Help! perplexing problem on volume revolution of region about a line how do we find the volume when the region bounded by $y = x^{\frac 1 2}$ and $y = \frac x 2$ is revolved about the line $y = 1$? I understand how to solve for the regions above and below $y = 1$, however because of the intersection of volumes from $x = 1$ to $x = 2$, the volume of the entire region both above and below $y = 1$ eludes me.
heres the graphs: https://www.desmos.com/calculator/njvae9suit
I would greatly appreciate an explained solution to solving a problem such as this. Thank you!
 A: You should do something like
$$\pi \left[\int_0^1 \left(\left(\frac{x}{2}-1\right)^2-\left(\sqrt{x}-1\right)^2\right) \, dx+ \int_1^4 \left(\left(\sqrt{x}-1\right)^2-\left(\frac{x}{2}-1\right)^2\right) \, dx\right]$$
A: Using your sketch, the point to note is that for $1 \leq x \leq 2$, one of the curves is below $y = 1$ and one of the curves is above $y=1$.
The radius of revolution will be defined by which vertical distance is more from $y = 1$. As you can see from the sketch that for $x = 1$, the distance to $y = \frac{x}{2}$ is more whereas for $x = 2$, distance to $y = \sqrt{x}$ is more. Say that the distance  from $y = 1$ to both curves equal at $x = a$ then the volume of revolution can be written as
$V = \displaystyle \int_0^1 \int_{1-\sqrt x}^{1-\frac{x}{2}} 2 \pi r \ dr \ dx + 
\int_1^a \int_{0}^{1-\frac{x}{2}} 2 \pi r \ dr \ dx + \int_a^2 \int_{0}^{\sqrt x - 1} 2 \pi r \ dr \ dx + \int_2^4 \int_{\frac{x}{2} - 1}^{\sqrt x - 1} 2 \pi r \ dr \ dx$
Now to find $a$, equate $1 - \frac{a}{2} = \sqrt a - 1$. Solving it we get $a = 6 - 2 \sqrt5$.
