Volume of a solid rotated around axis

Question

The task is finding the volume of the solid found by rotating $$ x=2y-y^2$$

From $$x=0$$

around the axis $$y=-1$$

Answer 1

In order to evaluate such volume we use the Pappus's centroid theorem.

We rotate the plane region $$A:=\{(x,y): y\in [0,2], 0\leq x\leq y(2-y)\}$$ around the line $y=-1$. By symmetry the centre of mass of $A$ is along the line $y=1$ and therefore its distance from the axis $y=-1$ is $r=1-(-1)=2$.

Hence the volume is $$V= 2\pi\cdot r\cdot|A|=\frac{16\pi}{3}$$ where $|A|$ is the area of $A$, that is $$|A|=\int_{y=0}^2 (2y-y^2) dy=\left[y^2-\frac{y^3}{3}\right]_0^2=\frac{4}{3}.$$

P.S. By using the disc formula you should evaluate $$\pi\int_{x=0}^!\left((2+\sqrt{1-x})^2-(2-\sqrt{1-x})^2\right)dx=\frac{16\pi}{3}.$$