I don't understand why finding the arc length is always difficult. I understand the formula and know pretty much all the integration methods, but whenever I try to find the arc length of a function like 8x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}= 27y^{3}from 1 to 8 it's unusually difficult.

I would start by solving for x

\begin{equation}

\begin{split}

&\frac{2}{3} x^{\frac{2}{3}} = y\\

&\frac{4}{9x^{\frac{1}{3}}} = y'\\

\end{split}

\end{equation}

Now I add this into the integration formula [itex]\int_a^b \sqrt{1+(y')^{2}} \, dx[/itex]

[tex]\begin{equation}

\begin{split}

L^{8}_{1} &= \int_1^8 \, \sqrt{1+ (\frac{4}{9x^{\frac{1}{3}}})^{2}} \, dx \\

&= \int_1^8 \, \sqrt{1+ (\frac{16}{81x^{\frac{2}{3}}})} \, dx \\

&= \int_1^8 \sqrt{\frac{81x^{\frac{2}{3}}+16}{81x^{\frac{2}{3}}}} \, dx \\

&= \int_1^8 \frac{\sqrt{81x^{\frac{2}{3}}+16}}{9x^{\frac{1}{3}}} \, dx

\end{split}

\end{equation}[/tex]

From here I would do a u-substitution of [itex]u=81x^{\frac{2}{3}}+16[/itex] and then take the derivative to find [itex]\frac{1}{54} du = \frac{1}{x^{\frac{1}{3}}} dx [/itex]

My new limits would be 97 and 340

\begin{equation}

\begin{split}

L^{8}_{1} &= \frac{1}{9*54} \int_{97}^{340} \sqrt{u} \, du \\

&= \frac{1}{486} \Bigg(\frac{3}{2} u^{\frac{3}{2}} |^{97}_{340} \Bigg) \\

&=7.28... \\

\end{split}

\end{equation}

I got the question right, by going slowly and typing it all out, derp. Anyway, is there a special trick for these types of problems? Or is it to just go slow and remember rules.

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# Arc Length integration always unusually difficult? Any special trick?

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