Instructor: Longfei Li
Math 243 Lecture Notes
15.2 Iterated Integrals
Let f be an integrable function of 2 variables over the rectangle region R = [a, b] × [c, d]. Integrate
f (x, y) w.r.t. y (fixing x) from c to d:
d
Z
A(x) =
f (x, y)dy
c
Then, integrate A(x) w.r.t. x from a to b,
Z b Z
b
Z
A(x)dx =
a
a
d
f (x, y)dy dx
c
Usually, the brackets are omitted, thus
Z bZ
Z b Z
d
d
f (x, y)dy dx
f (x, y)dydx =
a
a
c
c
This is called iterated integral. Similarly, if we integrate x first, and then integrate y, we get another
iterated integral
Z dZ b
f (x, y)dxdy
c
a
Fubini’s Theorem: If f is continuous on the rectangle R = {(x, y)|a ≤ x ≤ b, c ≤ y ≤ d}, then
Z bZ
ZZ
f (x, y)dA =
Z
dZ b
f (x, y)dydx =
a
R
d
c
f (x, y)dxdy
c
a
The proof of Fubini’s Theorem is beyond the scope of this course, but you can get an intuition from
the plot below
A(x) is the area of a cross section of S. The volume of S can be obtained by integrating the cross
section area along the x-axis:
ZZ
Z b
f (x, y)dA = V =
A(x)dx
R
a
1
and
d
Z
f (x, y)dy
A(x) =
c
Therefore,
Z bZ
ZZ
d
f (x, y)dydx
f (x, y)dA =
a
R
c
Similarly, we can see the intuition for the other iterated integral. The volume of S can be obtained by
integrating the cross section area along the y-axis:
ZZ
Z
dZ b
f (x, y)dA =
f (x, y)dxdy
R
c
a
Therefore,
ZZ
Z
dZ b
f (x, y)dA =
R
Z bZ
f (x, y)dxdy =
c
a
d
f (x, y)dydx
a
c
RR
Example: Evaluate R y sin(xy)dA, where R = [1, 2] × [0, π].
Solution: If we first integrate w.r.t x, then
ZZ
Z πZ 2
y sin(xy)dA =
y sin(xy)dxdy
R
0
1
Z π
=
− cos(xy)|x=2
x=1 dy
Z0 π
=
− cos(2y) + cos(y)dy
0
y=π
1
=0
= (− sin 2y + sin y)
2
y=0
If we first integrate w.r.t y, then
ZZ
Z
2Z π
y sin(xy)dA =
R
y sin(xy)dydx
1
0
Use integrate by parts. Let
u=y
dv = sin(xy)dy
then
du = dy
v=−
2
cos(xy)
x
So
π
Z
0
Z π
y cos(xy) y=π
cos(xy)
y sin(xy)dy = −
−
−
dy
x
x
y=0
0
π cos(πx) sin(xy) y=π
+
=−
x
x2
y=0
π cos(πx) sin(πx)
=−
+
x
x2
So
2
Z
ZZ
π cos(πx) sin(πx)
+
dx
x
x2
1
Z 2
Z 2
π cos(πx)
sin(πx)
=
−
dx +
dx
x
x2
1
1
−
y sin(xy)dA =
R
Let’s integrate the first term, using integration by parts:
u=−
1
x
du =
dv = π cos(πx)dx
1
dx
x2
v = sin(πx)
S0
Z 2
π cos(πx)
sin(πx) x=2
sin(πx)
−
dx = −
dx
−
x
x
x2
x=1
1
1
Z 2
Z 2
π cos(πx)
sin(πx)
sin(πx) x=2
−
⇒
dx +
dx
=
−
x
x2
x
x=1
1
1
ZZ
x=2
sin(πx) ⇒
y sin(xy)dA = −
=0
x
x=1
R
Remark: Different order of integration might have different difficulty to evaluate. Choose a wise order
to simplify your calculation.
Z
2
Another Property: If f (x, y) = g(x)h(y) on R = [a, b] × [c, d], then
ZZ
ZZ
Z bZ d
Z b
Z
f (x, y)dA =
g(x)h(y)dA =
g(x)h(y)dydx =
g(x)dx
R
R
a
c
a
Example:
3
1 1 1 3
x y dydx =
x dx
y 2 dy = x3 · y 3 = 3
3 0 3 0
0
0
0
0
RR
π
π
Example: If R = [0, 2 ] × [0, 2 ], evaluate R sin x cos ydA
Solution:
ZZ
Z π
Z π
2
2
sin x cos ydA =
sin xdx
cos ydy
R
0
0
x= π
y= π 2
2
= − cos x
sin y Z
1Z 3
2 2
Z
1
2
Z
x=0
=1
3
y=0
c
d
h(y)dy
Example: Find the volume of the solid lying under the elliptic paraboloid
the rectangle R = [−1, 1] × [−2, 2].
Solution:
x2 y 2
+
+ z = 1 and above
4
9
x2 y 2
1−
− dA
4
9
R
Z 1Z 2
2
x
y2
1−
=
− dydx
4
9
−1 −2
Z 1
166
92
− x2 dx =
=
27
−1 27
ZZ
V =
Example: Find the average of f (x, y) = x2 y, R has vertices (−1, 0), (−1, 5), (1, 5), (1, 0).
Solution:
R = [−1, 1] × [0, 5], and the area of R is 10. Therefore, the average is
ZZ
1
favg =
f (x, y)dA
Area(R) R
Z 1
Z 5
1
2
=
x dx
ydy
10 −1
0
5
=
6
4