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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.

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

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I like your hints! +1 –  amWhy Feb 1 '13 at 1:21
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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.

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