You're integrating w.r.t. y, so express x in terms of y:
x = 1 - y
At any particular y value (with y<0 in the blue shaded triangle), the width of the shaded rectangle is 3 - x = 3 - (1 - y) = 2 + y and the height is |dy|. The radius (distance from the axis of revolution) is |y|, so the element of volume swept out by revolving that rectangle about the x-axis is:
dV = 2πrh dr = 2π |y| (2 + y) |dy|
The absolute value bars are in the way. For y<0, you can replace |y| with -y and by integrating in the positive direction (from y=-2 to 0) you make dy positive and can drop the bars:
V = 2π ∫ (-y)(2 + y) dy = -2π ∫ (y² + 2y) dy . . . . integral from y=-2 to 0
You can handle that integral, right? That shows working the problem "in place". Honestly, I'd probably have flipped the thing about the x-axis, with y = x - 1 as the line and moving the "blue triangle" to the 1st quadrant. I make fewer mistakes without so many minus signs in the problem.