The volume of the solid, generated by revolving about $y = 2$ the region bounded by $y^2\leq 2x, x \leq 8$ and $y \geq 2$, is
(A) $2\sqrt2\pi$ (B) $28\pi /3$ (C) $84\pi$ (D) none of these.
My Steps:
Total Volume of solid formed by the revolution of the area
$\begin{align} A & =\displaystyle\int\int\limits_A 2\pi y\;dy\;dx\\ & = \displaystyle\int\limits_{x=2}^{x=8}\int\limits_{y=2}^{y=\sqrt{2x}} 2\pi y\;dy\;dx \\ & = \displaystyle\int\limits_{x=2}^{x=8} 2\pi \frac{y^2}{2}|_{y=2}^{y=\sqrt{2x}}\;dx \\ & = \displaystyle\int\limits_{x=2}^{x=8} \pi y^2|_{y=2}^{y=\sqrt{2x}}\;dx \\ & = \displaystyle\int\limits_{x=2}^{x=8} \pi (2x-4) \;dx \\ & = 2\pi \frac{x^2}{2}|_{x=2}^{x=8}-4\pi x|_{x=2}^{x=8} \;dx \\ & = 36\pi \end{align}$
Is this correct ? or did I make some mistake here ? Please help me confirm my solution.
The setup should be a single integral (not a double one):
$$\pi\int_2^8 (\sqrt{2x}-2)^2 dx$$
See the following picture:
$A,B,C$ enclosed the region you want to rotate. Then we find the volume of the cone-shaped object. This has nothing to do with dividing by $2$.