let the region r be the area enclosed by the function f(x)=3e^x, the horizontal line y = 9, and the y…

Explanation:

Step1: Find the intersection point

Set $3e^{x}=9$, then $e^{x} = 3$, so $x=\ln(3)$.

Step2: Use the disk - washer method

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the disk - washer method is $V=\pi\int_{a}^{b}[(k - g(x))^{2}-(k - h(x))^{2}]dx$. Here, $k = 9$, $g(x)=3e^{x}$, $h(x)=0$, and $a = 0$, $b=\ln(3)$. So $V=\pi\int_{0}^{\ln(3)}[(9 - 3e^{x})^{2}]dx$.

Step3: Expand the integrand

$(9 - 3e^{x})^{2}=81-54e^{x}+9e^{2x}$.

Step4: Integrate term - by - term

$\int(81-54e^{x}+9e^{2x})dx=81x-54e^{x}+\frac{9}{2}e^{2x}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[81x-54e^{x}+\frac{9}{2}e^{2x}\right]_{0}^{\ln(3)}$ $=\pi\left[\left(81\ln(3)-54\times3+\frac{9}{2}\times9\right)-\left(0 - 54+\frac{9}{2}\right)\right]$ $=\pi\left(81\ln(3)-162+\frac{81}{2}+54-\frac{9}{2}\right)$ $=\pi\left(81\ln(3)-108 + 36\right)$ $=\pi\left(81\ln(3)-72\right)$. Using a calculator, $V\approx\pi(81\times1.0986 - 72)\approx\pi(89.98 - 72)\approx56.571$.

Answer:

$56.571$