Question
Find the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1.
1=5sin(theta)
theta=arcsin(1/5)
r'=5cos(theta)
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.
1=5sin(theta)
theta=arcsin(1/5)
r'=5cos(theta)
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.
Answers
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).
∫[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).