question let the region r be the area enclosed by the function f(x)=2x², the horizontal line y = -2 and the…

Explanation:

Step1: Recall disk - method formula

The formula for the volume $V$ of a solid of revolution about a horizontal line $y = k$ using the disk method is $V=\pi\int_{a}^{b}[R(x)]^{2}dx$, where $R(x)$ is the distance from the curve $y = f(x)$ to the line $y = k$. Here, $f(x)=2x^{2}$, $k = - 2$, $a = 0$, $b = 2$, and $R(x)=(2x^{2}-(-2))=2x^{2}+2$.

Step2: Set up the integral

$V=\pi\int_{0}^{2}(2x^{2}+2)^{2}dx$. Expand $(2x^{2}+2)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$, so $(2x^{2}+2)^{2}=(2x^{2})^{2}+2\times(2x^{2})\times2+2^{2}=4x^{4}+8x^{2}+4$. Then $V=\pi\int_{0}^{2}(4x^{4}+8x^{2}+4)dx$.

Step3: Integrate term - by - term

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(4x^{4}+8x^{2}+4)dx=4\times\frac{x^{5}}{5}+8\times\frac{x^{3}}{3}+4x+C=\frac{4x^{5}}{5}+\frac{8x^{3}}{3}+4x+C$.

Step4: Evaluate the definite integral

$V=\pi\left[\frac{4x^{5}}{5}+\frac{8x^{3}}{3}+4x\right]_{0}^{2}$. Substitute the upper and lower limits: $V=\pi\left(\frac{4\times2^{5}}{5}+\frac{8\times2^{3}}{3}+4\times2\right)-\pi(0)$. Calculate $\frac{4\times32}{5}+\frac{8\times8}{3}+8=\frac{128}{5}+\frac{64}{3}+8$. Find a common denominator of 15: $\frac{128\times3}{15}+\frac{64\times5}{15}+\frac{8\times15}{15}=\frac{384 + 320+120}{15}=\frac{824}{15}$. So $V=\frac{824\pi}{15}\approx172.573$.

Answer:

$172.573$