The volume of solid of revolution that results when $ f(x)=x^2 + 1$, $g(x)=3-x^2$ revolved around $y=1$
What's the difference in the two solutions below?
Solution 1:
$$\pi \int_{-1}^{1} [1-g(x)]^2 - [1-f(x)]^2 \,dx $$ Solution 2:
$$\pi \int_{-1}^1 [g(x) - 1]^2 - [f(x) -1]^2 \,dx $$
Also, are the succeeding solutions the same as above ones?
Solution 1: $$\pi \int_{-1}^1 [1-f(x)]^2 - [1-g(x)]^2 \,dx $$ Solution 2: $$\pi \int_{-1}^1 [f(x) - 1]^2 - [g(x) -1]^2 \,dx $$
Within each pair of solutions 1 and 2, they are the same because $$[1-f(x)]^2 = [f(x)-1]^2.$$
Conceptually though, since $f(x)\ge 1$ and $g(x)\ge1$ in that range $-1\le x\le 1$, the $[f(x)-1]^2$ form may be easier to understand.
The first pair of solutions is different from the second pair of solution, and the first pair is the correct one.
By comparing $1 \le f(x)\le g(x)$ in the range $-1\le x \le 1$, $f(x)$ is the "inner edge of the washers" and $g(x)$ is the "outer edge of the washers", hence the integrand should be
$$[g(x)-1]^2 - [f(x)-1]^2 = [1-g(x)]^2 - [1-f(x)]^2.$$