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

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 ${x}^{2}-\left(2-{x}^{2}\right)$
Finding the Integral, ${\int }_{\mathrm{¬}1}^{1}2\pi \left(2-x\right)\left({x}^{2}-\left(2-{x}^{2}\right)\right)dx=2\pi \left(\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6}\right)+C$
Finding the limits, $\underset{x\to -1+}{lim}2\pi \left(\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6}\right)=2\pi \left(\frac{4\left(-1{\right)}^{3}}{3}-4\left(-1\right)-\frac{2\left(-1{\right)}^{4}}{3}+2\left(-1{\right)}^{2}-\left(-1{\right)}^{2}+\frac{\left(-1{\right)}^{4}}{6}\right)=19.897$
$\underset{x\to 1-}{lim}2\pi \left(\frac{4{x}^{3}}{3}-4x-\frac{2{x}^{4}}{3}+2{x}^{2}-{x}^{2}+\frac{{x}^{4}}{6}\right)=2\pi \left(\frac{4\left(1{\right)}^{3}}{3}-4\left(1\right)-\frac{2\left(1{\right)}^{4}}{3}+2\left(1{\right)}^{2}-\left(1{\right)}^{2}+\frac{\left(1{\right)}^{4}}{6}\right)=-13.614$
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.
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Explanation:
In the region $-1\le x\le 1$, $2-{x}^{2}>{x}^{2}$ (the $2-{x}^{2}$ curve is above the ${x}^{2}$ curve). So the height is $\left(2-{x}^{2}\right)-{x}^{2}$, not ${x}^{2}-\left(2-{x}^{2}\right)$.
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Step 1
The expression of the volume should be
${\int }_{-1}^{1}2\pi \left(2-x\right)\left(\left(2-{x}^{2}\right)-{x}^{2}\right)dx$
Step 2
Because, $2-{x}^{2}\ge {x}^{2}\phantom{\rule{1em}{0ex}}-1\le x\le 1$