I know that I can use either washers method or shells method. I have a few questions which I have googled but I did not receive much luck.
- How do I decide in which case which method is easier and what are the requirements of the method?
- Why is the radius in shells method "x" when rotating about the y-axis? In other words I do not see how the radius, "x", represents a distance.
- When I am rotating about another line that is negative how do I find the radius? If it the line is a positive number?
I am new here but I have looked around and I am still so confused on this.
$\endgroup$2 Answers
$\begingroup$(1) You use whichever is simpler. I can't give a rule, you just need to do a bunch of them and get a feel. If you are rotating around $y$ for washers you are integrating $x(y)dy$ and for shells you are integrating $y(x)dx$. It depends on the function you are given which is simpler.
(2) The element you are integrating in the shell method is a cylinder around $y$. The axis of the cylinder is the $y$ axis. The volume element is a shell from $x$ to $x+dx$ of height $y$. Think of a soda can centered on the $y$ axis. The radius of the can is $x$.
(3) When you are rotating around any other line you need to find the distance from that line to use as the radius. If the line is parallel to one of the axes you can just define a new set of axes that are translated from the original ones. Define $u=y-a$ where $a$ is the coordinate of the center of rotation. Now the rotation is around $u=0$ and the techniques you are used to will work if you write the equations in terms of $u$.
As an example, let $f(x)=|x|$.
If you want to find the volume of the cone obtained by rotating the region above $f(x)$ and below $y=1$, you would use cylindrical shells because the volume is contained inside one region.
If you want to find the volume of the shape obtained when rotating the region bound by $f(x)$, $y=1$, and $x=2$ about the $y$-axis, then you would use the washer method since the shape you get after rotating has a "hole" in it.
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