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

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: 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, $a = 0$, $b = 2$, and $r(x)=e^{x}+1$. So $V=\pi\int_{0}^{2}(e^{x}+1)^{2}dx$.

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

$V=\pi\left[\frac{1}{2}e^{2x}+2e^{x}+x\right]_{0}^{2}=\pi\left(\left(\frac{1}{2}e^{4}+2e^{2}+2\right)-\left(\frac{1}{2}+2 + 0\right)\right)=\pi\left(\frac{1}{2}e^{4}+2e^{2}+2-\frac{1}{2}-2\right)=\pi\left(\frac{1}{2}e^{4}+2e^{2}-\frac{1}{2}\right)$.

Step6: Calculate the numerical value

Using a calculator, $e^{2}\approx7.389$, $e^{4}\approx54.598$. Then $V=\pi\left(\frac{1}{2}\times54.598+2\times7.389-\frac{1}{2}\right)=\pi(27.299 + 14.778-0.5)=\pi(41.577)\approx130.537$.

Answer:

$130.537$