math237_a910 - SOLUTIONS FOR ASSIGNMENT 9 & 10...

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Unformatted text preview: SOLUTIONS FOR ASSIGNMENT 9 & 10 Problem Set 8, A4 (ii): Sketch the region of integration and evaluate the following iterated integral by reversing the order of integration: integraldisplay 1 parenleftBigg integraldisplay x y = x sin y y dy parenrightBigg dx The region D can be described in two equivalent ways: D = { ( x, y ) R 2 | x 1 , x y x } = { ( x, y ) R 2 | y 1 , y 2 x y } . The sketch of D is shown in (1). D x y Figure 1: Sketch for problem A4 (ii) It follows that: I = integraldisplay 1 parenleftBigg integraldisplay x y = x sin y y dy parenrightBigg dx = integraldisplay 1 parenleftBigg integraldisplay y x = y 2 sin y y dx parenrightBigg dy = integraldisplay 1 (sin y y sin y ) dy = cos y | 1 integraldisplay 1 y d ( cos y ) dy dy = 1 cos 1 parenleftbigg y cos y | 1 + integraldisplay 1 cos ydy parenrightbigg = 1 sin 1 . Thus integraldisplay 1 parenleftBigg integraldisplay x y = x sin y y dy parenrightBigg dx = 1 sin 1 . 1 Problem Set 8, A6: Let V denote the volume of the thetrahedron with vertices ( a, , 0) , (0 , b, 0) , (0 , , c ) and (0 , , 0) with a, b, c > . Show that V = 1 6 abc . We can find V in two equivalent ways. 1. Let f ( x, y ) = c (1 x a y b ) and D xy = { ( x, y ) R 2 | x a, y b (1 x a ) } be the closed and bounded region on which f is integrable. Then: V = integraldisplay a x =0 integraldisplay b (1- x a ) y =0 f ( x, y ) dydx = integraldisplay a x =0 integraldisplay b (1- x a ) y =0 c parenleftbigg 1 x a y b parenrightbigg...
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This note was uploaded on 07/28/2009 for the course MATH MATH137 taught by Professor Oancea during the Spring '08 term at Waterloo.

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math237_a910 - SOLUTIONS FOR ASSIGNMENT 9 & 10...

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