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Exam_solutions_1_ - MATH 141 1st Examination-Solution Key...

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MATH 141, 1st Examination—Solution Key Prof. Jonathan Rosenberg Wednesday, September 19, 2007 1. (25 points) Find the volume of the solid whose base is the region in the x - y plane where 4 x 2 + y 2 1 , such that each cross-section perpendicular to the x -axis is a solid square. Solution. For fixed x , the cross-section has height 2 1 - 4 x 2 and thus cross-sectional area A ( x ) = bracketleftbig 2 1 - 4 x 2 bracketrightbig 2 = 4(1 - 4 x 2 ) . The range of values of x is given by the inequality 4 x 2 1 , or x 2 1 4 , or - 1 2 x 1 2 . So the volume is integraldisplay 1 2 1 2 A ( x ) dx = integraldisplay 1 2 1 2 4(1 - 4 x 2 ) dx. This works out to 2 integraldisplay 1 2 0 4(1 - 4 x 2 ) dx = 8 integraldisplay 1 2 0 (1 - 4 x 2 ) dx = 8 bracketleftBig x - 4 3 x 3 bracketrightBig 1 2 0 = 8 bracketleftBig 1 2 - 4 3 · 1 8 bracketrightBig = 4 - 4 3 = 8 3 . 2. (25 points) Find the coordinates x, ¯ y ) of the centroid (center of gravity) of the region in the x - y plane bounded on the right by the parabola x = 4 - y 2 and on the left by the y -axis. (You can use symmetry to simplify the problem.)
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