I have been working on this problem for a while, which gives me a region R bounded by the curves of $ \ \ x = y^2 + 2$, $ \ \ y = x - 4$, and $ \ \ y = 0$.
I am to find the volume of R when it is rotated about the x-axis.
I tried both the washer and shell method for this one problem in order to solidify my understanding of this topic but instead, I raised more questions for myself because both methods produced different answers.
Washer : $$\int_2^6\pi((\sqrt{x-2})^2-(x-4)^2)\, dx = \frac{8\pi}{3}$$
Shell : $$\int_0^22\pi y((y+4)-(y^2+2))\, dy = \frac{16\pi}{3}$$
I am hopeful that my formula and answer for the washer method portion is correct. I am also aware that for the shell method portion, I would have gotten $$\frac{8\pi}{3}$$ if I did not multiply it by $2$. However, the shell method calls for this formula structure, which includes the 2—so what gives? I am confused and I hope someone can shed some light on this for me. Thanks!
I think you're too hung up on "canned methods". You should really try to understand what's going on.
Always, always start by sketching the curves.
https://www.desmos.com/calculator/erqould4cr
Compute the intersection points and all intercepts with the axis of rotation. Here you have to solve a simple quadratic but you can see that the parabola and the line intersect at $x = 3$ and $x=6$ and only the first intersection is relevant to the problem.
Then, think about the element of area that you're rotating about the $x$-axis. In this case, that's a slice of radius $|y|$ and height $dx$, which, when rotated about the $x$-axis, forms an infinitesimal cylinder. Between $x = 2$ and $x=3$, $y$ is given by the parabola, and between $x = 3$ and $x = 4$, $y$ is given by the line. Divide up the problem in that fashion and sum the respective volumes you get.
So the volume integral is $\displaystyle V = \pi(\int_2^4(\sqrt{x-2})^2dx + \int_4^6 (\sqrt{x-2})^2 - (x-4)^2 dx)$