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math237_a910

# math237_a910 - SOLUTIONS FOR ASSIGNMENT 9& 10 Problem...

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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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math237_a910 - SOLUTIONS FOR ASSIGNMENT 9& 10 Problem...

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