calculusintegrationvolume
A region $R$ is bounded above by the graph of $x^\frac{2}{3}+y^\frac{2}{3}=4$ and below by the x-axis. Find the volume of the region. Rotating region $R$ about the vertical line $x=8$ generates a solid of revolution $S$. I am confused with the picture below. Why is the radius considered to be $8-x$ shouldn't it be $x-8$?
You would only "use the absolute value of 1 plus the x" if the region of rotation was to the right of the vertical line of revolution AND the left boundary of the region of rotation were to the left of the y-axis. Perhaps most of the problems you did in the past were this case.
There are rules for these types of problems. If you are rotating about a vertical line, and the axis of rotation is the left boundary, you use $x-h$. But if you're rotating about the right boundary, you use $h-x$.
As for your problem, it is usually better to draw a picture. I don't know anyone who doesn't at least visualize these types of find-the-volume-of-rotation problems.
Since $1\le 1< 2$, use $\int_1^2 (x-1)f(x)dx$. You get
$$2\pi\int_1^2(x-1)x^2dx$$
However if you were to rotate it around the line h=2, not shown, you get
$$2\pi\int_1^2(2-x)x^2dx$$
Note that you would get a larger volume for the first rotation (h=1) because the taller part of it has to rotate more, covering more volume, but for the second rotation (h=2), the larger part is in the "center" and creates less volume.
Best Answer
The radius ought to be shown as $|x-8|$ or $|8-x|$ as a radius is a length quantity which cannot be negative, thus this would account for if $x<8$ or $x>8$ and give a positive value in either case.