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Unformatted text preview: 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 crosssection perpendicular to the xaxis is a solid square. Solution. For fixed x , the crosssection has height 2 √ 1 4 x 2 and thus crosssectional 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 4(1 4 x 2 ) dx = 8 integraldisplay 1 2 (1 4 x 2 ) dx = 8 bracketleftBig x 4 3 x 3 bracketrightBig 1 2 = 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...
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This note was uploaded on 09/12/2009 for the course MATH 141 taught by Professor Hamilton during the Spring '07 term at Maryland.
 Spring '07
 Hamilton
 Math

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