Use the arc length formula,

`L=intds`

`ds=sqrt(1+(dy/dx)^2)dx` , if y=f(x), a`<=` x`<=` b

Given `x^2+y^2=9`

`=>y^2=9-x^2`

`y=(9-x^2)^(1/2)`

`dy/dx=1/2(9-x^2)^(1/2-1)*(-2x)`

`=-x/sqrt(9-x^2)`

Plug in the above in ds,

`ds=sqrt(1+(-x/sqrt(9-x^2)))^2dx`

`ds=sqrt(1+x^2/(9-x^2))dx`

`ds=sqrt((9-x^2+x^2)/(9-x^2))dx`

`ds=3/sqrt(9-x^2)dx`

The limits are x=0 and x=2,

`L=int_0^2 3/sqrt(9-x^2)dx`

`=3int_0^2 1/sqrt(9-x^2)dx`

Now let's first evaluate the indefinite integral by using integral substitution,

Let `x=3sin(u)`

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Use the arc length formula,

`L=intds`

`ds=sqrt(1+(dy/dx)^2)dx` , if y=f(x), a`<=` x`<=` b

Given `x^2+y^2=9`

`=>y^2=9-x^2`

`y=(9-x^2)^(1/2)`

`dy/dx=1/2(9-x^2)^(1/2-1)*(-2x)`

`=-x/sqrt(9-x^2)`

Plug in the above in ds,

`ds=sqrt(1+(-x/sqrt(9-x^2)))^2dx`

`ds=sqrt(1+x^2/(9-x^2))dx`

`ds=sqrt((9-x^2+x^2)/(9-x^2))dx`

`ds=3/sqrt(9-x^2)dx`

The limits are x=0 and x=2,

`L=int_0^2 3/sqrt(9-x^2)dx`

`=3int_0^2 1/sqrt(9-x^2)dx`

Now let's first evaluate the indefinite integral by using integral substitution,

Let `x=3sin(u)`

`dx=3cos(u)du`

`int1/sqrt(9-x^2)dx=int1/sqrt(9-(3sin(u))^2)3cos(u)du`

`=int(3cos(u))/sqrt(9-9sin^2(u))du`

`=int(3cos(u))/(sqrt(9)sqrt(1-sin^2(u)))du`

`=int(3cos(u))/(3cos(u))du`

`=int1du`

`=u`

substitute back u and add a constant C to the solution,

`=arcsin(x/3)+C`

`L=3[arcsin(x/3)]_0^2`

`L=3[arcsin(2/3)-arcsin(0)]`

`L=3arcsin(2/3)`

`~~2.19`