# sol2 - Solutions to Test 2, Math 20C, Nov 24,2008 Center of...

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Solutions to Test 2, Math 20C, Nov 24,2008 Center of gravity problem: The denominator in the center of gravity formula is the area of the sector R. The whole circle has area 2Pi, so the sector R has area 2Pi/8 = Pi/4. We now look at the double integral in the numerator of the center of gravity formula. The line y=x intersects the circle at the point (1,1), since y^2 + y^2 = 2 implies y=1 (since y is positive). The outer integral goes from y=0 to y=1, and the inner integral goes from x=y to x=sqrt(2-y^2). The inner integral of x equals x^2/2 , slashed from x=y to x=sqrt(2-y^2); thus the inner integral equals 1 - y^2. The outer integral is thus y - y^3/3, slashed from y=0 to y=1. Therefore the outer integral equals 2/3. Thus the x-coordinate of the center of gravity is 2/3 divided by Pi/4, which is 8/(3Pi). Maximum box volume problem: Let z denote the height of the box, with x=length and y=width. The given constraint is 2x + 2y + z - 12 = 0.

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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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sol2 - Solutions to Test 2, Math 20C, Nov 24,2008 Center of...

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