Computing volume using shell resulted in negative value?

The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...

Use cylindrical shells to compute the volume of $y={x}^{2}$ and $y=2-{x}^{2}$, revolved about $x=2$.

${x}^{2}=2-{x}^{2}\to 2{x}^{2}=2\to \frac{2{x}^{2}}{2}=\frac{2}{2}\to {x}^{2}=1$

$x=1,x=-1$

Radius is $r=2-x$

Height is ${x}^{2}-(2-{x}^{2})$

Finding the Integral, ${\int}_{\mathrm{\neg}1}^{1}2\pi (2-x)({x}^{2}-(2-{x}^{2}))dx=2\pi (\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6})+C$

Finding the limits, $\underset{x\to -1+}{lim}2\pi (\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6})=2\pi (\frac{4(-1{)}^{3}}{3}-4(-1)-\frac{2(-1{)}^{4}}{3}+2(-1{)}^{2}-(-1{)}^{2}+\frac{(-1{)}^{4}}{6})=19.897$

$\underset{x\to 1-}{lim}2\pi (\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6})=2\pi (\frac{4(1{)}^{3}}{3}-4(1)-\frac{2(1{)}^{4}}{3}+2(1{)}^{2}-(1{)}^{2}+\frac{(1{)}^{4}}{6})=-13.614$

And finally, computing the volume from the obtained limits $V=-13.614-19.897=-33.510$.

So, if someone could let me know if (or where) I went wrong, that would be great.

The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...

Use cylindrical shells to compute the volume of $y={x}^{2}$ and $y=2-{x}^{2}$, revolved about $x=2$.

${x}^{2}=2-{x}^{2}\to 2{x}^{2}=2\to \frac{2{x}^{2}}{2}=\frac{2}{2}\to {x}^{2}=1$

$x=1,x=-1$

Radius is $r=2-x$

Height is ${x}^{2}-(2-{x}^{2})$

Finding the Integral, ${\int}_{\mathrm{\neg}1}^{1}2\pi (2-x)({x}^{2}-(2-{x}^{2}))dx=2\pi (\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6})+C$

Finding the limits, $\underset{x\to -1+}{lim}2\pi (\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6})=2\pi (\frac{4(-1{)}^{3}}{3}-4(-1)-\frac{2(-1{)}^{4}}{3}+2(-1{)}^{2}-(-1{)}^{2}+\frac{(-1{)}^{4}}{6})=19.897$

$\underset{x\to 1-}{lim}2\pi (\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6})=2\pi (\frac{4(1{)}^{3}}{3}-4(1)-\frac{2(1{)}^{4}}{3}+2(1{)}^{2}-(1{)}^{2}+\frac{(1{)}^{4}}{6})=-13.614$

And finally, computing the volume from the obtained limits $V=-13.614-19.897=-33.510$.

So, if someone could let me know if (or where) I went wrong, that would be great.