Volume of revolution generated by revolving region How can I find the volume of the solid when revolving the region bounded by $y=1-\frac{1}{2}x$, $y=0$, and $x=0$ about the line $ x=-2$? How could I set it up? 

Would it be $x=2-2y$
so radius $r(y) = 2-2y -(-2) $ => $r(y)= 4-2y$ 

$π\int (4-2y)^2 dy$ ?

What would be my limits of integration? Would it be from 0 to 2?
$\textbf{1)}$ Using the disc method, you have $R(y)=(2-2y)-(-2)=4-2y$ and $r(y)=0-(-2)=2$, 

so $V\displaystyle=\int_0^1\pi\big((4-2y)^2-2^2\big)dy$.

$\textbf{2)}$ Using the shell method, you have $r(x)=x-(-2)$ and $h(x)=(1-\frac{1}{2}x)-0$,

so $V\displaystyle=\int_0^22\pi(x+2)\big(1-\frac{1}{2}x\big) dx$.
First thing you should do is draw a picture of the volume. It looks like a volcano. Then you should decide the shape of the slice you want to use. Either a shell shape or washer shape would work, but not a disk. 

For a shell shape, you are integrating dx, and for a washer shape, you are integrating dy. Decide which one you want to use. For a shell, you have to find height and radius. They should be functions of x, no y's in there. For a washer, you have to find inner radius and outer radius. They should be functions of y, no x's. 

After you have the dimensions right, THEN you can use your volume formula and integrate.