SME
Statics · Problem type
Centroid under a curve by integration
Use this for “Find the centroid of the area under y = x² from x = 0 to x = 3.”
Chart
Shaded: ∫ from 0 to 3 of f(x) dx = 9
This is an example. Type your own values to start a new problem.
A
=
∫
0
k
x
dx
x̄
=
(
+
1)
×
+
2
ȳ
=
(
+
1)
×
×
a
2
×
(2
×
+
1)
Formulas
A
=
k
n
+ 1
×
a
n
+ 1
9
=
1
2 + 1
×
3
2 + 1
x̄
=
(
n
+
1)
×
a
n
+ 2
2.25
=
(2
+
1)
×
3
2 + 2
ȳ
=
(
n
+
1)
×
k
×
a
n
2 × (2 ×
n
+ 1)
2.7
=
(2
+
1)
×
1
×
3
2
2 × (2 × 2 + 1)
Assumptions
•
The area lies under y = kxⁿ and above the x-axis, from x = 0 to x = a, with k > 0 and n ≥ 0.
•
Cut it into upright strips y wide dx: x̄ = ∫x dA ÷ A, and ȳ = ∫(y ÷ 2) dA ÷ A.
•
Each integral is the power rule: ∫xᵐ dx = xᵐ⁺¹ ÷ (m + 1).
Step-by-step
WE KNOW
k = 1, n = 2, a = 3
FIND
area (A), centroid x (x̄), centroid y (ȳ)
Step 1 · Find area (A)
Use A = k ÷ (n + 1) × aⁿ⁺¹
A
=
a
∫
0
kxⁿ dx
:
raise
the
power
by
1
and
divide
by
the
new
power.
A
=
1
2 + 1
×
3
2 + 1
A
=
1
3
×
3
3
A
=
1
3
×
27
A
=
0.3333
×
27
A
=
9
Step 2 · Find centroid x (x̄)
Use x̄ = (n + 1) × a ÷ (n + 2)
∫x
·
kxⁿ
dx
from
0
to
a
is
kaⁿ⁺²
÷
(
n
+
2);
divided
by
A
,
k
and
aⁿ⁺¹
cancel.
x̄
=
(2
+
1)
×
3
2 + 2
x̄
=
3
×
3
4
x̄
=
9
4
x̄
=
2.25
Step 3 · Find centroid y (ȳ)
Use ȳ = (n + 1) × k × aⁿ ÷ (2 × (2 × n + 1))
Each
strip’s
center
is
at
y
2,
so
∫(
y
2
)y
dx
=
k
2
a
2
ⁿ⁺¹
÷
(2(
2
n
+
1));
divide
by
A
.
ȳ
=
(2
+
1)
×
1
×
3
2
2 × (2 × 2 + 1)
ȳ
=
3
×
1
×
9
2 × 5
ȳ
=
27
10
ȳ
=
2.7
Check
Put the numbers back in. Both sides should match.
9
=
1
2 + 1
×
3
2 + 1
✓
2.25
=
(2
+
1)
×
3
2 + 2
✓
2.7
=
(2
+
1)
×
1
×
3
2
2 × (2 × 2 + 1)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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