let the region r be the area enclosed by the function f(x)=x^(1/3) - 2, the horizontal line y = -5 and the…

Explanation:

Step1: Identify the radius function

The distance from the function $y = x^{\frac{1}{3}}-2$ to the axis of rotation $y=-5$ is $r(x)=(x^{\frac{1}{3}} - 2)-(-5)=x^{\frac{1}{3}}+3$.

Step2: Use the disk - method formula

The formula for the volume $V$ of a solid of revolution about a horizontal line using the disk - method is $V=\pi\int_{a}^{b}[r(x)]^{2}dx$. Here, $a = 0$, $b = 7$, and $r(x)=x^{\frac{1}{3}}+3$. So $V=\pi\int_{0}^{7}(x^{\frac{1}{3}} + 3)^{2}dx$.

Step3: Expand the integrand

Expand $(x^{\frac{1}{3}}+3)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. We get $(x^{\frac{1}{3}})^{2}+2\times x^{\frac{1}{3}}\times3+3^{2}=x^{\frac{2}{3}}+6x^{\frac{1}{3}} + 9$.

Step4: Integrate term - by - term

$\int(x^{\frac{2}{3}}+6x^{\frac{1}{3}} + 9)dx=\frac{3}{5}x^{\frac{5}{3}}+\frac{6\times3}{4}x^{\frac{4}{3}}+9x+C=\frac{3}{5}x^{\frac{5}{3}}+\frac{9}{2}x^{\frac{4}{3}}+9x+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{3}{5}x^{\frac{5}{3}}+\frac{9}{2}x^{\frac{4}{3}}+9x\right]_{0}^{7}=\pi\left(\frac{3}{5}\times7^{\frac{5}{3}}+\frac{9}{2}\times7^{\frac{4}{3}}+9\times7\right)$. Using a calculator: $\frac{3}{5}\times7^{\frac{5}{3}}\approx\frac{3}{5}\times16.266 = 9.7596$, $\frac{9}{2}\times7^{\frac{4}{3}}\approx\frac{9}{2}\times8.888=40.00$, $9\times7 = 63$. $V=\pi(9.7596 + 40.00+63)\approx\pi(112.7596)\approx354.677$.

Answer:

$354.677$