integral with disks We are looking for the volume of the space between $y=\frac{-1}{2}x+\frac{3}{2}$ and $y=x$ as rotated about the x-axis. I used the disk/washer method ($\pi r^2$) and broke it up into 3 pieces. I notice that there is a bit of overlap under the x-axis, so I have excluded it in my computation. Please see the attached picture for a more detailed view. image
$\pi (\int_{-2}^0 [(-1/2)x+(3/2)]^2)dx+\pi(\int_0^1[(-1/2)x+(3/2)]^2-[x]^2)dx+\pi\int_1^3(x^2-[(-1/2)x+(3/2)]^2-[x]^2)dx=$
$\frac{-\pi}{4}\int_0^{-2}(x^2-6x+9)dx+\frac{\pi}{4}\int_0^1(x^2-6x+9)dx+\frac{\pi}{4}\int_1^3(3x^2+6x-x)dx=$
$\frac{\pi}{4}[(-\frac{x^3}{3}+3x^2-9x)_0^{-2}+(-x^3-3x^2+9x)_0^1+(x^3+3x^2-9x)_1^3]=$
$\frac{\pi}{4}[8/3+12+18-1+3+9+27+27-27-1-3+9]=$
$\frac{\pi}{4}[8/3+73]=$
$\frac{\pi}{4}(227/3)=$
$59.42846...$

You split up the integral into three parts and these parts should be
$$ \pi \int_{-2}^{0}\left(\frac{-x}{2}+\frac 3 2\right)^2dx$$
$$\pi \int_{0}^1\left(\frac{-x}{2}+\frac 3 2\right)^2 - x^2dx$$
$$\pi \int_{1}^3 x^2 - \left(\frac{-x}{2}+\frac 3 2\right)^2dx$$
Notice that your third term has an extra $-x^2$ inside the integral which shouldn't be there, I guess that's a typo.
The way you calculate the first term is quite odd (why switch the bounds of the integral ?) but looks mostly fine.
The main problem is the computation of the second term since you drop the $-x^2$ inside the integral instead of what you wrote you should have
\begin{align}
\pi \int_{0}^1\left(\frac{-x}{2}+\frac 3 2\right)^2 - x^2dx &= \frac \pi 4 \int_{0}^1(-x + 3)^2 - 4x^2dx \\
&=\frac \pi 4 \int_{0}^1x^2 - 6x + 9 - 4x^2dx\\ 
&= \frac \pi 4 \int_{0}^1 -3x^2 - 6x + 9dx
\end{align}
For the third term you shouldn't obtain $3x^2+6x-x$ inside the integral rather you should have $3x^2 + 6x - 9$.
