A function f is defined as $f(x)=\ln(5-x)$, $x<5$.

1. Find the axis intercepts to determine the limits of integration.

At the $x$-axis, $y = 0$:
$\ln(5-x) = 0 \implies 5-x = 1 \implies x = 4$
At the $y$-axis, $x = 0$:
$y = \ln(5-0) = \ln 5$
The enclosed region is bounded between $y = 0$ and $y = \ln 5$.

2. Express $x$ in terms of $y$ to prepare for integration with respect to $y$.

$y = \ln(5-x)$
$e^y = 5-x$
$x = 5-e^y$

3. Set up the definite integral for the volume of revolution about the $y$-axis.

The volume $V$ generated by rotating the region completely about the $y$-axis is given by:
$V = \pi \int_{0}^{\ln 5} x^2 \, dy$
$V = \pi \int_{0}^{\ln 5} (5-e^y)^2 \, dy$

4. Evaluate the integral to find the volume correct to 2 decimal places.

Expanding the integrand:
$V = \pi \int_{0}^{\ln 5} (25 - 10e^y + e^{2y}) \, dy$
Integrating analytically:
$V = \pi \left[ 25y - 10e^y + \frac{1}{2}e^{2y} \right]_{0}^{\ln 5}$
$V = \pi \left( \left(25\ln 5 - 10(5) + \frac{1}{2}(25)\right) - \left(0 - 10(1) + \frac{1}{2}(1)\right) \right)$
$V = \pi \left( 25\ln 5 - 37.5 - (-9.5) \right)$
$V = \pi (25\ln 5 - 28)$
Evaluating to 2 decimal places:
$V \approx 38.44$
(Note: Alternatively, students may legitimately evaluate the definite integral directly from the setup using a graphing calculator without showing the analytical steps.)

1. Solve the equation for $x \ge 0$ by removing the inner modulus and using the property of the outer modulus.

For $x \ge 0$, $|x| = x$, so the equation becomes $f(x) = |f(x)|$.
This holds if and only if $f(x) \ge 0$.
$\ln(5-x) \ge 0$
$5-x \ge 1$
$x \le 4$
Combining with the condition $x \ge 0$, we get $0 \le x \le 4$.

2. Check for any solutions when $x < 0$ to ensure completeness.

For $x < 0$, $|x| = -x$.
The equation becomes $f(-x) = |f(x)|$, which means $\ln(5+x) = |\ln(5-x)|$.
Since $x < 0$, $5-x > 5$, which implies $\ln(5-x) > \ln(5) > 0$.
Thus, the equation simplifies to $\ln(5+x) = \ln(5-x)$, giving $5+x = 5-x \implies x=0$.
This contradicts $x < 0$, so there are no valid solutions for negative $x$.

The complete set of values is therefore $0 \le x \le 4$.

1. Write f(|x|) explicitly on the interval [-9/2, 0]

For $-\tfrac{9}{2} \le x \le 0$, $|x| = -x \ge 0$, so
$f(|x|) = \ln\big(5 - |x|\big) = \ln(5 + x).$
Note that $y = f(|x|)$ is an even function (symmetric about the $y$-axis), and for $x \ge 0$, $|x| = x$ so $f(|x|) = f(x) = \ln(5 - x)$.

2. Show the second integral equals the first by substitution / symmetry

In $\displaystyle\int_{-9/2}^{0} f(|x|)\,dx = \int_{-9/2}^{0}\ln(5+x)\,dx$, let $x = -t$, $dx = -dt$.
When $x = -\tfrac{9}{2}$, $t = \tfrac{9}{2}$; when $x = 0$, $t = 0$:
$\int_{-9/2}^{0}\ln(5+x)\,dx = \int_{9/2}^{0}\ln(5 - t)\,(-dt) = \int_{0}^{9/2}\ln(5-t)\,dt = \int_{0}^{9/2} f(x)\,dx.$

3. Take the difference and state a and b

Therefore
$\int_{0}^{9/2} f(x)\,dx - \int_{-9/2}^{0} f(|x|)\,dx = \int_{0}^{9/2} f(x)\,dx - \int_{0}^{9/2} f(x)\,dx = 0.$
Writing this in the form $a + b\ln 2$ gives $a = 0$ and $b = 0$.