sol3 - Solutions to Final Exam Math 20C Dec 9 2008(1 The...

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Solutions to Final Exam, Math 20C, Dec 9, 2008 (1) The formula for the volume of a cone is not needed for this problem. Piece A has the same shape as the big cone, but all three of its dimensions are reduced by a factor of 4/5, as we can see using similar triangles. Thus the volume of piece A is (4/5)^3 times the volume of the big cone. Since (4/5)^3 = 64/125 > 1/2, piece A has more than half the volume of the big cone, so piece A has greater volume than piece B (but not by much). (2) The inner integral is the integral of 1, from z = - sqrt(1 - x^2 - y^2) to z = sqrt(1 - x^2 - y^2). Thus the inner integral equals 2sqrt(1 - x^2 - y^2). It follows that the volume of the unit sphere is the double integral of 2sqrt(1 - x^2 - y^2) dx dy over the unit disk on the floor centered at the origin. Changing to polar coordinates, this becomes the double integral of 2sqrt(1-r^2) r dr dt (where t = theta). The inner integral goes from r=0 to r=1, and the outer integral goes from t=0 to t=2Pi. Thus the sphere's volume is 2Pi times the integral of 2r sqrt(1-r^2) dr
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This note was uploaded on 06/01/2009 for the course PHYS PHYS2A taught by Professor Hicks during the Fall '08 term at UCSD.

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sol3 - Solutions to Final Exam Math 20C Dec 9 2008(1 The...

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