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

Explanation:

Step1: Identify the radius function

The distance from the curve $y = e^{x}-1$ to the axis of rotation $y=-2$ is $r(x)=(e^{x}-1)-(-2)=e^{x}+1$.

Step2: Apply the disk - method formula

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

Step3: Expand the integrand

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

Step4: Integrate term - by - term

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

Step5: Evaluate the definite integral

$V=\pi\left[\frac{1}{2}e^{2x}+2e^{x}+x\right]_{0}^{3}$. $V=\pi\left(\frac{1}{2}e^{6}+2e^{3}+3\right)-\pi\left(\frac{1}{2}+2 + 0\right)$. $V=\pi\left(\frac{1}{2}e^{6}+2e^{3}+3-\frac{1}{2}-2\right)$. $V=\pi\left(\frac{1}{2}e^{6}+2e^{3}+\frac{1}{2}\right)$. Using a calculator, $e^{3}\approx20.086$ and $e^{6}\approx403.429$. $V=\pi\left(\frac{1}{2}\times403.429+2\times20.086+\frac{1}{2}\right)$. $V=\pi(201.7145 + 40.172+0.5)$. $V=\pi(242.3865)$. $V\approx761.537$.

Answer:

$761.537$