Math 21a Triple Integrals Spring, 2009

1 Evaluate the following triple integrals as iterated integrals.

(a)

∫∫∫E

xy dV , where E = [0, 1]× [0, 2]× [0, 3].

(b)

∫∫∫E

xy3z2 dV , where E = [−1, 1]× [−3, 3]× [0, 3].

(c)

∫∫∫E

y2z cos(xyz) dV , where E = [0, π]× [0, 1]× [0, 2].

Hint: Try using different orders of integration.

2 For each of the following regions E, write the triple integral∫∫∫

Ef(x, y, z) dV as an iterated

integral. There may be up to six different ways to do this, depending on whether you write itwith dx dy dz or dz dy dx or dx dz dy or. . .

(a) The tetrahedron bounded by the planes x+ y + z = 1, x = 0, y = 0, and z = 0.

(b) The (solid) sphere x2 + y2 + z2 = a2.

(c) The region between the paraboloid x = 1− y2 − z2 and the yz-plane.

(d) The region bounded by the surface z = 3xy + 1 and the planes z = 0, x = 0, x = 1,y = 0, and y = x.

(e) The region bounded by the cylinder x2 + z2 = 1 and the planes y = 0 and y + z = 2.

(f) The pyramid whose base is the square [−1, 1]× [−1, 1] in the xy-plane and whose vertexis the point (0, 0, 1).

3 Evaluate the following integrals.

(a)

∫∫∫E

1 dV , where E is the region in Problem 2(a).

(b)

∫∫∫E

z dV , where E is the region in Problem 2(d).

4 Find the volume of the pyramid described in Problem 2(f).

5 Consider a brick in the region [0, 1]×[0, 2]×[0, 1] whose density at a point (x, y, z) is ρ(x, y, z) =2 + xy − 2z. Find the mass of the brick.

Triple Integrals – Solutions

1 (a) The integral is ∫ 1

0

∫ 2

0

∫ 3

0

xy dz dy dx =

∫ 1

0

∫ 2

0

3xy dy dx

=

∫ 1

0

3x

(22

2− 0

)dx

=

∫ 1

0

6x dx = 3.

(b) The integral is∫ 3

0

∫ 3

−3

∫ 1

−1

xy3z2 dx dy dz =

∫ 3

0

∫ 3

−3

y3z2

(12

2− (−1)2

2

)dy dz = 0.

Note that we can put the three integrals in whatever order we want, and putting theintegral with respect to x first makes the computation easier.

(c) If we write the integral as ∫ π

0

∫ 1

0

∫ 2

0

y2z cos(xyz) dz dy dx,

we have to use integration by parts and the integral is lots of work. However, if we writeit as ∫ 1

0

∫ 2

0

∫ π

0

y2z cos(xyz) dx dz dy,

it is a lot easier. In the inner integral, we can note that

∂

∂xy sin(xyz) = y2z cos(xyz)

so we get∫ 1

0

∫ 2

0

∫ π

0

y2z cos(xyz) dx dz dy =

∫ 1

0

∫ 2

0

(y sin(πyz)− y sin(0)) dz dy

=

∫ 1

0

∫ 2

0

y sin(πyz) dz dy.

We then similarly have∂

∂zcos(xyz) = −y sin(xyz)

so ∫ 1

0

∫ 2

0

y sin(πyz) dz dy =−1

π

∫ 1

0

(cos(2πy)− cos(0)) dy

=−1

π

∫ 1

0

(cos(2πy)− 1) dy =1

π.

2 We only give one possible way to express each integral; there are others that are equallycorrect.

(a)

∫∫∫E

f(x, y, z) dV =

∫ 1

0

∫ 1−x

0

∫ 1−x−y

0

f(x, y, z) dz dy dx.

(b)

∫∫∫E

f(x, y, z) dV =

∫ a

−a

∫ √a2−x2

−√a2−x2

∫ √a2−x2−y2

−√a2−x2−y2

f(x, y, z) dz dy dx.

(c) This region is defined by the inequality 0 ≤ x ≤ 1 − y2 − z2. For 1 − y2 − z2 to benonnegative, we also need (y, z) to lie on the disk D bounded by y2 + z2 = 1. Thus weget ∫∫∫

E

f(x, y, z) dV =

∫∫D

∫ 1−y2−z2

0

f(x, y, z) dx dA

=

∫ 1

−1

∫ √1−y2

−√

1−y2

∫ 1−y2−z2

0

f(x, y, z) dx dz dy.

(d) This region is defined by the inequalities 0 ≤ z ≤ 3xy + 1, 0 ≤ x ≤ 1, and 0 ≤ y ≤ x, sowe get ∫∫∫

E

f(x, y, z) dV =

∫ 1

0

∫ x

0

∫ 3xy+1

0

f(x, y, z) dz dy dx.

(e) Being between the planes y = 0 and y + z = 2 says that 0 ≤ y ≤ 2− z, and being insidethe cylinder says that (x, z) is in the disk D bounded by x2 + z2 = 1. Thus we get∫∫∫

E

f(x, y, z) dV =

∫∫D

∫ 2−z

0

f(x, y, z) y dA =

∫ 1

−1

∫ √1−x2

−√

1−x2

∫ 2−z

0

f(x, y, z) dy dz dx.

(f) In the pyramid, z ranges from 0 at the base to 1 at the top. The cross-section of thepyramid given by a fixed value of z is the square [−1 + z, 1− z]× [−1 + z, 1− z]. Thuswe get ∫∫∫

E

f(x, y, z) dV =

∫ 1

0

∫ 1−z

−1+z

∫ 1−z

−1+z

f(x, y, z) dx dy dz.

3 (a) The integral is∫ 1

0

∫ 1−x

0

∫ 1−x−y

0

1 dz dy dx =

∫ 1

0

∫ 1−x

0

(1− x− y) dy dx

=

∫ 1

0

((1− x)2 − (1− x)2

2

)dx

=

∫ 1

0

(1− x)2

2dx =

1

6.

(b) The integral is∫ 1

0

∫ x

0

∫ 3xy+1

0

z dz dy dx =

∫ 1

0

∫ x

0

(3xy + 1)2

2dy dx

=1

2

∫ 1

0

∫ x

0

(9x2y2 + 6xy + 1) dy dx

=1

2

∫ 1

0

(3x5 + 3x3 + x) dx =7

8.

4 The volume of a region E is given by the integral∫∫∫

E1 dV . In this case, that integral is∫ 1

0

∫ 1−z

−1+z

∫ 1−z

−1+z

1 dx dy dz =

∫ 1

0

∫ 1−z

−1+z

(2− 2z) dy dz

=

∫ 1

0

(2− 2z)2 dz =4

3.

5 The mass is given by integrating the density over the region, so the mass is∫ 1

0

∫ 2

0

∫ 1

0

(2 + xy − 2z) dz dy dx =

∫ 1

0

∫ 2

0

(2 + xy − 1) dy dx

=

∫ 1

0

(4 + 2x− 2) dy dx

= 4 + 1− 2 = 3.