SME
Calculus III (Multivariable) · Problem type
A double integral over a region between curves
Use this for “Find ∬ x dA over the region between y = x² and y = 3x.”
Chart
Shaded between f and g from 0 to 3: area ∫|f − g| dx = 4.5
One slice at x = 1.5: height 2.25, width dx
This is an example. Type your own values to start a new problem.
∫
0
∫
x
x²
x
dy
dx
=
27
4
I
A
Area of the region
worked out (not typed)
x̄
Centroid’s x (average x)
worked out (not typed)
Formulas
I
=
k
4
12
27
4
=
3
4
12
A
=
k
3
6
9
2
=
3
3
6
x̄
=
I
A
3
2
=
27
4
÷
(
9
2
)
Assumptions
•
The curves meet where x² = kx, at x = 0 and x = k; with k > 0 the line is above the parabola between them.
•
Type I: for each x from 0 to k, y runs from x² up to kx (the slice drawn), then x runs from 0 to k.
•
The same region read the other way (type II) has x from y ÷ k to √y, for y from 0 to k²: the same I.
Step-by-step
WE KNOW
k = 3
FIND
double integral of x (I), area of the region (A), centroid’s x (average x) (x̄)
Step 1 · Find double integral of x (I)
Use I = k⁴ ÷ 12
Inside,
x
is
fixed:
∫
x
dy
from
y
=
x
2
to
y
=
kx
is
x(kx
−
x
2
)
=
kx²
−
x
3
.
Outside,
integrate
that
from
x
=
0
to
x
=
k
.
I
=
3
4
12
3
∫
0
(3
x
² −
x
3
) dx
=
[
x
3
−
x
4
4
]
from
0
to
3
=
27
4
I
=
27
4
Step 2 · Find area of the region (A)
Use A = k³ ÷ 6
The
area
is
∬
1
dA
over
the
same
region:
each
slice
is
kx
−
x
2
tall.
A
=
3
3
6
3
∫
0
(3
x
−
x
2
) dx
=
[1.5x²
−
x
3
3
]
from
0
to
3
=
9
2
A
=
9
2
Step 3 · Find centroid’s x (average x): x̄
Use x̄ = I ÷ A
Dividing
∬
x
dA
by
the
area
gives
the
average
x
over
the
region.
It
comes
out
k
2
:
the
region
is
balanced
about
its
middle
slice.
x̄
=
27
4
÷
(
9
2
)
x̄
=
3
2
Check
Put the numbers back in. Both sides should match.
27
4
=
3
4
12
✓
9
2
=
3
3
6
✓
3
2
=
27
4
÷
(
9
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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