MATH 251 – LECTURE 19
JENS FORSGÅRD
http://www.math.tamu.edu/~jensf/
This week: 13.4–6,8
webAssign: 13.4–6, due 3/7 11:55 p.m.
F: Quiz 12.7
Next week: 13.8–10
webAssign: 13.8–10, opens 3/7 12 a.m.
M W: Kevin
F: no lecture
Help Sessions:
M W 5.30–8 p.m. in BLOC 161
Office Hours:
BLOC 641C
M 12:30–2:30 p.m.
W 2–3 p.m.
or by appointment.
Triple integrals
Let f (x, y, z) be defined on a rectangular box
R = [a, b] × [c, d] × [r, s] = {a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}
We choose a partition P of the intervals [a, b], [c, d], and [r, s] using intermediate points
a = x0 < · · · < xm = b,
c = y0 < · · · < yn = d,
r = z0 < · · · < zl = s.
The volume of the box Rijk = [xi−1, xi] × [yj−1, yj ] × [zk−1, zk ] is ∆Vijk = ∆xi ∆yj ∆zk . Choosing points
∗
∗
) we form the triple Riemann sum
, zijk
(x∗ijk , yijk
m X
n X
l
X
∗
∗
, zijk
) ∆Vijk .
f (x∗ijk , yijk
i=1 j=1 k=1
Define the norm |P | of the partition the be the length of the longest diagonal of all boxes Rijk . Then
ZZZ
m X
n X
l
X
∗
∗
f (x, y, z) dV = lim
f (x∗ijk , yijk
, zijk
) ∆Vijk ,
R
|P |→0
i=1 j=1 k=1
provided that the limit exists.
Theorem 1 (Fubini’s Theorem). If f is continuous on the bow R, then
ZZZ
Z s Z d Z b
f (x, y, z) dx dy dz =
f (x, y, z) dx dy dz.
R
r
c
a
Triple integrals
Exercise 2. Evaluate the integral
RRR
R (x
2
+ yz) dV , where D = {0 ≤ x ≤ 2, −3 ≤ y ≤ 0, −1 ≤ z ≤ 1}.
Triple integrals
Exercise 4. Evaluate the integral
RRR
R (x
2
+ yz) dV , where D = {0 ≤ x ≤ 2, −3 ≤ y ≤ 0, −1 ≤ z ≤ 1}.
Triple integrals
Definition 3. A solid E is said to be of type 1 if it lies between to graphs of continuous functions in x and y.
That is,
S = {(x, y, z) | (x, y) ∈ D, φ1(x, y) ≤ z ≤ φ2(x, y)}
Then,
ZZZ
ZZ
Z
f (x, y, z) dV =
E
!
φ2 (x,y)
f (x, y, z) dz
D
φ1 (x,y)
dA.
Triple integrals
Exercise 4. Evaluate the integral
RRR
E
xyz dV , where E is the unit simplex.
Triple integrals
Definition 5. A solid E is said to be of type 1 if E = {(x, y, z) | (x, y) ∈ D, φ1(x, y) ≤ z ≤ φ2(x, y)}.
If E is of type 1, then
ZZZ
ZZ
Z
!
φ2 (x,y)
f (x, y, z) dV =
f (x, y, z) dz
E
D
dA.
φ1 (x,y)
Definition 6. A solid E is said to be of type 2 if E = {(x, y, z) | (y, z) ∈ D, ψ1(y, z) ≤ z ≤ ψ2(y, z)}.
If E is of type 2, then
ZZZ
ZZ
ψ2 (y,z)
Z
f (x, y, z) dV =
E
!
f (x, y, z) dx dA.
D
ψ1 (y,z)
Definition 7. A solid E is said to be of type 3 if E = {(x, y, z) | (x, z) ∈ D, χ1(x, z) ≤ z ≤ χ2(x, z)}.
If E is of type 1, then
ZZZ
ZZ
Z
χ2 (x,z)
f (x, y, z) dV =
E
!
f (x, y, z) dy dA.
D
χ1 (x,z)
Triple integrals
Exercise 8. Evaluate the integral
plane x = 2.
RRR
E
7x dV , where E is bounded by the paraboloid x = 2y 2 + 2z 2 and the
Triple integrals
Exercise 4. Evaluate the integral
plane x = 2.
RRR
E
7x dV , where E is bounded by the paraboloid x = 2y 2 + 2z 2 and the
Triple integrals
Exercise 9. Find the center of mass of the box R = [0, 1] × [−π, 0] × [0, π] when its density is ρ(x, y, z) =
1 + x sin(y) cos(z).
Triple integrals
Exercise 6. Find the center of mass of the box R = [0, 1] × [−π, 0] × [0, π] when its density is ρ(x, y, z) =
1 + x sin(y) cos(z).