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Find the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1.




I tried the integral between arcsin(1/5) and pi-arcsin(1/5) of (((5sin(theta))^2+(5cos(theta))^2))^1/2 which gives me 13.694 Webworks (a math homework website) says that my answer is wrong.

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  1. z=arcsin(1/5)=0.201358 is the angle about (0,0) where the circles intersect. So, the arc length on the outer circle is indeed

    ∫[z,π-z] 5 dθ = 13.6944

    But now you have to add the arc on the inner circle to complete the perimeter

    ∫[z,π-z] 1 dθ = 2.73888

    so, the whole perimeter is 16.4333

    You can also do it without calculus, knowing that s=rθ, and using the same angles. But you have to note that the subtended angle in the larger circle is twice the angle about (0,0).

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