# tricky integral

• Jun 10th 2012, 02:10 PM
tleaf555
tricky integral
I'm trying to find the arc length of (x^3)(y^2) = 1 from (1,1) to (9,1/27). I tried parameterizing the function as r(t) = (t,t^-3/2) with no luck. I can't seem to find a way to evaluate the integral using u-substitution or trig sub. Any ideas? Thanks.
• Jun 10th 2012, 02:48 PM
simamura
Re: tricky integral
You don't need parametrization. You can write it as 2 functions: $\displaystyle y=x^{-\frac32}$ and $\displaystyle y=-x^{-\frac32}$.
Note that this function is same when reflected about x-axis, so you can find arc length of $\displaystyle y=x^{-\frac32}$ and then just double result.
• Jun 10th 2012, 03:12 PM
Reckoner
Re: tricky integral
Quote:

Originally Posted by simamura
You don't need parametrization. You can write it as 2 functions: $\displaystyle y=x^{-\frac32}$ and $\displaystyle y=-x^{-\frac32}$.
Note that this function is same when reflected about x-axis, so you can find arc length of $\displaystyle y=x^{-\frac32}$ and then just double result.

It looks like the original poster is only trying to find the length of a portion of the upper arc. Notice that he wanted the arc length between the points $\displaystyle (1,1)$ and $\displaystyle \left(9,\frac1{27}\right)$.

Finding the arc length of $\displaystyle y=x^{-3/2}$ leads to the integral $\displaystyle \int_1^9\sqrt{1+\frac9{4x^5}}\,dx$, which cannot be evaluated in terms of elementary functions. Hence the poster's question.

We can find a numeric solution, however. The length appears to be approximately 8.2373.
• Jun 10th 2012, 03:27 PM
simamura
Re: tricky integral
You're right. Integral can be only calculated approximately.
• Jun 10th 2012, 03:59 PM
tleaf555
Re: tricky integral
Thanks for confirming my suspicions. I think perhaps the instructor made a typo on the assignment. I tried all sorts of crazy stuff like a trig substitution that led to a integration by parts. It just didn't seem like it was leading anywhere, except maybe driving me crazy lol.