Find the volume for a solid of revolution using the shell method.

Question

Find the volume of the solid obtained by rotating the region bound by $y = x^2$ and $y^2 = x$ about the line $y = -1$.

The shell method is preferred but the disks/washers method would also be helpful.

Answer 1

The curves intersect at $(0,0)$ and $(1,1)$. For $0 \leq x \leq 1, \sqrt{x} \geq x^2.$

For the shell method, we add up cylindrical shell volumes: $dy\cdot 2\pi r h$. So

$$V = 2 \pi \int_0^1 dy (1 + y)(\sqrt{y} - y^2).$$

For the washer method, we add up washer volumes: $dx \cdot \pi (r_{out}^2 - r_{in}^2)$. So:

$$V = \pi \int_0^1 dx [(1 + \sqrt{x})^2 - (1 + x^2)^2].$$

For both I get

$$V = \frac{29 \pi}{30}.$$