Triple integrals
Math 21a
1 Evaluate the integral
RRR
E
Fall 2016
(xz − y 3 ) dV , where
E = {(x, y, z) : −1 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1}.
Solution. We write the triple integral as an iterated integral, and evaluate:
z=1
Z 1Z
Z 1 Z 2Z 1
xz 2
3
3 xz − y dz dy dx =
− y z
dy dx
−1 2 2
0
−1 0
z=0
Z
Z 2
Z 1
4 y=2
x
xy
y
= −11
− y 3 dy dx =
− dx
4 y=0
0 2
−1 2
x=1
Z 1
x2
x − 4 dx =
− 4x
= −8.
=
2
−1
x=−1
RRR
x dV , where
2 Evaluate the integral
E
E = {(x, y, z) : 0 ≤ x ≤ 1, −x ≤ y ≤ x2 , y − x ≤ z ≤ y + x}.
Solution. We write the triple integral as an iterated integral, and evaluate. Given the way
the solid E is described to us, the easy order of integration is dz dy dx.
Z 1 Z x2 Z y+x
Z −1 Z x4
z=y+x
x dz dy dx =
xz|z=y−x
dy dx
0
−x
y−x
−x
0
Z
1Z
x2
Z
1
y=x2
2x2 y y=−x dx
0
0
−x
x=1
Z 1
2x5 x4 9
4
3
=
2x + 2x dx =
+ = .
5
2 x=0 10
0
=
2
2x dy dx =
3 The plane cut out by the equation x + 2y + 3z = 6 passes through the points (6, 0, 0),
(0, 3, 0), and (0, 0, 2). Find the volume of the tetrahedron bounded by this plane and the
three coordinate planes.
Solution. Let’s choose the order of integration to be dz dy dx – this doesn’t matter for this
problem. Here’s how to figure out the bounds of integration:
• Over the entire tetrahedron, the minimum x-coordinate is 0 and the maximum xcoordinate is 6.
• For fixed x, y goes from 0 to
6−x
.
2
• For fixed x and y, z goes from 0 to
6−x−2y
.
3
So here is the iterated integral: the volume is
Z 6 Z 6−x Z 6−x−2y
Z 6 Z 6−x
Z 6 2
2
3
2
6 − x − 2y
x
1 dz dy dx =
dy dx =
− x + 3 dx = 6.
3
0
0
0
0
0 12
0
We can verify this answer using geometry. Recall that the volume of a pyramid is onethird the base times height, which in this case is 13 (9)(2) = 6.
RRR
4 Evaluate the triple integral
xy dV , where E is bounded by the parabolic cylinders
E
2
2
y = x and x = y , and the planes z = 0 and z = x + y.
Solution. Here’s how to figure out the bounds. Recall that the innermost bounds can depend
on the other two variables, that the middle bounds can depend on the outermost variable,
and the outermost bounds must be constants.
• Since z is given to us as between 0 and x + y, we’ll use z as the innermost integral.
The order of dx dy doesn’t matter, so we’ll use dz dx dy.
• If we project the entire solid to the xy-plane, its shadow is the region that is bounded
√
above by the curve x = y and bounded below by the curve x = y 2 , for 0 ≤ y ≤ 1.
This gives us the other bounds.
We have
Z 1 Z √y Z x+y
1
Z
√
Z
y
xy dz dx dy =
0
y2
Z
xy(x + y) dx dy =
0
y2
0
0
1
3
y 5/2 y 3 y 7 y 6
+
−
− dy = .
3
2
3
2
26
5 Let RRR
E be the solid enclosed by the paraboloids z = x2 + y 2 and z = 8 − (x2 + y 2 ).
Express
z dV as an iterated integral in the order dz dy dx.
E
Solution.
• We’ll use z as the innermost bound, and it goes from x2 + y 2 to 8 − (x2 + y 2 ).
• If we project the entire solid to the xy-plane, we get the disk of radius 2 centered
at√the origin. We√know how to give bounds for such regions, e.g., −2 ≤ x ≤ 2 and
− 4 − x2 ≤ y ≤ 4 − x2 .
So here’s the iterated integral:
Z
2
−2
√
Z
4−x2
√
− 4−x2
Z
8−x2 −y 2
z dz dy dx
x2 +y 2
How might one begin to evaluate this integral? The presence of terms like x2 +y 2 and disks
might remind you of polar coordinates, and you would want convert (x, y) into polar coordinates and integrate the resulting expression. But this is just the cylindrical coordinate system,
and we’ll discuss triple integrals in cylindrical and spherical coordinates in more detail next
time!
R1RzR1
6 Rewrite the iterated integral 0 0 y2 f (x, y, z) dx dy dz
(a) In the order dx dz dy.
(b) In the order dy dx dz.
(c) In the order dz dy dx.
Solution.
(a) Here we leave the innermost bounds unchanged and swap from dy dz to dz dy. In the
yz-plane, the bounds describe a triangle:
1.0
0.8
0.6
0.4
0.2
0.0
0.0
So we have
1
Z
1
Z
0.2
0.4
0.6
0.8
1.0
1
Z
f (x, y, z) dx dz dy.
0
y2
y
(b) Starting from the original integral, we leave the outermost bounds unchanged and swap
from dx dy to dy dx, fixing z. We sketch the region:
1
y=z
x = y^2
0
0
0.2
So we have
Z 1 Z 1 Z min(√x,z)
0.4
1
Z
0.6
Z
z2
0.8
√
Z
1
x
f (x, y, z) dy dx dz =
0
0
Z
1
Z
f (x, y, z)dy dx +
0
0
0
0
!
z
f (x, y, z) dy dx .
z2
0
(c) It’s hard to swap non-adjacent integrals, so let’s try switching from dx dz dy (from part
a) to dz dx dy, and then to dz dy dx. If we swap from dx dz to dz dx, holding y fixed, we
get
Z Z Z
1
1
1
f (x, y, z) dz dx dy.
y2
0
y
Now swappng dx dy, we get
Z
1
√
Z
x
Z
1
f (x, y, z) dz dy dx.
0
0
y
If you’re curious, here’s what the 3D solid looks like: