So you have:
x=1+ (y-2)^2 and you are rotating around x=2
For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2
The answer is 16/3 but I keep getting 32/3 pi. What am i doing wrong? It also has to be done by shels
So you have:
x=1+ (y-2)^2 and you are rotating around x=2
For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2
The answer is 16/3 but I keep getting 32/3 pi. What am i doing wrong? It also has to be done by shels
"For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2"
No, it is not. You are rotating around y= 2, not the x-axis. And it makes no sense to integrate a function of y from x= 1 to x= 3. Nor does it make sense to integrate from y= 1 when you are rotating around y= 2. You must mean to integrate from y= 2 to y= 3.
You don't give a right boundary but x= 2 is implied. For that I get neither 16/3 nor (32/3)pi. Doing it by both by shells and by disks, I get a volume of pi/4.
The answer cannot be 16/3, since the volume must be less than the volume of a sphere with radius 1 unit.
The radius of each cylinder is
The height of each cylinder is
Surface area of each cylinder is
Therefore the volume of revolution is
which comes out at
The volume of revolution lies between the volumes of the purple spheres in the 2nd attachment.