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 $x^{2}-2 = 2x - 2$. $x^{2}-2x=0$, so $x(x - 2)=0$. The solutions are $x = 0$ and $x = 2$.

Step2: Use the disk - washer method

The outer radius $R(x)=4-(x^{2}-2)=6 - x^{2}$, the inner radius $r(x)=4-(2x - 2)=6 - 2x$. The volume formula is $V=\pi\int_{a}^{b}(R^{2}(x)-r^{2}(x))dx$, where $a = 0$, $b = 2$. $V=\pi\int_{0}^{2}((6 - x^{2})^{2}-(6 - 2x)^{2})dx$. Expand the integrands: $(6 - x^{2})^{2}=36-12x^{2}+x^{4}$, $(6 - 2x)^{2}=36 - 24x+4x^{2}$. Then $(6 - x^{2})^{2}-(6 - 2x)^{2}=36-12x^{2}+x^{4}-(36 - 24x+4x^{2})=x^{4}-16x^{2}+24x$.

Step3: Integrate

$\int(x^{4}-16x^{2}+24x)dx=\frac{1}{5}x^{5}-\frac{16}{3}x^{3}+12x^{2}+C$. Evaluate the definite integral: $V=\pi\left[\frac{1}{5}x^{5}-\frac{16}{3}x^{3}+12x^{2}\right]_{0}^{2}$. $V=\pi\left(\frac{1}{5}(2)^{5}-\frac{16}{3}(2)^{3}+12(2)^{2}\right)$. $V=\pi\left(\frac{32}{5}-\frac{128}{3}+48\right)$. $V=\pi\left(\frac{96 - 640+720}{15}\right)=\pi\left(\frac{176}{15}\right)\approx36.861$.

Answer:

$36.861$