y 0 5 10 − 2x 5 5 =k 1 5 (10−2x) 5 3 5 dx = k 0 0 10− 2 x − 5 y ky (10−2x−5y ) 1 5 (10−2x) dx 0 10 − 2x 5 4 1 (5 − x)3 dx = k (5 − x)4 75 75 5 =k 0 2 5−x− 5 3 25 625 = k 75 3 Similarly, (omitting the details of the evaluations, which are very much like the one above): 1 5 (10−2x) 5 Myz = 0 1 5 (10−2x) 5 Mxz = Mxy = 0 Thus x = 0 Myz m 25 k 3 10− 2 x − 5 y yky dz dy dx = 20 k 3 zky dz dy dx = 50 k 3 0 0 5 xky dz dy dx = 0 0 0 10− 2 x − 5 y 1 5 (10−2x) 10− 2 x − 5 y 0 = 1, y = Mxz m = 4 , and z = 5 Mxy m = 2. dy dx...
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## This document was uploaded on 03/11/2014 for the course MATH 323 at Minnesota State University Moorhead .

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