length problem

[synx]

I have to find the length of the equation x^(2/3)+y^(2/3)=1

I solve for y and get, y=+/- (x^(2/3)+1)^(3/2)

now dy/dx = (x^(2/3) +1)/x^(1/3)

the limits of integration should obviously be -1 to 1

so i pretty much get stuck on trying to integrate L = int( sqrt [(dy/dx)^2 + 1])

Am I doing this right?

 

[galactus]

The algebra is the booger here. It whittles down to an easy integral.
Arc length formula:

\(\displaystyle \L\\\int\sqrt{1+(\frac{dy}{dx})^{2}}dx\)

\(\displaystyle \L\\\frac{dy}{dx}=\frac{-\sqrt{1-x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}\)

Square it and you get:

\(\displaystyle \L\\\frac{1-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\)

\(\displaystyle \L\\\int_{0}^{1}\sqrt{1+\left(\frac{1-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\right)}dx\)

\(\displaystyle \L\\\int_{0}^{1}\sqrt{\frac{1}{x^{\frac{2}{3}}}dx\)

\(\displaystyle \L\\\int_{0}^{1}{x^{\frac{-1}{3}}}dx\)


\(\displaystyle \L\\4\int_{0}^{1}x^{\frac{-1}{3}}dx\)

EDIT: yes, Soroban reminded me of something I overlooked. Because of the 'astroid'(new term for me), one must multiply by 4.

 

[soroban]

Hello, synx!

Here's another approach.
It may be beyond you presently, or you may not be allowed to use it.
I thought I'd explain it anyway . . .

\(\displaystyle \text{Find the length of the curve: }\L\,x^{\frac{2}{3}}\,+\,y^{\frac{2}{3}} \:=\:1\)

The curve is an astroid (not asteroid), a hypocycloid of four cusps.

          |
          *
         *|*
       *  |  *
  --*-----+-----*--
       *  |  *
         *|*
          *
          |

Its parametric equations are: \(\displaystyle \:\begin{array}{cc}x\:=\:\cos^3\theta \\ y \:=\:\sin^3\theta\end{array}\;\;\Rightarrow\;\;
\begin{array}\frac{dx}{d\theta} & \,= & -3\sin\theta\cos^2\theta \\ \frac{dy}{d\theta} & \,=\, & 3\sin^2\theta\cos\theta\end{array}\)


The parametric Arc Length formula is: \(\displaystyle \L\:L \;=\;\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2\,+\,\left(\frac{dy}{dt}\right)^2} \, dt\)


Under the radical, we have:
. . . . \(\displaystyle (-3\sin\theta\cos^2\theta)^2\,+\,(3\sin^2\theta\cos\theta)^2\)
. . \(\displaystyle = \;9\sin^2\theta\cos^4\theta\,+\,9\sin^4\theta\cos^2\theta\)
. . \(\displaystyle =\;9\sin^2\theta\cos^2\theta\left(\cos^2\theta\,+\,\sin^2\theta\right)\)
. . \(\displaystyle =\;9\sin^2\theta\cos^2\theta\)


The radical becomes: \(\displaystyle \:\sqrt{9\sin^2\theta\cos^2\theta} \;=\;3\sin\theta\cos\theta\)


We can integrate in the first quadrant and multiply by 4.

The integral becomes: \(\displaystyle \L\:L \;= \;4\,\times\,3\int^{\;\;\;\frac{\pi}{2}}_0\sin\theta\cos\theta\,d\theta\)

. . \(\displaystyle \L=\;\left\, 6\sin^2\theta\right|^{^{^{\frac{\pi}{2}}}}_{__{0}}\;=\;6\sin^2\left(\frac{\pi}{2}\right)\,-\,6\sin^20 \;= \;\fbox{6}\)

 

[galactus]

I like that, Soroban. A new '-oid' I haven't heard of.