SME
Calculus III (Multivariable)
Multiple integrals
Use this for “Evaluate ∫ from 0 to 2 ∫ from 0 to 3 of (xy + 1) dy dx.”
Diagram
f(x, y) = xy + 1.
16 prisms on bases 0.5 × 0.75, each as tall as f at its base’s middle: they add to 15.
∬ f dA = 15.
Heights are squeezed to fit: one unit of z is drawn 0.43 as long as one of x or y.
This is an example. Type your own values to start a new problem.
∫
∫
(
p
xy
+
q
)
dy
dx
=
I
m
Inner result’s x coefficient
worked out (not typed)
k
Inner result’s constant
worked out (not typed)
f̄
Average value over the rectangle
worked out (not typed)
Formulas
m
=
p
×
d
2
−
c
2
2
4.5
=
1
×
3
2
−
0
2
2
k
=
q
×
(
d
−
c
)
3 = 1 × (3 − 0)
I
=
m
×
b
2
−
a
2
2
+
k
×
(
b
−
a
)
15
=
4.5
×
2
2
−
0
2
2
+
3
×
(2
−
0)
f̄
=
I
(
b
−
a
) × (
d
−
c
)
2.5
=
15
(2 − 0) × (3 − 0)
Assumptions
•
The region is the rectangle a ≤ x ≤ b, c ≤ y ≤ d, with a < b and c < d.
•
The inner integral runs over y with x held fixed, so it leaves a function of x alone; the outer integral then runs over x.
•
Fubini: on a rectangle, integrating in x first and then y gives the same I.
•
Where f is below the floor (f < 0), its volume counts minus.
Step-by-step
WE KNOW
p = 1, q = 1, a = 0, b = 2, c = 0, d = 3
FIND
inner result’s x coefficient (m), inner result’s constant (k), value of the double integral (I), average value over the rectangle (f̄)
Step 1 · Find inner result’s x coefficient (m)
Use m = p × (d² − c²) ÷ 2
Hold
x
fixed
and
integrate
in
y:
pxy
becomes
pxy²
2
.
Between
y
=
c
and
y
=
d
that
leaves
p
(
d
2
−
c
2
)
2
times
x.
m
=
1
×
3
2
−
0
2
2
m
=
9
2
m
=
4.5
Step 2 · Find inner result’s constant (k)
Use k = q × (d − c)
The
constant
q
becomes
qy,
which
grows
by
q
(
d
−
c
)
from
c
to
d
.
So
the
inner
result
is
the
line
g
(x)
=
mx
+
k
.
k
=
1
×
(3
−
0)
3
∫
0
(xy + 1)
dy
=
4.5x
+
3
k
=
3
Step 3 · Find value of the double integral (I)
Use I = m × (b² − a²) ÷ 2 + k × (b − a)
Now
integrate
the
inner
result
g
(x)
=
mx
+
k
in
x
from
a
to
b
:
mx
becomes
mx²
2
and
k
becomes
kx.
I
=
4.5
×
2
2
−
0
2
2
+
3
×
(2
−
0)
2
∫
0
(4.5
x
+ 3)
dx
=
[
4.5x²
2
+
3x]
from
0
to
2
I
=
9
+
6
I
=
15
Step 4 · Find average value over the rectangle (f̄)
Use f̄ = I ÷ ((b − a) × (d − c))
The average height of the surface is the volume under it over the area of the base.
f̄
=
15
(2 − 0) × (3 − 0)
f̄
=
15
6
f̄
=
2.5
Check
Put the numbers back in. Both sides should match.
4.5
=
1
×
3
2
−
0
2
2
✓
3 = 1 × (3 − 0) ✓
15
=
4.5
×
2
2
−
0
2
2
+
3
×
(2
−
0)
✓
2.5
=
15
(2 − 0) × (3 − 0)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
A double integral over a region between curves
Use this for “Find ∬ x dA over the region between y = x² and y = 3x.”
A double integral in polar coordinates
Use this for “Evaluate ∬ (x² + y²) dA over the part of the ring 1 ≤ r ≤ 2 with π/4 ≤ θ ≤ 3π/4 (45° to 135°).”
Mass and center of mass of a box
Use this for “A block 2 m by 3 m and 4 m tall has density rising evenly from 0 at the base to 600 kg/m³ at the top. Find its mass and center of mass.”