# Please help with the attached picture.

### 1 Answer | Add Yours

Being given the cardioid:

`r = 1-sin(theta)`

the infinitesimal length of the arc is

`ds = sqrt(r^2+((dr)/(d theta))^2)`

Since `r^2 =1+sin^2(theta) -2*sin(theta)` and `(dr)/(d theta) =-cos(theta)` then

`ds =sqrt(2-2sin(theta)) =sqrt(2)*sqrt(1-sin(theta))`

The length of the complete arc `s` is

`s =int_0^(2*pi)ds =sqrt(2)int_0^(2*pi)sqrt(1-sin(theta))*d(theta)`

We make the substitution `sin(theta) =y rArr cos(theta)*d(theta)=dy`

Thus `d(theta) =dy/(sqrt(1-y^2)) =dy/(sqrt((1-y)(1+y)))`

Hence the arc length is

`s =sqrt(2)*int_0^(2*pi)sqrt(1-y)/sqrt(1-y^2)*dy =sqrt(2)*int_0^(2*pi)1/sqrt(1+y)dy =sqrt(2)*2*sqrt(y+1)(0-> 2*pi) =2sqrt(2)(sqrt(2*pi+1) -1)`

The graph is below attached.

Observation: If you want to compute the overall length of the complete cardioid you need to integrate between `(0 -> 6*pi)` . The value of the above integral is in this case

`s =2sqrt(2)(sqrt(6*pi+1) -1)`

**Answer: The arc length of the given function is `s =2sqrt(2)(sqrt(2pi+1)-1)` **