Difficult simplification for Arc length integral

[nicnicman]

Homework Statement

Find the length of the curve
x = 3 y^{4/3}-\frac{3}{32}y^{2/3}, \quad -64\le y\le 64

Homework Equations

Integral for arc length (L):
L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^{2}} dx

The Attempt at a Solution

Using symmetry of the interval and the above integral for arc length I got
L = 2\int_0^{64} \sqrt{1 + (4y^{\frac{1}{3}-}\frac{1}{16y^{\frac{1}{3}}})^{2}} dy

Unfortunately, I'm having trouble integrating this beast. Using Wolfram Alpha as a last resort, I found the integrand can be simplified to this:
4y^{\frac{1}{3}}+\frac{1}{16y^{\frac{1}{3}}}
However, I can't see how Worlfram made this simplification. (A step-by-step solution was not available.)

Could someone show me how the the original integrand could be simplified to this form? After that I should be fine.

 

[az_lender]

The expression 4y^1/3 - (1/16)y^-1/3 isn't dy/dx, it's dx/dy. So the differential in your expression for L should be dy rather than dx. But, OK, to answer your actual question:

The expression under the radical sign can be simplified if you square the parenthesized expression by the rules of high-school algebra. It comes out to 16y^2/3 - 1/2 + (1/256)y^-2/3, so when you add 1 to it, the -1/2 becomes a +1/2, and the square root is just the same as what was formerly in the parentheses...except now the minus is a plus.

 

[nicnicman]

Thanks for the reply.

Yeah I know the dx and dy are backwards...it's just the generic formula we were given. Sorry for any confusion.

I did expand the expression under the radical and came up with the same expansion you did, but I still could not see a good way to integrate it.

That's when I turned to Wolfram and found the simplification shown in the original post. However, I don't see how to get to that simplification.

 

[Dick]

Thanks for the reply.

Yeah I know the dx and dy are backwards...it's just the generic formula we were given. Sorry for any confusion.

I did expand the expression under the radical and came up with the same expansion you did, but I still could not see a good way to integrate it.

That's when I turned to Wolfram and found the simplification shown in the original post. However, I don't see how to get to that simplification.

Just follow az-lender's lead and factor 16y^(2/3) + 1/2 + (1/256)y^(-2/3) into a perfect square.

 

[Mark44]

Thanks for the reply.

Yeah I know the dx and dy are backwards...it's just the generic formula we were given.

It's likely that you were given three formulas: one where y is a function of x, another where x is a function of y, and a third where x and y are functions of a parameter such as t. It's important to know which one to apply to a given situation - you can't just mix and match things.

Sorry for any confusion.

I did expand the expression under the radical and came up with the same expansion you did, but I still could not see a good way to integrate it.

That's when I turned to Wolfram and found the simplification shown in the original post. However, I don't see how to get to that simplification.

 

[nicnicman]

Okay thanks for the help everyone