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 $\displaystyle (2-x)$
The height of each cylinder is $\displaystyle h=2|y-2|=2\sqrt{x-1}$
Surface area of each cylinder is $\displaystyle 2{\pi}rh$
Therefore the volume of revolution is $\displaystyle 2{\pi}\displaystyle\int_{1}^2rh}dx=4{\pi}\int_{1}^ 2(2-x)\sqrt{x-1}}dx$
which comes out at $\displaystyle \displaystyle\frac{16}{15}{\pi}$
The volume of revolution lies between the volumes of the purple spheres in the 2nd attachment.