The curve $C$ has equation $y=\ln\left(2-\lvert x\rvert\right)$. The region $R$ is bounded by $C$, the lines $y=\ln 2$, $x=1$ and $x=-1$. Find the exact area of $R$. The curve $C$ is defined parametrically by $x=(1+t)^{2/3},\qquad y=\ln\left(t^2\right)\qquad\text{for }t\le -1.$ Find the volume of the solid generated when the region enclosed by $C$, the lines $x=0$ and $x=1$ and the $x$-axis is rotated $2\pi$ radians about the $x$-axis, giving your answer correct to 2 decimal places.

1. Express the exact area using a definite integral with respect to $x$, taking advantage of symmetry.

The curve $C$ is symmetric about the $y$-axis because $\lvert -x \rvert = \lvert x \rvert$.
For $0 \le x \le 1$, the function simplifies to $y = \ln(2-x)$.
The region $R$ is bounded above by $y = \ln 2$ and below by $y = \ln(2-x)$ between $x = -1$ and $x = 1$.
By symmetry, the total area is twice the area in the right half-plane:
$\text{Area} = 2 \int_0^1 \left( \ln 2 - \ln(2-x) \right) \, \mathrm{d}x$

2. Evaluate the integral of $\ln(2-x)$ using substitution and integration by parts.

Let $I = \int_0^1 \ln(2-x) \, \mathrm{d}x$.
Use the substitution $u = 2-x$, so $\mathrm{d}u = -\mathrm{d}x$.
When $x = 0, u = 2$; when $x = 1, u = 1$.
$I = \int_2^1 \ln u \, (-\mathrm{d}u) = \int_1^2 \ln u \, \mathrm{d}u$
Using integration by parts (let $p = \ln u \implies \mathrm{d}p = \frac{1}{u}\mathrm{d}u$, and $\mathrm{d}q = \mathrm{d}u \implies q = u$):
$I = \left[ u \ln u \right]_1^2 - \int_1^2 u \left(\frac{1}{u}\right) \, \mathrm{d}u$
$I = (2 \ln 2 - 1 \ln 1) - \left[ u \right]_1^2$
$I = 2 \ln 2 - (2 - 1) = 2 \ln 2 - 1$

3. Combine the results to calculate the final exact area.

$\text{Area} = 2 \int_0^1 \ln 2 \, \mathrm{d}x - 2 \int_0^1 \ln(2-x) \, \mathrm{d}x$
$\text{Area} = 2 \left[ x \ln 2 \right]_0^1 - 2(2 \ln 2 - 1)$
$\text{Area} = 2 \ln 2 - 4 \ln 2 + 2$
$\text{Area} = 2 - 2 \ln 2$

1. Find the limits for the parameter $t$ corresponding to the boundary lines $x=0$ and $x=1$.

Given $x = (1+t)^{2/3}$ and the domain $t \le -1$:
When $x=0$, $(1+t)^{2/3} = 0 \implies t = -1$.
When $x=1$, $(1+t)^{2/3} = 1 \implies 1+t = \pm 1$.
Since $t \le -1$, we reject $1+t = 1$ (which gives $t=0$) and accept $1+t = -1 \implies t = -2$.

2. Differentiate the parametric equation for $x$ with respect to $t$.

$x = (1+t)^{2/3}$
$\frac{dx}{dt} = \frac{2}{3}(1+t)^{-1/3}$

3. Set up the definite integral for the volume of revolution.

The volume of the solid generated when rotated $2\pi$ radians about the $x$-axis is given by $V = \int_{t_1}^{t_2} \pi y^2 \frac{dx}{dt} \,dt$:
$V = \pi \int_{-1}^{-2} \left[\ln\left(t^2\right)\right]^2 \left(\frac{2}{3}(1+t)^{-1/3}\right) \,dt$

4. Evaluate the definite integral using a graphing calculator.

(Note: To safely compute this on a graphing calculator without non-real errors from negative fractional powers, one can reverse the limits and explicitly factor the negative base: $V = \pi \int_{-2}^{-1} \left[\ln\left(t^2\right)\right]^2 \left(\frac{2}{3}(-(1+t))^{-1/3}\right) \,dt$, or use the equivalent Cartesian integral $V = 4\pi \int_0^1 \left(\ln(1+x^{3/2})\right)^2 \,dx$.)

Evaluating the integral using a graphing calculator:
$V \approx 1.8049\dots$
$V = 1.80 \text{ (to 2 d.p.)}$