so the shortcut mentioned in exercise 1 doesn’t apply here.
Z 1 Z 2Z 3
(x2 + y 2 + z 2 ) dz dy dx
−1 0
1
Section 5.4 selected answers
!
Z 1Z 2
3 3
z
Math 131 Multivariate Calculus
dy dx
=
x2 z + y 2 z + 3 z=1
D Joyce, Spring 2014
−1 0
Z 1 Z 2
1
2
2
2
2
3x + 3y + 9 − x − y −
=
dy dx
Exercises 1, 2, 5, 6, 11, 13, 17.
3
−1 0
Z 1 Z 2
26
2
2
ZZZ
=
2x + 2y +
dy dx
3
−1 0
xyz dV .
1.
2 !
Z 1
3
[−1,1]×[0,2]×[1,3]
2y
26y
2x2 y +
=
dx
+
This region is a product of intervals, so each in3
3 y=0
−1
tegral is over a constant integral.
Z 1
68
2
=
4x +
dx
Z 1 Z 2Z 3
3
−1
1
xyz dz dy dx
4 3 68 −1 0
1
x + x = 48
=
Z 1 Z 2 Z 3
3
3 −1
=
x
y
z dz dy dx
−1
0
1
Z 1 Z 2 2 3 6.
z Z 3 Z z Z xz
y
x
=
dy dx
2 1
0
−1
(x + 2y + z) dy dx dz
Z 1 Z 2
1
0
1
xz
Z 3Z z
4y dy dx
x
=
2
0
−1
dx dz
=
(xy
+
y
+
yz)
Z 1
2
1
0
y=1
2
=
x 2y dx
Z 3Z z
0
−1
(x2 z + x2 z 2 + xz 2 − x − 1 − z) dx dz
=
1
Z 1
z
Z1 3 0 3
8x dx = 4x2 = 0
=
x z x3 z 2 x2 z 2 x2
−1
−1
=
+
+
−
− x − xz dz
3
3
2
2
1
x=0
Z 3 4
You might have noticed that the integrand xyz is
z
z5 z4 z2
2
=
+
+
−
−z−z
dz
an odd function in the variable x and that x is
3
3
2
2
1
integrated over the symmetric interval [−1, 1], so,
Z 3 4
5z
z 5 3z 2
of course, the integral is 0. In general, if f is an
=
+
−
− z dz
6
3
2
1
odd function,
that
is,
f
(−x)
=
−f
(x)
for
all
x,
Z a
3
z5 z6
z 3 z 2 then
f (x) dx = 0.
=
+
−
− = some number (574/9?)
6
18
2
2 1
−a
ZZZ
11. Integrate f (x, y, z) = 2x − y + z over the region W bounded by the cylinder z = y 2 (this is
The has the same limits of integration as in exer- a parabolic cylinder), the xy-plane, and the planes
cise 1. But the integrand here is an even function, x = 0, x = 1, y = −1, and y = 2.
(x2 + y 2 + z 2 ) dV .
2.
[−1,1]×[0,2]×[1,3]
1
Both x and y are limited to constant intervals, depend on z. You can solve the equation for
x to get
Z √9−3z2
√
but the range of z depends on the value of y, so
2
z should vary in the innermost integral. Here’s a x = ± 9 − 3z , so the next integral is x=−√9−3z2 .
triple integral that works.
Finally, y varies from 0 to 3 − x. Therefore, the
integral is
Z 1 Z 2 Z y2
(2x − y + z) dz dy dx
Z 1 Z √9−3z2 Z 3−x
0
−1 0
(x + y) dy dx dz
√
−1
− 9−3z 2
0
13. Integrate f (x, y, z) = 8xyz over the region W
bounded by the cylinder y = x2 , the plane y+z = 9, If you choose a different order for the variables,
you’ll get a different looking integral.
and the xy-plane.
The parabolic cylinder y = x2 is cut off at the Math 131 Home Page at
bottom by the plane z = 0 and cut off at the top
http://math.clarku.edu/~djoyce/ma131/
by the diagonal plane y+z = 9. Note that these two
planes meet in the line in the xy-plane where y = 9.
That means the base of the solid is the parabolic
segment in that plane between x2 ≤ y ≤ 9. The
two curves y = 9 and y = x2 meet in that plane at
(x, y, z) = (±3, 9, 0).
There are various orders you can do the integration, and the limits are different for the different
orders, but the easiest is probably to take x to be
the outer variable, y the middle variable, and z the
inner variable. Then −3 ≤ x ≤ 3, x2 ≤ y ≤ 9, and
0 ≤ z ≤ 9 − y. That gives the integral
Z 3 Z 9 Z 9−y
8xyz dz dy dx
−3
x2
0
17. Integrate f (x, y, z) = (x + y) over the region
W bounded by the cylinder x2 + 3z 2 = 9 and the
planes y = 0 and x + y = 3.
First note that in the xz-plane, the equation
2
x + 3z 2 = 9 describes an ellipse where the extreme
values of x are ±3 while the extreme values of z
are ±1. The base of the elliptical cylinder is that
ellipse. The top of the region W is an elliptical slice
of the cylinder cut by the diagonal plane x + y = 3.
That diagonal ellipse meets the base at the point
(x, y, z) = (3, 0, 0).
You can choose either x or z to vary in the outer
integral.
Let’s take z. The the outer integral is
Z
1
. For the next integral the values of x will
z=−1
2