Find the the equation of the surface revolution obtained by rotating the following curve around y axis.
$$ x^{3/2}+y^{3/2}=1 $$
I tried to form the f(y) from the equation above and form the circle equation but according to the graphing softwares my answer seems to be wrong.
Any suggestion on how to find the final equation?
Well, when you rotate a function around the y-axis you get:
$$\mathcal{A}\left(\text{a},\text{b}\right):=\int_\text{a}^\text{b}x\cdot\sqrt{1+\left(\text{y}'\left(x\right)\right)^2}\space\text{d}x\tag1$$
So, in your case we get (using $\text{y}\left(x\right)=\left(1-x^\frac{3}{2}\right)^\frac{2}{3}$):
$$\mathcal{A}\left(\text{a},\text{b}\right)=\int_\text{a}^\text{b}x\cdot\sqrt{1+\left(\frac{\text{d}}{\text{d}x}\left(\left(1-x^\frac{3}{2}\right)^\frac{2}{3}\right)\right)^2}\space\text{d}x=\int_\text{a}^\text{b}x\cdot\sqrt{1+\left(-\frac{\sqrt{x}}{\sqrt[3]{1-x^\frac{3}{2}}}\right)^2}\space\text{d}x=$$ $$\int_\text{a}^\text{b}x\cdot\sqrt{1+\frac{x}{\left(1-x^\frac{3}{2}\right)^\frac{2}{3}}}\space\text{d}x\tag2$$