let the region r be the area enclosed by the function f(x)=2x^3, the horizontal line y = 2, and the y…

Explanation:

Step1: Find the intersection point

Set $2x^{3}=2$, then $x^{3}=1$, so $x = 1$.

Step2: Use the disk - washer method

The radius of the cross - section is $r(x)=2 - 2x^{3}$. The volume formula for a solid of revolution about the line $y = 2$ using the disk method is $V=\pi\int_{a}^{b}[r(x)]^{2}dx$, where $a = 0$ and $b = 1$. So $V=\pi\int_{0}^{1}(2 - 2x^{3})^{2}dx$.

Step3: Expand the integrand

$(2 - 2x^{3})^{2}=4-8x^{3}+4x^{6}$.

Step4: Integrate term - by - term

$\int(4-8x^{3}+4x^{6})dx=4x - 2x^{4}+\frac{4}{7}x^{7}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\left(4x - 2x^{4}+\frac{4}{7}x^{7}\right)\big|_{0}^{1}\right]=\pi\left(4\times1-2\times1^{4}+\frac{4}{7}\times1^{7}\right)=\pi\left(4 - 2+\frac{4}{7}\right)=\pi\left(2+\frac{4}{7}\right)=\frac{18\pi}{7}\approx8.079$.

Answer:

$8.079$