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

Explanation:

Step1: Recall the disk - washer method formula

The formula for the volume $V$ of the solid of revolution about the horizontal line $y = k$ using the disk - washer method is $V=\pi\int_{a}^{b}[R(x)]^{2}dx$, where $R(x)$ is the radius of the cross - sectional disk. Here, $a = 0$, $b = 4$, $k=-3$ and $R(x)=3x^{\frac{1}{3}}-(-3)=3x^{\frac{1}{3}} + 3$.

Step2: Set up the integral

$V=\pi\int_{0}^{4}(3x^{\frac{1}{3}} + 3)^{2}dx$. Expand the integrand $(3x^{\frac{1}{3}}+3)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. So, $(3x^{\frac{1}{3}}+3)^{2}=9x^{\frac{2}{3}}+18x^{\frac{1}{3}} + 9$.

Step3: Integrate term - by - term

$\int(9x^{\frac{2}{3}}+18x^{\frac{1}{3}} + 9)dx=9\int x^{\frac{2}{3}}dx+18\int x^{\frac{1}{3}}dx+9\int dx$. Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: $9\times\frac{3}{5}x^{\frac{5}{3}}+18\times\frac{3}{4}x^{\frac{4}{3}}+9x+C=\frac{27}{5}x^{\frac{5}{3}}+\frac{27}{2}x^{\frac{4}{3}}+9x+C$.

Step4: Evaluate the definite integral

$V=\pi\left[\frac{27}{5}x^{\frac{5}{3}}+\frac{27}{2}x^{\frac{4}{3}}+9x\right]_{0}^{4}$. $V=\pi\left(\frac{27}{5}(4)^{\frac{5}{3}}+\frac{27}{2}(4)^{\frac{4}{3}}+9(4)\right)-\pi(0)$. $(4)^{\frac{5}{3}}=4^{\frac{3}{3}}\times4^{\frac{2}{3}} = 4\times4^{\frac{2}{3}}$ and $(4)^{\frac{4}{3}}=4^{\frac{3}{3}}\times4^{\frac{1}{3}}=4\times4^{\frac{1}{3}}$. $V=\pi\left(\frac{27}{5}\times4\times4^{\frac{2}{3}}+\frac{27}{2}\times4\times4^{\frac{1}{3}}+36\right)$. Using a calculator: $V\approx\pi\left(\frac{27}{5}\times4\times2.52+ \frac{27}{2}\times4\times1.59+36\right)$. $V\approx\pi(54.432 + 85.86+36)$. $V\approx\pi(176.292)$. $V\approx553.982$.

Answer:

$553.982$