Tutorial06-solutions

# 8 06 04 02 02 1 1 0 e s e 3 4 3 4 0 06 0 1 3 1 08 2 1

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Unformatted text preview: 0 1 = 3 1 0.8 2 1 3 0.4 0 32 4+ 3 1 3 = ln(4 + 0 Compute the volume of the solid bounded by the surfaces = 6, and + = 8. The solid is shown in Figure 1. Its volume is 4 0 = ln ) √ √ − = 8− 6− = 2 5 4 , 64 . 3 :) = 4, :) For the triple integral √ 4− 2 2 2 √ 0 2+ 2 0 ∗ sketch the solid whose volume is taken and rewrite the integral as 3 ∗ ∗ ∗ ∗ ∗ . Figure 1: Region of integration in Exercise 3 Figure 2: Region of integration in Exercise 4 4 Figure 3: Region of integration in Exercise 5 S The solid is a quarter of the cone (see Figure 2), the integral would be √ 2 0 E 3 :) 0 23 Evaluate the triple integral bounded by the surfaces S 2− 2 2 It is 1 0 = 23 0 0 , =, = V = 1, , where the volume V is = 0. 1 . 364 :) Calculation with change of variables Evaluate the area enclosed by the curve ( 2 + 2 )2 = ( 3 − 3 2 ), where > 0 is a parameter. Hint : You can re-write the equations deﬁning the curve using polar coordinates, and then apply the result of Exercise 2. Next, you will have to take some care in ﬁnding the integration limits α and...
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## This document was uploaded on 02/10/2014.

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