<p>Can someone explain to me by analytic means how to find the volume of the solid generated by revolving the given region about the y-axis:</p>
<p>1.the region enclosed by the triangle with vertices (1,0), (2,1), and (1,1)</p>
<p>2.the region enclosed by the triangle with vertices (0,1),(1,0), and (1,1)</p>
<p>For the most part, I am confused as to what method I should use for these types of problems, washer cross sections or cylindrical shells? When do I use when over the other (also, #1 uses washer cross sections and #2 uses cylindrical shells).</p>
<p>Ok, well here is a way to think of how to do them. First draw the region, then draw the reflected region. Decide either if you want a width of dx or dy. Each problem can be done both ways, although sometimes you cant integrate it one way.
Pick say dx, and now say I can only put vertical lines, and I need to have many lines next to eachother to encompass the solid. I need to use cylindrical shells.
If you pick dy, you would say I am basically stacking disks on top of eachother of different radii, and so I need to use the methods of disks and put everything in terms of y.
Make sure that you draw the picture. And do like this distance is r and this is x, so the difference is like r-x. That helps a lot, and makes tricky problems very simple.</p>
<p>Ok to use methods of cylindrical shells, each shell is of a width dx. That means that everything has to be in terms of x. So you x points of integration will be x=0 to x=1. The r is just x. The height is the slant line of the triangle minus 1. Because the lower bound is the line y=1. The slope is like x+1. So I think the height is one. so int(2pi * x * (X+1 - 1) dx) from 0 to 1.
The other way, will have to be pi r^2 dy. So then each term will have to be in terms of y. The y integration points are y=1 to y=2. It will be an outer radius^2 - inner radius^2 dy. THe outer r = 2. The inner radius = x. However you cant integrate the x with dy. So y=x+1. So x = y-1.
int(pi * [(2)^2 - (y-1)^2] dy) from y = 1 to y=2.</p>