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

Explanation:

Step1: Recall the disk - washer method formula

The formula for the volume $V$ of a solid of revolution about the horizontal line $y = k$ using the disk - washer method when rotating the region between $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $V=\pi\int_{a}^{b}[f(x)-k]^2dx$. Here, $f(x)=x^{\frac{1}{3}}-2$, $k = - 3$, $a = 0$, and $b = 5$. So the radius $r(x)=(x^{\frac{1}{3}}-2)-(-3)=x^{\frac{1}{3}} + 1$.

Step2: Set up the integral for the volume

The volume $V$ is given by the integral $V=\pi\int_{0}^{5}(x^{\frac{1}{3}} + 1)^2dx$. Expand the integrand: $(x^{\frac{1}{3}}+1)^2=(x^{\frac{1}{3}})^2 + 2x^{\frac{1}{3}}+1=x^{\frac{2}{3}}+2x^{\frac{1}{3}} + 1$.

Step3: Integrate term - by - term

We know that $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ for $n\neq - 1$. $\int_{0}^{5}(x^{\frac{2}{3}}+2x^{\frac{1}{3}} + 1)dx=\int_{0}^{5}x^{\frac{2}{3}}dx+2\int_{0}^{5}x^{\frac{1}{3}}dx+\int_{0}^{5}1dx$. $\int_{0}^{5}x^{\frac{2}{3}}dx=\left[\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}\right]{0}^{5}=\left[\frac{x^{\frac{5}{3}}}{\frac{5}{3}}\right]{0}^{5}=\frac{3}{5}x^{\frac{5}{3}}\big|{0}^{5}=\frac{3}{5}\times5^{\frac{5}{3}}$. $2\int{0}^{5}x^{\frac{1}{3}}dx=2\left[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]{0}^{5}=2\left[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right]{0}^{5}=\frac{3}{2}\times2x^{\frac{4}{3}}\big|{0}^{5}=3\times5^{\frac{4}{3}}$. $\int{0}^{5}1dx=x\big|_{0}^{5}=5$.

Step4: Calculate the volume

$V=\pi\left(\frac{3}{5}\times5^{\frac{5}{3}}+3\times5^{\frac{4}{3}} + 5\right)$. Using a calculator: $\frac{3}{5}\times5^{\frac{5}{3}}=\frac{3\times5^{\frac{5}{3}}}{5}=3\times5^{\frac{2}{3}}\approx3\times2.924 = 8.772$. $3\times5^{\frac{4}{3}}\approx3\times6.694 = 20.082$. $V=\pi(8.772+20.082 + 5)=\pi(33.854)\approx106.392$.

Answer:

$106.392$