Calculus Volume of Revolution

Question

I have to find the volume of revolution of the region bounded by $y=1-x^2$ and $y=-1+x^2$ and $x=-2$ and $x=2$ with respect to the $y$ axis. I have manipulated the equation so it is $y=1+\sqrt{1-x}$ and $y=2x-x^2$ and that I have to solve for $x$ and I think use cylindrical shells. However, I am not sure what to do beyond this.

Answer 1

Algebraic manipulation is not needed, but a picture definitely is, as well as some interpretation of what is intended.

Out of reflex, I will use the symmetry, and just rotate the top half of the region, and then multiply by $2$. Draw a thin strip of width "$dx$" at $x\ge 0$, and rotate it.

The cylindrical shell we get is a little different from $x=0$ to $x=1$ than from $x=1$ to $x=2$. For from $0$ to $1$ the "top" curve is $y=1-x^2$, while later it is $y=x^2-1$. So we split the integration at $1$, and get volume $$2\int_0^1 2\pi x(1-x^2)\,dx + 2\int_1^2 2\pi x(x^2-1)\,dx .$$

Remark: We can also do it by slicing (washers), integrating with respect to $y$. Then expressing $x$ in terms of $y$ is useful. The calculation is not too bad, but definitely more complicated than with the shells approach.