The area of the region which is inside the curve r = -4 sinθ but outside r = 2 will be 4.21 square units
When the two curves intersect then they bound the region is known as the area bounded by the curve.
The region outside the circle r = 2 and inside the circle r = - 4 sinθ.
Then the intersection point will be given as
[tex]\rm 2 = -4 \sin \theta\\\\\theta = 3.66519 ,5.7595[/tex]
Then by the integration, we have
[tex]\rightarrow \dfrac{1}{2} \int _{3.6652}^{5.7595} [(-4\sin \theta)^2 - (2)^2] dx\\\\\\\rightarrow \dfrac{1}{2} \int _{3.6652}^{5.7595} [16 \sin ^2 \theta - 4]dx\\\\\\\rightarrow \dfrac{1}{2} \int _{3.6652}^{5.7595} [ 8(1- \cos 2\theta -) - 4]dx\\\\\\\rightarrow \dfrac{1}{2} \int _{3.6652}^{5.7595} [4 - \cos 2\theta] dx\\\\\\[/tex]
[tex]\rightarrow \dfrac{1}{2} \left [ 4\theta + \dfrac{1}{2} \sin 2\theta \right ]_{3.6652}^{5.7595} \\\\\\\rightarrow \dfrac{1}{2} \left [ 4(5.7595 - 3.6652) + 0.5 (0.19969-0.12759) \right ][/tex]
[tex]\rightarrow \dfrac{1}{2} \left [ 4(2.0943) + 0.5 (0.0721) \right ]\\\\\rightarrow \dfrac{1}{2} \left [ 8.41325 \right ]\\\\\rightarrow 4.2066 \approx 4.21[/tex]
Thus, the area of the region is 3.75 square units.
More about the area bounded by the curve link is given below.
brainly.com/question/24563834
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