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2019
P1 Q10
Topical
Areas & Volumes
19 P1 Q10
2019 H2 Mathematics Paper 1 Question 10
Definite Integrals: Areas and Volumes
Answers
(i)
Line of symmetry:
y
=
0
{y=0}
y
=
0
(ii)
θ
=
0
,
π
,
or
2
π
{\theta = 0, \pi, \textrm{ or } 2 \pi}
θ
=
0
,
π
,
or
2
π
(iii)
θ
1
=
0
,
θ
2
=
π
{\theta_1 = 0, \; \theta_2 = \pi}
θ
1
=
0
,
θ
2
=
π
(iv)
6
a
2
π
units
2
{6 a^2 \pi \textrm{ units}^2}
6
a
2
π
units
2
Full solutions
(i)
Line of symmetry:
y
=
0
■
y=0 \; \blacksquare
y
=
0
■
(ii)
When
C
{C}
C
meets the
x
-
{x\textrm{-}}
x
-
axis,
y
=
0
{y=0}
y
=
0
a
(
2
sin
θ
−
sin
2
θ
)
=
0
2
sin
θ
−
2
sin
θ
cos
θ
=
0
sin
θ
(
1
−
cos
θ
)
=
0
\begin{gather*} a(2 \sin \theta - \sin 2 \theta) = 0 \\ 2 \sin \theta - 2 \sin \theta \cos \theta = 0 \\ \sin \theta (1 - \cos \theta) = 0 \end{gather*}
a
(
2
sin
θ
−
sin
2
θ
)
=
0
2
sin
θ
−
2
sin
θ
cos
θ
=
0
sin
θ
(
1
−
cos
θ
)
=
0
sin
θ
=
0
or
cos
θ
=
1
θ
=
0
,
π
,
2
π
θ
=
0
,
2
π
\begin{align*} \sin \theta &= 0 & \textrm{ or } && \cos \theta &= 1 \\ \theta &= 0, \pi, 2\pi &&& \theta &= 0, 2\pi \end{align*}
sin
θ
θ
=
0
=
0
,
π
,
2
π
or
cos
θ
θ
=
1
=
0
,
2
π
∴
θ
=
0
,
π
,
or
2
π
■
\therefore \theta = 0, \pi, \textrm{ or } 2 \pi \; \blacksquare
∴
θ
=
0
,
π
,
or
2
π
■
(iii)
d
x
d
θ
=
a
(
−
2
sin
θ
+
2
sin
2
θ
)
\frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-2\sin \theta + 2 \sin 2 \theta)
d
θ
d
x
=
a
(
−
2
sin
θ
+
2
sin
2
θ
)
Area
=
∫
−
3
a
a
y
d
x
=
∫
π
0
a
(
2
sin
θ
−
sin
2
θ
)
a
(
−
2
sin
θ
+
2
sin
2
θ
)
d
θ
=
∫
π
0
a
2
(
−
4
sin
2
θ
+
6
sin
θ
sin
2
θ
−
2
sin
2
2
θ
)
d
θ
=
∫
0
π
a
2
(
4
sin
2
θ
−
6
sin
θ
sin
2
θ
+
2
sin
2
2
θ
)
d
θ
■
\begin{align*} & \textrm{Area} \\ & = \int_{-3a}^{a} y \; \mathrm{d}x \\ & = \int_{\pi}^{0} a ( 2 \sin \theta - \sin 2 \theta ) \; a(-2\sin \theta + 2 \sin 2 \theta) \; \mathrm{d}\theta \\ & = \int_{\pi}^{0} a^2 ( -4 \sin^2 \theta + 6 \sin \theta \sin 2 \theta - 2 \sin^2 2 \theta) \; \mathrm{d}\theta \\ & = \int_{0}^{\pi} a^2 ( 4 \sin^2 \theta - 6 \sin \theta \sin 2 \theta + 2 \sin^2 2 \theta) \; \mathrm{d}\theta \; \blacksquare \end{align*}
Area
=
∫
−
3
a
a
y
d
x
=
∫
π
0
a
(
2
sin
θ
−
sin
2
θ
)
a
(
−
2
sin
θ
+
2
sin
2
θ
)
d
θ
=
∫
π
0
a
2
(
−
4
sin
2
θ
+
6
sin
θ
sin
2
θ
−
2
sin
2
2
θ
)
d
θ
=
∫
0
π
a
2
(
4
sin
2
θ
−
6
sin
θ
sin
2
θ
+
2
sin
2
2
θ
)
d
θ
■
(iv)
Total area enclosed by
C
=
2
∫
0
π
a
2
(
4
sin
2
θ
−
6
sin
θ
sin
2
θ
+
2
sin
2
2
θ
)
d
θ
=
2
a
2
∫
0
π
(
2
−
2
cos
2
θ
+
3
(
cos
3
θ
−
cos
θ
)
+
1
−
cos
4
θ
)
d
θ
=
2
a
2
∫
0
π
(
3
−
2
cos
2
θ
+
3
cos
3
θ
−
3
cos
θ
−
cos
4
θ
)
d
θ
=
2
a
2
[
3
θ
−
sin
2
θ
+
sin
3
θ
−
3
sin
θ
−
1
4
sin
4
θ
]
0
π
=
6
a
2
π
units
2
■
\begin{align*} & \textrm{Total area enclosed by } C \\ & = 2 \int_{0}^{\pi} a^2 ( 4 \sin^2 \theta - 6 \sin \theta \sin 2 \theta + 2 \sin^2 2 \theta) \; \mathrm{d}\theta \\ & = 2a^2 \int_{0}^{\pi} \left( 2-2\cos 2\theta +3 (\cos 3 \theta - \cos \theta ) + 1 - \cos 4 \theta \right) \; \mathrm{d}\theta \\ & = 2a^2 \int_{0}^{\pi} \left( 3 - 2\cos 2\theta +3 \cos 3 \theta - 3\cos \theta - \cos 4 \theta \right) \; \mathrm{d}\theta \\ & = 2a^2 \left[ 3\theta - \sin 2\theta + \sin 3 \theta - 3 \sin \theta - \frac{1}{4} \sin 4 \theta \right]_{0}^{\pi} \\ & = 6 a^2 \pi \textrm{ units}^2 \; \blacksquare \end{align*}
Total area enclosed by
C
=
2
∫
0
π
a
2
(
4
sin
2
θ
−
6
sin
θ
sin
2
θ
+
2
sin
2
2
θ
)
d
θ
=
2
a
2
∫
0
π
(
2
−
2
cos
2
θ
+
3
(
cos
3
θ
−
cos
θ
)
+
1
−
cos
4
θ
)
d
θ
=
2
a
2
∫
0
π
(
3
−
2
cos
2
θ
+
3
cos
3
θ
−
3
cos
θ
−
cos
4
θ
)
d
θ
=
2
a
2
[
3
θ
−
sin
2
θ
+
sin
3
θ
−
3
sin
θ
−
4
1
sin
4
θ
]
0
π
=
6
a
2
π
units
2
■
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