Calculus STAHP!

Jimmy King

I'm working with using the length of an arc to find the surface area. Not a problem conceptually. I've got one example in my book that I'm so not following the algebra on, though.

After getting the derivative for the function of the curve and inserting it into the surface area formula we have this:
[img]http://latex.codecogs.com/gif.latex?2\pi\int_1^8{x\sqrt{1 + \tfrac{1}{9x^{4/3}}}} dx[/img]

But then through some sort of sorcery it becomes this?


How did the book get there? What rule have I forgotten about or applied incorrectly? The integrand is what I already knew I wanted it to be, because that allows me to use u substitution to get my u-du form, but I don't see what math got us here. Inside the radical it looks like everything was just multiplied by 9x^(4/3), but that doesn't explain what happened outside of the radical.

musanman

First get a common denominator under the sqrt so you have ((9x^(4/3)+1)/(9x^(4/3)) as the radicand.

Then when you take the square root of the fraction you get a 1/(3x^(2/3)) to come out. The 1/3 can come out of the integral, and x / x^(2/3) is x^(1/3).

Dunadan019

1 = (9x^(4/3))/(9x^(4/3))

so the inside expression is (9x^(4/3))/(9x^(4/3) + 1/(9x^(4/3))

pull out 1/(9x^(4/3)) and take the square root of it. it will give you 1/(3x^(2/3) which can have the 3 brought outside the integral and x/x^(2/3) = x^1/3

Jimmy King

Ah hah. Thanks guys. Stupid algebra.