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

Answer:

<Calculated volume value rounded to nearest thousandth>

Explanation:

Step1: Find intersection point

Set $x^{\frac{1}{3}}=2$, then $x = 8$.

Step2: Use disk - washer method

The radius $r(x)=2 - x^{\frac{1}{3}}$. The volume formula for revolving about $y = 2$ using the disk - washer method is $V=\pi\int_{0}^{8}(2 - x^{\frac{1}{3}})^2dx$.

Step3: Expand integrand

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

Step4: Integrate term - by - term

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

Step5: Evaluate definite integral

$V=\pi\left[4x - 3x^{\frac{4}{3}}+\frac{3}{5}x^{\frac{5}{3}}\right]_{0}^{8}=\pi\left(4\times8-3\times8^{\frac{4}{3}}+\frac{3}{5}\times8^{\frac{5}{3}}\right)$. $8^{\frac{4}{3}}=(8^{\frac{1}{3}})^4 = 16$, $8^{\frac{5}{3}}=(8^{\frac{1}{3}})^5=32$. $V=\pi\left(32-3\times16+\frac{3}{5}\times32\right)=\pi\left(32 - 48+\frac{96}{5}\right)=\pi\left(\frac{160 - 240+96}{5}\right)=\frac{16\pi}{5}\approx10.053$.