Triple Integrals

Practice problems.
1. Evaluate the triple integral
a) ∫ ∫ ∫
E
x 3 y 2 z dx dy dz
where E = { (x, y, z) | 1 ≤ x ≤ 2, 0 ≤ y ≤ x, 0 ≤ z ≤ y 2 }
b) ∫ ∫ ∫
E
2x dx dy dz
where E = { (x, y, z) | 0 ≤ y ≤ 2, 0 ≤ x ≤ √ 4 − y 2 , 0 ≤ z ≤ y }
c) ∫ ∫ ∫
E
6xy dx dy dz
where E lies under the plane z = x + y + 1 and above the region in the xy-plane bounded
by the curves y = √ x, y = 0 and x = 1.
d) ∫ ∫ ∫
E
xy dx dy dz
where E is the solid tetrahedron with vertices (0,0,0), (1, 0, 0), (0, 2, 0) and (0, 0, 3).
2. Find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x + 3y +
6z = 12.
3. Find the average value of the function f(x, y, z) = xyz over the cube with side length 4 that
lies in the first octant with one vertex in the origin and edges parallel to the coordinate axes.
4. Find the mass and the center of mass of the solid E given in problem 1c) and that has the
density function ρ(x, y, z) = 2.
Solutions.
1. a) ∫ 2
∫ x
1 0
∫ y 2
0 x 3 y 2 z dx dy dz = ∫ 2
1 x3 dx ∫ x
0 y2 dy ∫ y 2
0 z dz = ∫ 2
1 x3 dx ∫ x
0 y2 dy z2
2 |y2 0 = ∫ 2
1 x3 dx ∫ x
0
y 6
2 dy = ∫ 2
1 x3 dx y7
14 |x 0 = ∫ 2
1 x10
14
dx =
x11
14(11) |2 1 = 13.29.
b) Here you have to evaluate the integral with respect to y last since all the other variables
have y in the bounds. ∫ 2
0 dy ∫ √ 4−y 2
0 2x dx ∫ y
0 dz = ∫ 2
0 dy ∫ √ 4−y 2
0 2x dx y = ∫ √
2
0 y dy x2 4−y
|
2
0 =
∫ 2
0 y(4 − y2 ) dy = (2y 2 − y4
4 )|2 0 = 8 − 4 = 4.
c) Sketch the region in xy-plane first. The x-bounds are 0 ≤ x ≤ 1. The y-bounds are
0 ≤ y ≤ √ x. The z-bounds are determined by the plane z = x + y + 1 and the xy-plane
and so 0 ≤ z ≤ x + y + 1. So ∫ 1 ∫ √ x ∫ x+y+1
0 0 0 6xy dx dy dz = ∫ 1
0 6xdx ∫ √ x
0 y dy ∫ x+y+1
0 dz =
∫ 1
0 6xdx ∫ √ x
0 y dy(x + y + 1) = ∫ 1
y2
0 6xdx (x + y3
+ y2
2 3 2 )|√ x
0 = ∫ 1
0 6xdx (x x + x3/2 + x) = 65.
2 3 2 28