question let the region r be the area enclosed by the function f(x)=x² - 2 and g(x)=2x - 2. find the volume…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $x^{2}-2 = 2x - 2$. Rearranging gives $x^{2}-2x=0$, factoring $x(x - 2)=0$. The solutions are $x = 0$ and $x=2$.

Step2: Use the washer - method formula

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the washer - method is $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer radius and $r(x)$ is the inner radius. Here, $a = 0$, $b = 2$, $R(x)=(2x - 2)-(-3)=2x + 1$ and $r(x)=(x^{2}-2)-(-3)=x^{2}+1$.

Step3: Set up the integral

$V=\pi\int_{0}^{2}((2x + 1)^{2}-(x^{2}+1)^{2})dx=\pi\int_{0}^{2}(4x^{2}+4x + 1-(x^{4}+2x^{2}+1))dx=\pi\int_{0}^{2}(4x^{2}+4x + 1 - x^{4}-2x^{2}-1)dx=\pi\int_{0}^{2}(-x^{4}+2x^{2}+4x)dx$.

Step4: Integrate term - by - term

$\int(-x^{4}+2x^{2}+4x)dx=-\frac{1}{5}x^{5}+\frac{2}{3}x^{3}+2x^{2}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[-\frac{1}{5}x^{5}+\frac{2}{3}x^{3}+2x^{2}\right]_{0}^{2}=\pi\left(-\frac{1}{5}(2)^{5}+\frac{2}{3}(2)^{3}+2(2)^{2}\right)=\pi\left(-\frac{32}{5}+\frac{16}{3}+8\right)$. First, find a common denominator of 15: $-\frac{32}{5}+\frac{16}{3}+8=-\frac{96}{15}+\frac{80}{15}+\frac{120}{15}=\frac{-96 + 80+120}{15}=\frac{104}{15}$. So $V=\frac{104\pi}{15}\approx21.782$.

Answer:

$21.782$