question let the region r be the area enclosed by the function f(x)=2x^(1/3), the horizontal line y = 2, and…

Explanation:

Step1: Find the intersection point

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

Step2: Use the disk - washer method

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

Step3: Expand the integrand

$(2 - 2x^{\frac{1}{3}})^{2}=4-8x^{\frac{1}{3}} + 4x^{\frac{2}{3}}$.

Step4: Integrate term - by - term

$\int_{0}^{1}(4-8x^{\frac{1}{3}} + 4x^{\frac{2}{3}})dx=\left[4x-8\times\frac{3}{4}x^{\frac{4}{3}}+4\times\frac{3}{5}x^{\frac{5}{3}}\right]_{0}^{1}$. $=4\times1 - 6\times1^{\frac{4}{3}}+\frac{12}{5}\times1^{\frac{5}{3}}$. $=4 - 6+\frac{12}{5}$. $=\frac{20 - 30+12}{5}=\frac{2}{5}$.

Step5: Calculate the volume

$V=\pi\times\frac{2}{5}=\frac{2\pi}{5}\approx1.257$.

Answer:

$1.257$