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 rewrite 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.
 Spring '13

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