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: Identify the radius function

The distance from the function $y = f(x)=x^{\frac{1}{3}}- 2$ to the axis of rotation $y=-3$ is $r(x)=(x^{\frac{1}{3}} - 2)-(-3)=x^{\frac{1}{3}}+1$.

Step2: Apply the disk - method formula

The formula for the volume $V$ of the solid of revolution about a horizontal line using the disk - method is $V=\pi\int_{a}^{b}[r(x)]^{2}dx$, where $a = 0$, $b = 5$ and $r(x)=x^{\frac{1}{3}}+1$. So $V=\pi\int_{0}^{5}(x^{\frac{1}{3}} + 1)^{2}dx$.

Step3: Expand the integrand

Expand $(x^{\frac{1}{3}}+1)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. We get $(x^{\frac{1}{3}}+1)^{2}=x^{\frac{2}{3}}+2x^{\frac{1}{3}}+1$.

Step4: Integrate term - by - term

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

Step5: Evaluate the definite integral

$V=\pi\left[\frac{3}{5}x^{\frac{5}{3}}+\frac{3}{2}x^{\frac{4}{3}}+x\right]_{0}^{5}$. $V=\pi\left(\frac{3}{5}(5)^{\frac{5}{3}}+\frac{3}{2}(5)^{\frac{4}{3}}+5\right)$. Using a calculator: $\frac{3}{5}(5)^{\frac{5}{3}}=\frac{3}{5}\times5\times5^{\frac{2}{3}} = 3\times5^{\frac{2}{3}}\approx3\times2.924 = 8.772$. $\frac{3}{2}(5)^{\frac{4}{3}}=\frac{3}{2}\times5\times5^{\frac{1}{3}}\approx\frac{3}{2}\times5\times1.710=12.825$. $V=\pi(8.772 + 12.825+5)=\pi(26.597)\approx83.578$.

Answer:

$83.578$