Volume of solid revolution against a line

Question

If an Area is bounded by two curves $y=\sqrt{x}$ and $y=\frac{x}{2}$, and it is revolved wrt to the line $x=-1$, then the Volume can be calculated using Ring method as this way? : $$ V = \pi \int_{0}^{2} (2y+1)^{2} - (y^{2}+1)^{2} dy $$ By inverting the two bounding curves first. The key is knowing the radius and the orientation of the Riemann rectangles, is this correct..?

Answer 1

Yes it is correct, as an alternative by shell method

$$V = 2\pi \int_{0}^{4} (x+1)(\sqrt x-x/2) dx=2\pi\left[\frac25x^{\frac52}-\frac{x^3}3+\frac23x^{\frac32}-\frac{x^2}4\right]_0^4=\frac{104\pi}5$$