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Mathematics
Problem
The curves
r
=
1
+
2
sin
θ
r=1+2\sin\theta
r
=
1
+
2
sin
θ
and
r
=
2
r=2
r
=
2
intersect at
θ
=
π
/
6
\theta=\pi/6
θ
=
π
/6
and
θ
=
5
π
/
6
\theta=5\pi/6
θ
=
5
π
/6
. What is the exact area of the region they enclose for
0
≤
θ
≤
2
π
0\le\theta\le 2\pi
0
≤
θ
≤
2
π
? a)
2
π
3
+
5
3
2
\dfrac{2\pi}{3}+\dfrac{5\sqrt3}{2}
3
2
π
+
2
5
3
b)
5
3
2
−
π
3
\dfrac{5\sqrt3}{2}-\dfrac{\pi}{3}
2
5
3
−
3
π
c)
π
3
+
5
3
\dfrac{\pi}{3}+5\sqrt3
3
π
+
5
3
d)
2
π
3
+
5
3
\dfrac{2\pi}{3}+5\sqrt3
3
2
π
+
5
3
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