See attachment--(actually I'd reccomend shel method)
The region in the first quadrant bounded by the graphs of and is revolved about the y-axis. Find the volume of the resulting solid.
So basically it is where is the outer Radius and is the inner radius. Is this correct thus far? My question is how do I depict from the graph what the outer and inner radius is? And how do I determine the limits? If I can figure out these aspects I should be able to take the integral from there.
With all Volumes of revolutioni is always easiest to put in a representative rectangle and actually create a washer or disk-- later when you learn the shell method th same thing.
If you revolve fns of x about a vertical axis and use disks/washers you have to invert and work wrt y
If you revolve fns of x about a horizontal axis axis and use disks/washers you work wrt x
With the shell method this will be just the opposite
Alright I was wondering just that as I have a shell method problem next and its revolved about the y axis so I just work in terms of x then? And also do you always work in the first quadrant or can it be in another because in this problem it would seem i have to work in the 4th quadrant. Sorry for posting this here, thought it was somewhat relevant in method wise.
Certainly there is no restriction on the region or the axis of rotation.
I would have to see the exact problem.
As an example you could rotate sin(x) 0 < x < 2pi about the
line say x= -1
Then you would have (shell method)
2pi integral[(x+1)sin(x)dx] fro 0 to 2pi
I'll be signing off now so I won't be available to reply further for a while
well the problem is finding the volume of the solid generated by revolving 2x-y=12 , x-2y=3 , x=4 about the y-axis. But it would seem I would have to use the 4th quadrant for this but the graph generated looks strange to me. You probably won't see this as you're signing off, I'll look into solving it more in a bit.