# quiz6_sol - x y z | x ≥ y ≥ z ≥ The intersection of...

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Date: Mar 02, 2011 MA3160-08 Quiz 6 No Calculators! Justify all answers! Name (print): Solutions 1. (5pts.) Evaluate Z 1 y = 0 Z 1 x = y e x 3 dxdy by reversing the order of integration. Solution Z 1 y = 0 Z 1 x = y e x 3 dxdy = Z 1 x = 0 Z x 2 y = 0 e x 3 dydx = Z 1 x = 0 h y e x 3 i x 2 y = 0 dx = Z 1 x = 0 x 2 e x 3 dx = h e x 3 / 3 i 1 x = 0 = e - 1 The last integral is solved by substituting w = x 3 and dw = 3 x 2 dx . Therefore, Z 1 y = 0 Z 1 x = y e x 2 dxdy = (e - 1) / 3 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 X Axis Y Axis y=x 2 or x= y Region R 2. (5pts.) Find the volume of the region under the graph of z = 1 - x - y in the ﬁrst octant. (The ﬁrst octant is the region { ( x , y , z ) : x 0 , y 0 , z 0 } .) Solution The ﬁrst octant is the region
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Unformatted text preview: { ( x , y , z ) | x ≥ , y ≥ , z ≥ } . The intersection of the surface (plane) x + y + z = 1 with the xy-plane is the line y = 1-x . Therefore, Volume = Z 1 Z y = 1-x y = (1-x-y ) dydx = Z 1 " y-xy-y 2 2 # y = 1-x y = dx = Z 1 " (1-x )-x (1-x )-(1-x ) 2 2 # y = 1-x y = dx = Z 1 ( 1 2-x + x 2 2 ) dx = " x 2-x 2 2 + x 3 6 # 1 = 1 6 Volume = 1 6...
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## This note was uploaded on 04/04/2011 for the course MA 3160 taught by Professor Staff during the Spring '08 term at Michigan Technological University.

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