I'm having trouble visualizing the volume that we need to calculate. The question is:
Find the volume of revolution of the shape created by $y=x^2$ and $y^2=x$ revolving around the $x$ axis.
The area defined by these curves is pretty strange. I drew them, obviously $y^2=x$ and $y=x^2$ intersect at $(1,1)$ but I fail to see how $y=-\sqrt{x}$ adds anything to the shape. I'm having trouble visualizing what exactly needs to be calculated and I'd appreciate any input. Maybe there is something wrong with the question? This is very basic introductory to calculus and should not be difficult.
Also, they didn't study improper integrals yet, and the fact we didn't get any bounds on $x$ kind of implies to me that there is something wrong, since this seems to demand an improper integral
The area looks kind of like a thin leaf in the first quadrant.
The graph of $y = \sqrt{x}$ is greater than $y = x^2$ in the first quadrant on the interval $[0,1]$.
So take your differential volume to be the ring of thickness $dx$ with outer radius $\sqrt{x}$ and inner radius $x^2$. Then your differential area is $\pi(\sqrt{x}^2 - (x^2)^2) \mathrm{dx}$.
Integrate from $x=0$ to $x=1$ and you're done.