Volume of revolution of a region bounded by the $x$- axis

Question

So I have this problem which says:

''The plane region bounded above by the curves $x=y^2$ and $x+y=2$ and below by the $x$-axis is rotated about the line $y=2$. The volume of the obtained solid of revolution is...''

I'm a little confused as to how to set up the region. I sketched the region but I am not really sure how to do it. The function definition changes at $1$ so I don't think the method of rings is appropriate here. So I then tried to use the shell method with:

$$A(y)=2\pi(y-2)(2-y-y^2)$$ which I thought was right but if I integrate that, I get 0 which I don't think is right. That doesn't make sense.

Could anyone tell me where I went wrong and help me set up the integral? Thanks a lot...

Answer 1

Your strategy is right.

The correct integral is

$$\int_{0}^1 2\pi(2-y)(2-y-y^2)\, dy$$

which evaluates to $\displaystyle{\frac{23\pi}{6}}$.

Answer 2

The shaded area is the region that we are looking for.

If we rotate the shaded region about the line $y=2$, then using the Shell Method, we get $$V=\int_{0}^1 2\pi(2-y)\cdot \big[(2-y)-y^2\big]dy=\frac{23\pi}{6}.$$