calculus- arc length and surfaces of revolution

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I will like to know an easier method of solving the following problem.
Find the arc length of the function over the indicated interval.

y= 3/2 x^(2/3) 1?x?8

S= integral from a to b ?(1+[f '(x))]^2) dx

so y ' = x^(-1/3) = 1/x^(1/3)

= integral from 1 to 8 times the square root of (x^2/3 + 1)/ x^2/3

then I apply substitution:
u= x^2/3 + 1
du= 2/3x^-1/3 dx
dx= 3/2 du
2/3 times integral from 1 to 8 of u^1/2 du = 3/2[u^3/2 / 3/2] from 1 to 8 = (x^2/3 + 1)^3/2 from 1 to 8= 5 ?5 -2?2

if anyone could please tell me how I could solve this exercise more easily I will greatly appreciate it.
 
If you ended up with the right answer (as you should -- it seems you did everything correct) then I'm sorry to inform you that this is the only way to do it!

The only easier way would be to use your calculater to integrate for you. This should be possible since you are in Calc III ?

If you own a TI - 83 or 84 this can be done by going to "Math" then selecting "fnint"

Type it in in this format....

fnint( f(x) , x , a , b )

Where f(x) is what you are integrating, x is the variable you are integrating, a is the lower bound and b is the upper bound.

Hope this helps some =/
 
Hi, thank you for your help.

Actually I am in calc 2 and we are not allowed to use a calculator. The professor mentioned something about solving y in terms of x and then applying the substitution, but I am not sure how solve it that way for this kind of problem.

Thanks again!
 
The professor means solve \(\displaystyle y=\frac{3}{2}x^{\frac{2}{3}}\) for x, change the limits of integration From x=1 to 8 to y=3/2 to 6 and do it

that way by integrating in terms of y. The integral is easier.
 
galactus said:
The professor means solve \(\displaystyle y=\frac{3}{2}x^{\frac{2}{3}}\) for x, change the limits of integration From x=1 to 8 to y=3/2 to 6 and do it

that way by integrating in terms of y. The integral is easier.
Click to expand...

Ah hah! That does make it much simpler. Thanks galactus.
 
\(\displaystyle My \ Way:\)

\(\displaystyle Arc \ Length \ = \ s \ = \ \int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx\)

\(\displaystyle Find \ s \ of \ f(x) \ = \ \frac{3x^{2/3}}{2} \ on \ interval \ [1,8].\)

\(\displaystyle f'(x) \ = \ \bigg(\frac{3}{2}\bigg)\bigg(\frac{2}{3}\bigg)x^{-1/3} \ = \ x^{-1/3}, \ [x^{-1/3}]^{2} \ = \ x^{-2/3}\)

\(\displaystyle Hence, \ s \ = \ \int_{1}^{8}\sqrt{1+x^{-2/3}}dx \ = \ \int_{1}^{8}\frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}dx\)

\(\displaystyle Now, \ let \ u \ = \ x^{2/3}+1, \ \implies \ x^{2/3} \ = \ u-1, \ and \ du \ = \ \frac{2}{3}x^{-1/3}dx,\)

\(\displaystyle \implies \ dx \ = \ \frac{3\sqrt{u-1}}{2}du, \ [x^{2/3}]^{1/2} \ = \ \sqrt{u-1}\)

\(\displaystyle Ergo, \ we \ have \ \int_{2}^{5}\frac{\sqrt{u}}{\sqrt{u-1}}\bigg(\frac{3}{2}\bigg)\sqrt{u-1}du \ = \ \frac{3}{2}\int_{2}^{5}u^{1/2}du\)

\(\displaystyle = \ \bigg(\frac{3}{2}\bigg)\bigg[\frac{2u^{3/2}}{3}\bigg]_{2}^{5} \ = \ u^{3/2}\bigg]_{2}^{5} \ = \ 5^{3/2}-2^{3/2} \ \dot= \ \ 8.352 \ units.\)
 
Thank you all very much fo your help! I understand it now.

Thanks again! :D
 
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