let the region r be the area enclosed by the function f(x)=e^x - 1, the horizontal line y = -4 and the…

Explanation:

Step1: Identify the radius function

The distance from the curve $y = e^{x}-1$ to the axis of rotation $y=-4$ is $r(x)=(e^{x}-1)-(-4)=e^{x}+3$.

Step2: Use the disk - method formula

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

Step3: Expand the integrand

Expand $(e^{x}+3)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. We get $(e^{x}+3)^{2}=e^{2x}+6e^{x}+9$.

Step4: Integrate term - by - term

$\int(e^{2x}+6e^{x}+9)dx=\frac{1}{2}e^{2x}+6e^{x}+9x+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{1}{2}e^{2x}+6e^{x}+9x\right]_{0}^{2}=\pi\left(\left(\frac{1}{2}e^{4}+6e^{2}+18\right)-\left(\frac{1}{2}+6 + 0\right)\right)$. $=\pi\left(\frac{1}{2}e^{4}+6e^{2}+18-\frac{1}{2}-6\right)=\pi\left(\frac{1}{2}e^{4}+6e^{2}+11.5\right)$. Using a calculator, $e^{2}\approx7.389$, $e^{4}\approx54.598$. $V=\pi\left(\frac{1}{2}\times54.598+6\times7.389 + 11.5\right)=\pi(27.299+44.334+11.5)=\pi(83.133)$. $V\approx261.177$.

Answer:

$261.177$