I will refer to the curves as r(t) (where t - polar angle) to avoid confusion with Cartesian coordinates. I.e.
circle: r=3sin(t), cardioid r=1+sin(t)
Area in polar coordinates
1 / 2 ∫ r 2 ( t ) d t 1/2\int r^2(t)dt 1/2 ∫ r 2 ( t ) d t
Draw simple chart
Region area
A = 1 / 2 ∫ t 1 t 2 ( 3 sin ( t ) ) 2 − 1 / 2 ∫ t 1 t 2 ( 1 + sin ( t ) ) 2 d t = ∫ t 1 t 2 4 sin 2 ( t ) − sin ( t ) − 1 / 2 d t A = 1/2\int_{t_1}^{t_2} (3\sin(t))^2 - 1/2\int_{t_1}^{t_2} (1+\sin(t))^2dt\\
= \int_{t_1}^{t_2} 4\sin^2(t) - \sin(t) -1/2 dt A = 1/2 ∫ t 1 t 2 ( 3 sin ( t ) ) 2 − 1/2 ∫ t 1 t 2 ( 1 + sin ( t ) ) 2 d t = ∫ t 1 t 2 4 sin 2 ( t ) − sin ( t ) − 1/2 d t
Find integration bounds as intersections of curves
3 sin ( t ) = 1 + sin ( t ) , 0 ≤ t ≤ 2 π sin ( t ) = 1 / 2 t 1 = π / 6 , t 2 = 5 π / 6 3\sin(t) = 1+ \sin(t), 0 \le t \le 2\pi\\
\sin(t) = 1/2\\
t_1 = \pi/6, t_2=5\pi/6 3 sin ( t ) = 1 + sin ( t ) , 0 ≤ t ≤ 2 π sin ( t ) = 1/2 t 1 = π /6 , t 2 = 5 π /6
Integrate
A = ∫ π / 6 5 π / 6 4 sin 2 ( t ) − s i n ( t ) d t − 1 / 2 d t = [ 2 t − sin ( 2 t ) + cos ( t ) − 1 / 2 t ] π / 6 5 π / 6 = 3 / 2 ( 5 π / 6 − π / 6 ) − ( sin ( 5 π / 3 ) − sin ( π / 3 ) ) + ( cos ( 5 π / 6 ) − cos ( π / 6 ) ) = π + 2 sin ( π / 3 ) − 2 cos ( π / 6 ) = π + 2 3 / 2 − 2 3 / 2 = π ≈ 3.1416 A= \int_{\pi/6}^{5\pi/6}4\sin^2(t) - sin(t)dt - 1/2dt\\
= \Big[2t -\sin(2t) +\cos(t) - 1/2t\Big]_{\pi/6}^{5\pi/6}\\
= 3/2(5\pi/6 -\pi/6) - (\sin(5\pi/3) - \sin(\pi/3)) + (\cos(5\pi/6) - \cos(\pi/6))\\
= \pi +2\sin(\pi/3) - 2\cos(\pi/6)\\ = \pi + 2\sqrt{3}/2 - 2\sqrt{3}/2\\ = \pi \approx 3.1416 A = ∫ π /6 5 π /6 4 sin 2 ( t ) − s in ( t ) d t − 1/2 d t = [ 2 t − sin ( 2 t ) + cos ( t ) − 1/2 t ] π /6 5 π /6 = 3/2 ( 5 π /6 − π /6 ) − ( sin ( 5 π /3 ) − sin ( π /3 )) + ( cos ( 5 π /6 ) − cos ( π /6 )) = π + 2 sin ( π /3 ) − 2 cos ( π /6 ) = π + 2 3 /2 − 2 3 /2 = π ≈ 3.1416
Answer: 3.1416
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