Find the volume of solid generated by revolving the described region about $y$ axis

The region enclosed by the line $y =x$, $y =-x/2$ and $x =2$

So, this is how I setup the integral

Volume = $\displaystyle 2\pi \int_{0}^{2}(2-x) (3x/2) dx$

When I calculate this the answer comes out to be $4\pi$

However, the answer given to me is $8\pi$

Can anyone please check and tell me what is wrong with my solution ?

Thank you.

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If you draw the figure in the $xy$ plane, you have a triangle with a vertex at the origin. Then, using cylindrical shells with radius $x$ and height $3x/2$, your integral is $$V=2\pi\int_0^2x\frac32 x dx=8\pi$$

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