Find the surface area of this function when rotated around the y axis. Stuck Say I have this equation:
$$y = \frac{1}{3} x^{\frac{3}{2}}$$ and the limits of the integration provided are $0 \leq x \leq 12$
So since we're rotating around the y axis, I'm going to change the limits of integration to use y.
$$0 \leq y \leq 8 \sqrt{3}$$
Also, arc length is:
$$\sqrt{1 + \frac{dy}{dx}^2}$$
and $\frac{dy}{dx} = \frac{1}{2} x^{\frac{1}{2}}$ and so $\frac{dy^2}{dx} = \frac{1}{4}x$
So I have this SA equation:
$$SA = 2 \pi \int_0^{8 \sqrt{3}} x \sqrt{1 + \frac{1}{4}x}dx$$
$$ = 2 \pi \int_0^{8 \sqrt{3}} x \sqrt{\frac{4 + x}{4}x}dx $$
taking out the $\sqrt{4}$ from the denominator:
$$ = \pi \int_0^{8 \sqrt{3}} x \sqrt{4 + x} $$
Then I get $u = x + 4$ and $ x = u - 4$
(this is odd... I guess this is another way to usub?)
new limits of integration are:
when $x = 0$, $u = 0$
when $x = 8 \sqrt{3}$, $u = 8 \sqrt{3} + 4$
$$ = \pi \int_0^{8 \sqrt{3} + 4} \sqrt{u} (u -4) du$$
$$ = \pi \int_0^{8 \sqrt{3} + 4} u^{\frac{3}{2}} - 4 \sqrt{u} du$$
$$ = \pi * \frac{2}{5}u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}} \biggr]_0^{8 \sqrt{3} + 4}$$
This comes out to be pretty ugly... did I make a mistake?

Yes, you did make a mistake in the very beginning, when you kinda decided to switch to integration with respect to $y$ … but you didn't actually switch. Note that your integral
$$2\pi\int_0^{8\sqrt{3}}x\sqrt{1+\frac{1}{4}x}\,dx$$
is with respect to $x$, but you're using the $y$-values from $0\le y\le8\sqrt{3}$ as your limits of integration. That doesn't make sense — effectively you're saying that $x$ ranges from $0$ to $8\sqrt{3}$, which isn't true.
Once you decided that

I'm going to change the limits of integration to use $y$,

you should actually do that. From the given equation you express $x$ as a function of $y$, $x=\ldots\text{something in terms of }y\ldots$, and then the arclength will be $ds=\sqrt{1+\left(\frac{dx}{dy}\right)^2}$, which you will use for setting up an integral with respect to $dy$.
