Given two curves, find volume of the solid that results when the region enclosed by the curves is revolved about the given axis. Curves are:
$-y^2 + 2 $
$x = y$
Axis is $x = -2$.
Having found the intersection points and from the graph; a portion exists below the x axis i.e it is negative. As per the answer, while integrating, we don't take the absolute of the negative portion. Why DON'T we have to take the absolute value of the negative portion? Normally when negative, we would take the absolute value to get the volume. Thanks
We assume that you have drawn a reasonably good picture of the curves $x=-y^2+2$ and $x=y$. (The post says one of the curves is $-y^2+2$. That is not an equation, we assume $x=-y^2+2$ is intended.)
We use "slicing" with slices perpendicular to the $y$-axis. Let $A(y)$ be the area of cross-section of a slice at height $y$. Then our volume is $$\int_{y=-2}^1 A(y)\,dy.$$ We need to calculate $A(y)$.
Note that any cross-section is a circular disk (more sloppily, circle) with a circular hole.
The outer circle has radius $(-y^2+2)+2$, or if you prefer a more algebraic calculation, the outer radius is $(-y^2+2)-(-2)$. So the outer radius is $4-y^2$.
The inner radius is $y+2$.
It follows that $A(y)=\pi\left((4-y^2)^2-(y+2)^2\right)$.
Remark: We have not mentioned absolute values, and so have not answered your question. This is because of a conviction that a volume calculation must be done by paying careful attention to the geometry.