let the region r be the area enclosed by the function f(x)=2x³, the horizontal line y = 1, and the y - axis…

Explanation:

Step1: Find the intersection point

Set $2x^{3}=1$, then $x = \frac{1}{\sqrt[3]{2}}$.

Step2: Use the disk - washer method formula

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

Step3: Expand the integrand

Expand $(1 - 2x^{3})^{2}=1-4x^{3}+4x^{6}$.

Step4: Integrate term - by - term

$\int(1-4x^{3}+4x^{6})dx=x - x^{4}+\frac{4}{7}x^{7}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[x - x^{4}+\frac{4}{7}x^{7}\right]_{0}^{\frac{1}{\sqrt[3]{2}}}=\pi\left(\frac{1}{\sqrt[3]{2}}-\frac{1}{2\sqrt[3]{2}}+\frac{4}{7\times2\sqrt[3]{2}}\right)$. Simplify the expression: [ \begin{align*} V&=\pi\left(\frac{14 - 7+ 4}{14\sqrt[3]{2}}\right)\ &=\frac{11\pi}{14\sqrt[3]{2}}\approx1.047 \end{align*} ]

Answer:

$1.047$