Volume of a Parabola Rotated Around a Line The question is as follows: Find the volume of the solid generated by revolving the region bounded by the parabola $y = x^2$ and the line $y = 1$ about the line $y = -1$.
My attempt: Using the washer method, I set the outer radius to $1 + x^2$ and the inner radius to be $1$. This gives me the integral:
$V = \pi\int_{-1}^1(1+x^2)^2-1^2dx = \pi\int_{-1}^12x^2+x^4dx$, which when evaluated gives the answer of $26\pi/15$. This however, is wrong. The correct answer is $64\pi/15$.
I put it through Wolfram Alpha and it gave me the same answer: https://www.wolframalpha.com/input/?i=rotate+the+region+between+0+and+x%5E2+with+-1%3Cx%3C1+around+the+line+y+%3D+-1
Any help would be appreciated.
EDIT: I tried using the shell method and got the correct answer, using $y+1$ as the radius and $\sqrt y$ as the height. So now the question is how would you do this using the washer method.
 A: *

*Your wolfram alpha link has the wrong bounds, using $y=0$ instead of $y=1$ as one of the bounds. Even with the correct bounds, wolfram|alpha seems to be screwing up the integral and missing the central "missing" piece.

*Using the washer method, the outer radius is $R_{outer} = 1 - (-1) = 2$ and the inner radius is $R_{inner} = x^2 - (-1) = x^2 + 1$. Thus
\begin{align*}
 V &= \int_{-1}^1 \pi(  R_{outer}^2 - R_{inner}^2) ~\mathrm{d} x \\
 &= \pi \int_{-1}^1 2^2 - (x^2+1)^2 ~\mathrm{d}x
\end{align*}
which evaluates to the correct result. 

A: In the image you can see that the outer radius is $2$, not $1$ and the inner radius is $1+x^2$. Furthermore, you were subtracting in the wrong order.
You may avoid a bit of arithmetic by taking advantage of the symmetry by doubling the integral from $0\le x\le1$ rather than integrating from $-1$ to $1$.
As a double-check you can find the volume by the cylindrical shell method.
\begin{eqnarray}
V&=&\int_0^1 2\pi rh\,dy\text{ where }r=1+y\text{ and }h=2x=2\sqrt{y}\\
&=&4\pi\int_0^1y^{1/2}+y^{3/2}dy\\
&=&4\pi\left[\frac{2}{3}y^{3/2}+\frac{2}{5}y^{5/2}\right]_0^1\\
&=&\frac{64\pi}{15}
\end{eqnarray}

A: 
Another way is to find center of mass for parabola by integration
$$\bar y= \dfrac{\int_{-1}^{1}  ydx}{\int_{-1}^{1}  dx}= \dfrac23 $$
Parabola area is also two-thirds surrounding box ares.
Then use Pappu's thm to find area of raised parabola swept out by rotation:
$$ 2 \pi(1+\frac23)\cdot \frac23 \cdot(2\cdot 1)= \frac{40\pi}{9}.$$
