$y = x^2$ and $y^2=x$ rotated about the line $y=1$. There is something fundamental that I am overlooking. I am able to solve if rotated about x or y axis but when its about another line I am lost.
I am not sure why my answer is wrong I have tried several times:
$x^2=\sqrt{x} \to x^4=x \to x(x^3-1)\to x=0, x=1$
given $x=.5, \sqrt{x} \gt x^2$
thus
$$\pi \int_0^1 (\sqrt{x})^2 - (x^2)^2$$
$$pi \int_0^1 x-x^4 \to \frac{1}{2}x^2 - \frac{1}{5}x^5 \Big\vert_0^1 = \frac{5\pi}{10}- \frac{2\pi}{10} = \frac{3\pi}{10}$$
But the answers is $\frac{11\pi}{30}$
If you feel more confortable working with rotations around the lines $x=0$ and $y=0$, then translate those functions. The volume of the region that you are interested in is equal to the volume of the rotation of the graphs of $y=x^2-1$ and $y=\sqrt x-1$ around the line $y=0$. So, it is equal to$$\pi\int_0^1\left(x^2-1\right)^2-\left(\sqrt x-1\right)^2\,\mathrm dx,$$which is indeed equal to $\dfrac{11\pi}{30}$.