calculus - Finding the surface area of a solid of revolution

I'm given the function

$x=\frac{1}{15}(y^2+10)^{3/2}$

and I need to find the area of the solid of revolution obtained by rotating the function from $y=2$ to $y=4$ about the $x-axis$.

I've tried applying the formula:

$2\pi\int_{a}^{b}f(x)\sqrt{1+(f'(x))^2}dx$

where $f(x)=\sqrt{(15x)^{2/3}-10}$ and

$f'(x)=\frac{15^{2/3}}{3x^{1/3}\sqrt{{(15x)}^{2/3}-10}}$

but it still isn't the correct answer. Is there any way to do the problem without having to find $f(x)$ but by just working with $f(y)$?

Thanks for the help

Answer

You calculated wrongly $f(x)$: the exact value is
$$f(x)=\sqrt{(15x)^{3/2}-10}$$
But why not use the formula
$$A=2\pi\int_2^4y\sqrt{g'^{\,2}(y)+1}\,\mathrm dy$$
since $x$ is given as a function $g(y)$ ?