Volume of solid of revolution about x-axis via Disks

Question

Good morning, I am attempting to solve the following problem from the back of my book, but am struggling with the correct approach to setup as I cannot find Youtube videos with examples of how to work with curves involving absolute values.

The volume of the region bounded by $y = |x − 3|$ and $y = 3$ revolved about the $x$-axis.

What I have so far:

Attempting to approach this via disk method: $\pi$ * [integral from b to a of $y_1^2 - y_2^2]$

The absolute value however throws me off. Does this mean I need to consider both $x+3$ and $x-3$?

Any guidance on how to setup this integral would be greatly appreciated.

Regards, J

Answer 1

The region revolved looks like this:

 ^y
 |
3#######
 |#####
 | ###
 o--#--->x
    3

Consider each side of the absolute value separately. For this region's left half with $0\le x\le3$, the lower bound is $y=3-x$ and the upper bound is $y=3$. For the right half with $3\le x\le6$, the lower bound changes to $y=x-3$.

The integrals can then be set up for the disc method. I show the left half's integral below; the right half is for you to complete, as well as the evaluation. $$\pi\int_0^3(3^2-(3-x)^2)\,dx$$