Volume- washer method of 1/(1+x^2)

[sam2557]

Much help would be appreciated here. The equation is 1/(1+x^2). It is bounded by y=0, x=0, x=2, and is rotating about x=2. I know solving the volume would be much easier through cylindrical shells. The equation through shells would be the integration of 2pi(2-x)(1/1+x^2) from 0 to 2. When I calculated the volume through this method my answer was near 8.5. However, when I solved this through the washer method, my answer was no where close. I would appreciate it if someone can help me in solving this equation through the washer method.

 

[craig16]

You could try the washer method but it wouldn't really make sense because if you look at the graph, taking horizontal slices would mean that you would have to change your limits of integration and solve for two integrals. One from y=0 to y=1/5 and the radius would just be constant, and the other from y=1/5 to y=1 with radius being something like
x= sqrt( (1/y) - 1 )

 

[Mark44]

Much help would be appreciated here. The equation is 1/(1+x^2).

An equation would be y = 1/(1 + x^2). IOW, two expressions connected with =.

It is bounded by y=0, x=0, x=2, and is rotating about x=2. I know solving the volume would be much easier through cylindrical shells. The equation through shells would be the integration of 2pi(2-x)(1/1+x^2) from 0 to 2. When I calculated the volume through this method my answer was near 8.5. However, when I solved this through the washer method, my answer was no where close. I would appreciate it if someone can help me in solving this equation through the washer method.

Your integral for the volume using shells is correct. I can't vouch for the value you got, as I haven't calculated the integral. It's possible you made a mistake when you found the volume using disks, since there are two integrals required - one for y between 0 and 1/5, and the other for y between 1/5 and 1. At y = 1/5 the typical volume elements change from disks to washers.