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

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 distance from the axis of revolution $y = k$ to the curve. Here, the axis of revolution is $y=-2$, and the function is $y = 2x^{3}$, so $R(x)=(2x^{3}-(-2))=2x^{3}+2$, and $a = 0$, $b = 2$.

Step2: Set up the integral

We have $V=\pi\int_{0}^{2}(2x^{3}+2)^{2}dx$. Expand $(2x^{3}+2)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$, where $a = 2x^{3}$ and $b = 2$. So $(2x^{3}+2)^{2}=(2x^{3})^{2}+2\times(2x^{3})\times2+2^{2}=4x^{6}+8x^{3}+4$.

Step3: Integrate term - by - term

$\int(4x^{6}+8x^{3}+4)dx=4\int x^{6}dx+8\int x^{3}dx + 4\int dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $4\times\frac{x^{7}}{7}+8\times\frac{x^{4}}{4}+4x+C=\frac{4x^{7}}{7}+2x^{4}+4x+C$.

Step4: Evaluate the definite integral

$V=\pi\left[\frac{4x^{7}}{7}+2x^{4}+4x\right]_{0}^{2}$. Substitute the upper limit $x = 2$ and lower limit $x = 0$: [ \begin{align*} V&=\pi\left(\frac{4\times2^{7}}{7}+2\times2^{4}+4\times2-0\right)\ &=\pi\left(\frac{4\times128}{7}+2\times16 + 8\right)\ &=\pi\left(\frac{512}{7}+32 + 8\right)\ &=\pi\left(\frac{512}{7}+40\right)\ &=\pi\left(\frac{512+280}{7}\right)\ &=\pi\times\frac{792}{7}\ &\approx\frac{22}{7}\times\frac{792}{7}\ &\approx354.514 \end{align*} ]

Answer:

$354.514$