26.5. EXERCISES 533
Example 26.4.4 Find the volume of the bounded region determined by x ≥ 0,y ≥ 0,z ≥ 0,and 1
7 x+ y+ 14 z = 1, and x+ 1
7 y+ 14 z = 1.
When z = 0, the plane 17 x+y+ 1
4 z = 1 intersects the xy plane in the line whose equationis 1
7 x+ y = 1, while the plane, x+ 17 y+ 1
4 z = 1 intersects the xy plane in the line whoseequation is x+ 1
7 y = 1. Furthermore, the two planes intersect when x = y as can be seenfrom the equations, x+ 1
7 y = 1− z4 and 1
7 x+ y = 1− z4 which imply x = y. Thus the two
dimensional picture to look at is depicted in the following picture.
x+ 17 y+ 1
4 z = 1
y+ 17 x+ 1
4 z = 1R1R2
y = x
You see in this picture, the base of the region in the xy plane is the union of the twotriangles, R1 and R2. For (x,y)∈ R1, z goes from 0 to what it needs to be to be on the plane,17 x+ y+ 1
4 z = 1. Thus z goes from 0 to 4(1− 1
7 x− y). Similarly, on R2, z goes from 0 to
4(1− 1
7 y− x). Therefore, the integral needed is
∫R1
∫ 4(1− 17 x−y)
0dzdV +
∫R2
∫ 4(1− 17 y−x)
0dzdV
and now it only remains to consider∫
R1dV and
∫R2
dV. The point of intersection of theselines shown in the above picture is
( 78 ,
78
)and so an iterated integral is
∫ 7/8
0
∫ 1− x7
x
∫ 4(1− 17 x−y)
0dzdydx+
∫ 7/8
0
∫ 1− y7
y
∫ 4(1− 17 y−x)
0dzdxdy =
76
26.5 Exercises1. Find the volume of the region determined by the intersection of the two cylinders,
x2 + y2 ≤ 16 and y2 + z2 ≤ 16.
2. Find the volume of the region determined by the intersection of the two cylinders,x2 + y2 ≤ 9 and y2 + z2 ≤ 9.
3. Find the volume of the region bounded by x2 + y2 = 4,z = 0,z = 5− y
4. Find∫ 2
0∫ 6−2z
0∫ 3−z
12 x
(3− z)cos(y2)
dydxdz.
5. Find∫ 1
0∫ 18−3z
0∫ 6−z
13 x
(6− z)exp(y2)
dydxdz.
6. Find∫ 2
0∫ 24−4z
0∫ 6−z
14 y
(6− z)exp(x2)
dxdydz.
7. Find∫ 1
0∫ 10−2z
0∫ 5−z
12 y
sinxx dxdydz.
Hint: Interchange order of integration.