Surface area of revolution about the y-axis- I'm trying to find the surface area of revolution when $$y=\frac{1}{3}x^3$$ is revolved around the y-axis. At the moment, I'm having difficulty setting up the integral.
I have: $\displaystyle 2\pi\int(3y)^\frac{1}{3}\sqrt{1+(3y)^\frac{-4}{3}} dy$
I would like to know if this looks correct and, if it is, I would like to have some idea about how to solve it as I'm finding it difficult to deal with the exponents and make a substitution.
 A: It looks basically right. You did not specify the problem fully, and the limits of integration are missing.
As to the integration, let $w=3y$. The thing inside the square root is 
$1+\dfrac{1}{w^{4/3}}$.   Rewrite this as $\dfrac{w^{4/3}+1}{w^{4/3}}$.  Take the square root. We get 
$$\frac{1}{w^{2/3}}\sqrt{w^{4/3}+1}.$$
Remembering the $w^{1/3}$ term in your integral, we want a constant times
$$\int \frac{1}{w^{1/3}} \sqrt{w^{4/3}+1}\,dw.$$
Now let $u=w^{2/3}$. We end up with a constant times
$$\int \sqrt{u^2+1}\,du.$$
This is a standard, albeit somewhat unpleasant integral.
A: Hint: Try to substituted $3y=t$ first and then use the second of the following fact:

Theorem: The integral
$$\int x^m(a+bx^n)^pdx$$
can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:
$1.$ $p$ is an integer( $p>0$ use the  Newton's binomial theorem and when $p<0$ then $x=t^k$ which $\text{lcm}(n,m)$).
$2.$ $\dfrac{m+1}{n}$ is an integer. So set $a+bx^n=t^{\alpha}$ wherein $\alpha$ is the denominator of $p$.
$3.$ $\dfrac{m+1}{n}+p$ is an integer.

Here we have $p=1/2,n=-4/3,m=1/3$
