Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, using the disk/washer method. Sketch the region, the solid, and a typical disk or washer.y = (x - 2)^2 ; y=2x-1 ; about the line y = -1

Answer 1

Check the picture below.

to check where the functions intersect, we can simply set them equal to each other so (x-2)² = 2x - 1, anyhow, not to bore you to death with that, it's as the picture shows it, at x = 1 and x = 5, so those are our bounds.

to get the R² part of the washer, what I do is using the "area under the curve" method of f(x) - g(x), where g(x) is the axis of rotation and f(x) is the farthest radius, so to get R² I'd process (x-2)² - (-1), that'd give me R and then I'd square that.

to get r², I do pretty much the same thing, f(x) - g(x) where g(x) is the axis of rotation and f(x) is the closest radius, in this namely (2x - 1) - (-1), and that'd give "r" and then we can square that to get r².

`\begin{cases} R = 2x\\ r = x^2-4x+5 \end{cases}\hspace{5em} \begin{array}{llll} \stackrel{R^2}{4x^2}~~ - ~~\stackrel{r^2}{(x^4-8x^3+26x^2-40x+25)} \\\\\\ -x^4+8x^3-22x^2+40x-25 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \pi \int\limits_(1)^(5)(-x^4+8x^3-22x^2+40x-25)dx \\\\\\ \pi \left[ -\cfrac{x^5}{5}+2x^4-\cfrac{22x^3}{3} +20x^2-25x\right]_(1)^(5)\implies \cfrac{1408\pi }{15}`