I'm working on a problem of finding volume between two functions using the shell method. The functions given are $f(x) = 2x - x²$ and $g(x) = x$. It is reflected across the x-axis.
I solved this previously using washers but the problem asks to solve using two methods. I believe this is a dy problem, thus I am trying to convert the two functions into $f(y)$ and g(y), but I don't understand how to convert $f(x)$ into an $f(y)$. It doesn't seem like it can be solved in terms of y. Maybe I am missing something after looking at this too long.
How would I solve this problem using shells? I am approaching this correctly?
Since the function describes a "vertical parabola", integration in the $ \ y-$ direction will involve using the "right" and "left" halves on this parabola, which will become the "upper" and "lower" halves when viewed from the $ \ y-$ axis. Solving for $ \ y \ $ gives us $ \ x = 1 \pm \sqrt{1-y} \ . $ [The "split" occurs because the parabola represents a single function of $ \ x \ , $ but not of $ \ y \ . $ ]
The parabola and the line meets at the origin and at $ \ (1,1) \ . \ $ So integration along the $ \ y-$ axis is going to require two integrals, one between the line $ \ x = y \ $ and the "lower arm" of the parabola, $ \ 1 \ - \ \sqrt{1-y} \ \ $ on the interval $ \ 0 \ \le \ y \ \le \ 1 \ , $ and the second, between the "upper arm" of the parabola, $ \ 1 \ + \ \sqrt{1-y} \ \ $ and the lower arm on the interval $ \ 1 \ \le \ y \ \le \ 2 \ . $
In ordinary practice, we would just pick the method that gives us one integral we can actually evaluate (when the situation permits that). But we make students do these volume calculations both ways to see how they work (and that they give the same result), and because sometimes the more complicated method is the only way we can calculate the integral(s).