SME
Calculus II · Problem type
Area inside a polar curve
Use this for “Find the area inside the cardioid r = 2 + 2 cos θ.”
Diagram
r = 2 + 2 cos θ: a cardioid, θ from 0° to 360°.
Shaded: A = ½∫ r² dθ from 0° to 360° = 18.85
This is an example. Type your own values to start a new problem.
r
=
a
+
b
cos
θ
A
=
½
∫
2π
0
r²
dθ
=
Formulas
A
=
π
×
(
a
2
+
b
2
2
)
18.8496
=
π
×
(
2
2
+
2
2
2
)
Assumptions
•
a ≥ b ≥ 0, so r is never negative and one turn, θ from 0 to 2π, traces the curve once.
•
A thin wedge of angle dθ is nearly a sector of area ½r² dθ, so A = ½∫ r² dθ with θ in radians.
•
Over a full turn cos θ integrates to 0 and cos² θ to π; r = a − b cos θ is the same curve turned, with the same area.
Step-by-step
WE KNOW
a = 2, b = 2
FIND
area inside (A)
Step 1 · Find area inside (A)
Use A = π × (a² + b² ÷ 2)
Square
r:
a
2
+
2ab
cos
θ
+
b
2
cos²
θ.
Over
a
full
turn
these
give
2πa²,
0
and
πb²;
half
of
the
sum
is
the
area.
A
=
π
×
(
2
2
+
2
2
2
)
A
=
(4
+
4
2
)
×
π
A
=
(4
+
2)
×
π
A
=
6
×
π
A
=
18.8496
Check
Put the numbers back in. Both sides should match.
18.8496
=
π
×
(
2
2
+
2
2
2
)
✓
Topic check
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