I need to find the volume of the solid obtained by rotating the region bounded by the curves $y=x, y=0, x=2$, and $x=4$ around the line $x=1$
I know I need to integrate $\pi*((\text{outer radius})^2 - (\text{inner radius})^2)$ but I can't figure out how to get the radii. Can someone please explain?
On
You can avoid integration* - the region is simple, one right-angled triangle atop a square of area 2x2=4. On sweeping the square about $x=1$, the average radius of the sweep is $(1+3)/2$ =2, giving you a volume of $2*\pi*2*4$ = $16\pi$.
*Oops, a fix: The average radius for the triangle is not 2 - The danger of using geometric arguments carelessly.
The correct radius to use is the radius to the abscissa of the centroid of the triangle $10/3$ from $x=1$, which gives $r=7/3$. The volume of the sweep is $(1/2)*2*2*7/3*2\pi$, which is $28\pi / 3$.
The total volume is $16\pi + 28\pi /3$, which gives v=$76\pi / 3$.
You'll need to do two cases: when $y \in [0, 2]$, and when $y \in [2, 4]$. In both cases, the outer border will be at $x = 4$. In the former, the inner border is at $x = 2$; in the latter, the inner border changes to $x = y$. So we obtain: $$ V = \pi\int_0^2 [(4 - 1)^2 - (2 - 1)^2] \, dy + \pi\int_2^4 [(4 - 1)^2 - (y - 1)^2] \, dy $$