Find volume of revolution Find the volume:
$$x=y^2; x=1-y^2; \text{ rotated around }x=3$$
Use the disk/washer method.
My inner radius is $3+y^2$ and outer radius is $3-y^2$ and the limits of integration are $\frac1{\sqrt2}$ and $\frac{-1}{\sqrt2}$
My final answer is $14\pi\frac{\sqrt2}3$
The answer is supposed to be $10\pi\frac{\sqrt2}3$

The inner radius is $3-(1-y^2)$, that is, $2+y^2$. By symmetry the volume is
$$2\int_{y=0}^{1/\sqrt{2}}\pi\left((3-y^2)^2-(2+y^2)^2\right)\,dy.$$
Expand and integrate.
Remark: The inner radius of $3+y^2$ used in the OP does not yield volume $14\pi\frac{\sqrt{2}}{3}$. 
