let the region r be the area enclosed by the function f(x)=x^1 - 2, the x - axis, and the y - axis. find the…

Explanation:

Step1: Determine the intersection point

Set $f(x)=x - 2=0$, then $x = 2$. The region $R$ is bounded by $x = 0$, $y = 0$ and $y=x - 2$ for $0\leq x\leq2$. We use the disk - washer method. The outer radius $R(x)$ and inner radius $r(x)$ for the cross - sections perpendicular to the axis of rotation $y = 1$. The distance from the function $y=x - 2$ to the line $y = 1$ is $1-(x - 2)=3 - x$. The inner radius $r(x)=1$ (distance from the $x$ - axis to $y = 1$) and the outer radius $R(x)=3 - x$.

Step2: Apply the volume formula

The volume $V$ of the solid of revolution using the washer method is given by $V=\pi\int_{a}^{b}(R^{2}(x)-r^{2}(x))dx$. Here, $a = 0$, $b = 2$, $R(x)=3 - x$, and $r(x)=1$. So $V=\pi\int_{0}^{2}((3 - x)^{2}-1^{2})dx=\pi\int_{0}^{2}(9-6x+x^{2}-1)dx=\pi\int_{0}^{2}(x^{2}-6x + 8)dx$.

Step3: Integrate the function

We know that $\int(x^{2}-6x + 8)dx=\frac{1}{3}x^{3}-3x^{2}+8x+C$. Then $\pi\left[\frac{1}{3}x^{3}-3x^{2}+8x\right]_{0}^{2}=\pi\left(\frac{1}{3}(2)^{3}-3(2)^{2}+8(2)\right)=\pi\left(\frac{8}{3}-12 + 16\right)=\pi\left(\frac{8}{3}+4\right)=\pi\left(\frac{8 + 12}{3}\right)=\frac{20\pi}{3}\approx20.944$.

Answer:

$20.944$