The area of the region common to both circles is -6.67 square units
How to find the area common to both circles?
Since r = -2sinθ and r = 1, we find their points of intersection.
So, r = r
-2sinθ = 1
sinθ = -1/2
θ = sin⁻¹(-1/2)
θ = 7π/6
So, the circles intersect at (1, 7π/6)
To find the area of the region common to both circles, we integrate from r = 0 to r = 1 and θ = 0 to θ = 7π/6
[tex]A = \int\limits^1 _0 \int\limits^\frac{7\pi }{6} _0( {r - r}) \, dr \,d\alpha \\A = \int\limits^1 _0 \int\limits^\frac{7\pi }{6} _0( {-2sin\alpha - 1}) \, dr \,d\alpha \\A = -\int\limits^1 _0 \int\limits^\frac{7\pi }{6} _0 {2sin\alpha \, dr \,d\alpha - \int\limits^1 _0 \int\limits^\frac{7\pi }{6} _01} \, dr \,d\alpha \\A = 2[cos\alpha]_{0} ^{\frac{7\pi }{6} } [r]_{0} ^{1} r - [r]_{0} ^{1}} [\alpha ]_{0} ^{\frac{7\pi }{6}}} \\[/tex]
[tex]A = 2[cos{\frac{7\pi }{6} - cos0] [1 - 0] - ([1 - 0][\frac{7\pi }{6} - 0]\\[/tex]
A = 2(-¹/₂ - 1) - (1 × 7π/6)
A = 2(-3/2) - 7π/6
A = -3 - 7π/6
A = (-18 - 7π)/6
A = (-18 - 21.99)/6
A = -39.99/6
A = -6.67 square units
So, the area of the region common to both circles is -6.67 square units
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