SME
Calculus II · Problem type
Trig substitution: the area under a circle
Use this for “Evaluate the integral of √(9 − x²) from 0 to 2” (r = 3).
Diagram
x² + y² = 4
Center (0, 0)
radius 2.
θ
=
sin⁻¹(
1
2
)
=
30°
=
0.5236
rad
triangle
1
2
×
1
×
√
2
2
−
1
2
=
0.866
sector ½ × 2² × 0.5236 = 1.047
∫
from
0
to
1
of
√
r
2
−
x
2
dx
=
0.866
+
1.047
=
1.913
This is an example. Type your own values to start a new problem.
∫
0
r
2
−
x²
dx
=
I
θ
Angle at the upper limit
worked out (not typed)
T
Area of the triangle
worked out (not typed)
S
Area of the sector
worked out (not typed)
Formulas
θ
=
sin⁻¹(
b
r
)
π
6
=
sin⁻¹(
1
2
)
T
=
1
2
×
b
×
√
r
2
−
b
2
0.866
=
1
2
×
1
×
√
2
2
−
1
2
S
=
1
2
×
r
2
×
θ
1.0472
=
1
2
×
2
2
×
π
6
I
=
T
+
S
1.9132 = 0.866 + 1.0472
Assumptions
•
Put x = r sin θ, so dx = r cos θ dθ and √(r² − x²) = r cos θ. The integral becomes one of r² cos² θ dθ.
•
Back in x, that gives (x ÷ 2)√(r² − x²) + (r² ÷ 2) sin⁻¹(x ÷ r) + C, which is 0 at x = 0.
•
The two terms at x = b are a triangle and a sector of the circle x² + y² = r², with θ in radians.
•
The radius r is at least 1 and the upper limit b runs from 0 to r.
Step-by-step
WE KNOW
r = 2, b = 1
FIND
angle at the upper limit (θ), area of the triangle (T), area of the sector (S), value of the integral (I)
Step 1 · Find angle at the upper limit (θ)
Use θ = sin⁻¹(b ÷ r)
Substitute
x
=
r
sin
θ
.
At
x
=
0,
θ
=
0;
at
x
=
b
,
sin
θ
=
b
r
.
This
θ
is
the
angle
between
the
upright
radius
and
the
radius
to
the
arc
above
b
.
θ
=
sin⁻¹(
1
2
)
∫
√
4 −
x
2
dx
=
(
x
2
)
√
4 −
x
2
+
2
×
sin⁻¹(
x
2
)
+
C
θ
=
π
6
(≈
0.52)
Step 2 · Find area of the triangle (T)
Use T = ½ × b × √(r² − b²)
The
first
term
at
x
=
b
,
(
b
2
)
√
r
2
−
b
2
,
is
a
right
triangle:
base
b
along
the
x-axis,
height
√
r
2
−
b
2
up
to
the
arc.
T
=
1
2
×
1
×
√
2
2
−
1
2
T
=
1
2
×
√
4 − 1
T
=
1
2
×
√
3
T
=
1
2
×
1.732
T
=
0.866
Step 3 · Find area of the sector (S)
Use S = ½ × r² × θ
The
second
term,
(
r
2
2
)
θ
,
is
the
area
of
a
sector
with
angle
θ
in
radians.
S
=
1
2
×
2
2
×
π
6
S
=
π
3
S
=
1.0472
Step 4 · Find value of the integral (I)
Use I = T + S
The
region
under
the
arc
from
0
to
b
is
the
triangle
and
the
sector
together.
I
=
0.866
+
1.0472
∫
from
0
to
1
of
√
4 −
x
2
dx
=
1.9132
I
=
1.9132
Check
Put the numbers back in. Both sides should match.
π
6
=
sin⁻¹(
1
2
)
✓
0.866
=
1
2
×
1
×
√
2
2
−
1
2
✓
1.0472
=
1
2
×
2
2
×
π
6
✓
1.9132 = 0.866 + 1.0472 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Integration by parts, partial fractions, trig substitution
Main lesson
By parts with a sine: x sin(kx)
Use this for “Evaluate the integral of x sin(2x) from 0 to π/2.”
Partial fractions: two linear factors
Use this for “Evaluate the integral of 1 ÷ ((x − 3)(x + 1)) from 4 to 6.”
Which integration method?
Use this for “Which method evaluates ∫ x·cos x dx?”