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

[violin_writer]

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?

 

[Char. Limit]

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

 

[violin_writer]

Tell me if I'm mistaken but that's... the equation squared plus its derivative squared... right sure making sure.

 

[Char. Limit]

You are not mistaken. That really is the correct formula for the arc-length of a polar function.