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
Best 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 $g(y)$ ?