I am aware that the general formula to calculate surface area of revolution is: $$2\pi\int_a^b x \sqrt{1+\left(\frac{dy}{dx}\right)^2}dy $$
Here, we can rearrange the equation so it becomes: $x = (4-y)^{2/3}y^{3/2}$
so, $\frac{dx}{dy} = −^{−1/3}(4−^{2/3})^{1/2}$
and $\frac{dx}{dy}^2 = −^{−2/3}(4−^{2/3})$
Assuming I've done everything correct, my integral should look like:
$2\pi\int_a^b (4−^{2/3})^{3/2}y^{-2/3}$
Have I done everything correct so far? What does my final integral evaluate into?