SME
Calculus III (Multivariable) · Problem type
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°).”
Diagram
Region: 1 ≤ r ≤ 2, 45° ≤ θ ≤ 135°; dA = r dr dθ; area ½(β − α)(r₂² − r₁²) = 2.36
This is an example. Type your own values to start a new problem.
∬
(x²
+
y²)
dA
=
I
r₁
≤
r
≤
r₂
α
°
≤
θ
≤
β
°
Δθ
Angle swept in radians
worked out (not typed)
A
Area of the region
worked out (not typed)
Formulas
Δθ
=
(
β
−
α
)
×
π
180
π
2
=
(135
−
45)
×
π
180
I
=
Δθ
×
r₂
4
−
r₁
4
4
15π/8
=
π
2
×
2
4
−
1
4
4
A
=
Δθ
×
r₂
2
−
r₁
2
2
3π/4
=
π
2
×
2
2
−
1
2
2
Assumptions
•
In polar coordinates x² + y² = r² and dA = r dr dθ, so the integrand becomes r² × r = r³.
•
The inner integral runs over r with θ fixed; the outer one multiplies by the angle swept, β − α in radians.
•
A whole ring is α = 0° to β = 360°, and a disk is r₁ = 0.
Step-by-step
WE KNOW
r₁ = 1, r₂ = 2, α = 45°, β = 135°
FIND
angle swept in radians (Δθ), integral of x² + y² (I), area of the region (A)
Step 1 · Find angle swept in radians (Δθ)
Use Δθ = (β − α) × π ÷ 180
The
θ
integral
needs
radians:
multiply
the
degrees
swept
by
π
180
.
Δθ
=
(135
−
45)
×
π
180
Δθ
=
90
×
π
180
Δθ
=
90π
180
Δθ
=
π
2
(≈
1.57)
Step 2 · Find integral of x² + y² (I)
Use I = Δθ × (r₂⁴ − r₁⁴) ÷ 4
Integrate
r
3
from
r₁
to
r₂
to
get
r₂
4
−
r₁
4
4
.
Nothing
depends
on
θ,
so
the
outer
integral
multiplies
by
Δθ.
I
=
π
2
×
2
4
−
1
4
4
2
∫
1
r
3
dr
=
[
r
4
4
]
from
1
to
2
=
3.75
I
=
3.75
×
(
π
2
)
I
=
15π/8
(≈
5.89)
Step 3 · Find area of the region (A)
Use A = Δθ × (r₂² − r₁²) ÷ 2
The
area
is
∬
1
dA:
integrate
r
dr
from
r₁
to
r₂
,
then
multiply
by
Δθ.
A
=
π
2
×
2
2
−
1
2
2
2
∫
1
r
dr
=
[
r
2
2
]
from
1
to
2
=
1.5
A
=
1.5
×
(
π
2
)
A
=
3π/4
(≈
2.36)
Check
Put the numbers back in. Both sides should match.
π
2
=
(135
−
45)
×
π
180
✓
15π/8
=
π
2
×
2
4
−
1
4
4
✓
3π/4
=
π
2
×
2
2
−
1
2
2
✓
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A double integral over a region between curves
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Mass and center of mass of a box
Topic check
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