Find the arc length of a polar function from 0 to 2pi

violin_writer
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Homework Statement


This is another problem my teacher game me.

Given the Polar function r=6*sin(t/2) where the variable t is the angle theta in radians, and that t is between 0 and 2*Pi inclusive. Find the distance around the perimeter of the graph. Hint: this is arc length , round to the nearest integer.

Homework Equations



I know to do this to calculate the arc length... it think integrate the equation from 0 to 2\pi

The Attempt at a Solution


integrate(6 sin(t/2) dt, 0, 2pi)
= -12 cos(t/2) ] 0, 2pi
= -12 cos(2pi/2) - (-12 cos(0/2))
= -12 cos(pi) - (-12 cos(0))
= -12 (-1) - (-12 (1)) = 24

I also used f-int in my calculator and got 24. Am I missing a step, what am I doing wrong?
 
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You know the formula for the arclength of a polar function, right?

S = \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta
 
Tell me if I'm mistaken but that's... the equation squared plus its derivative squared... right sure making sure.
 
You are not mistaken. That really is the correct formula for the arc-length of a polar function.
 
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