曲线 $C$ 的方程为 $y=\ln\left(2-\lvert x\rvert\right)$。区域 $R$ 由 $C$、直线 $y=\ln 2$、$x=1$ 和 $x=-1$ 围成。求 $R$ 的面积精确值。 曲线 $C$ 的参数方程定义为 $x=(1+t)^{2/3},\qquad y=\ln(t^2)\qquad\text{其中 }t\le -1.$ 求当由 $C$、直线 $x=0$、$x=1$ 以及 x 轴围成的区域绕 x 轴旋转 $2\pi$ 弧度时所产生的旋转体体积，答案保留两位小数。

1. Set up the integral for the exact area, using symmetry about the y-axis to handle the absolute value.

The curve $C$ is symmetric about the $y$-axis. For $x \ge 0$, $|x| = x$, so the curve equation simplifies to $y = \ln(2-x)$.

The line $y = \ln 2$ is the upper boundary since $\ln(2-x) \le \ln 2$ for $x \in [0, 1]$.

Therefore, the exact area of $R$ is given by:
$\text{Area} = 2 \int_0^1 \left( \ln 2 - \ln(2-x) \right) \, dx$

2. Use substitution to simplify the integral of $\ln(2-x)$.

Consider the integral $\int_0^1 \ln(2-x) \, dx$.
Use the substitution $u = 2-x$, which gives $\frac{du}{dx} = -1 \implies du = -dx$.

When $x = 0$, $u = 2$.
When $x = 1$, $u = 1$.

Applying the substitution:
$\int_0^1 \ln(2-x) \, dx = \int_2^1 \ln(u) \, (-du) = \int_1^2 \ln u \, du$

3. Evaluate the integral of $\ln u$ using integration by parts.

Evaluate $\int_1^2 \ln u \, du$ using integration by parts.
Let $w = \ln u \implies \frac{dw}{du} = \frac{1}{u}$.
Let $dv = 1 \, du \implies v = u$.

$\int_1^2 \ln u \, du = \left[ u \ln u \right]_1^2 - \int_1^2 u \left(\frac{1}{u}\right) \, du$
$= (2\ln 2 - 1\ln 1) - \int_1^2 1 \, du$
$= 2\ln 2 - [u]_1^2$
$= 2\ln 2 - (2 - 1) = 2\ln 2 - 1$

4. Calculate the final exact area of region $R$.

Substitute the evaluated integral back into the area equation:
$\text{Area} = 2 \left[ x \ln 2 \right]_0^1 - 2 \int_0^1 \ln(2-x) \, dx$
$= 2(\ln 2) - 2(2 \ln 2 - 1)$
$= 2\ln 2 - 4\ln 2 + 2$
$= 2 - 2\ln 2$

1. Find the Cartesian relationship by expressing $t$ in terms of $x$.

Given $x = (1+t)^{2/3}$, raising both sides to the power of $\frac{3}{2}$ yields:
$x^{3/2} = \sqrt{(1+t)^2} = |1+t|$
Since the domain is $t \le -1$, we know $1+t \le 0$, so $|1+t| = -(1+t)$.
Thus, $x^{3/2} = -1 - t \implies t = -1 - x^{3/2}$.

2. Express $y$ in terms of $x$.

Substitute $t = -1 - x^{3/2}$ into the parametric equation for $y$:
$y = \ln(t^2) = \ln\left((-1 - x^{3/2})^2\right)$
$y = \ln\left((1 + x^{3/2})^2\right) = 2\ln(1 + x^{3/2})$

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

The region is bounded by $x=0$, $x=1$, and the x-axis ($y=0$). The volume of revolution about the x-axis is given by:
$V = \pi \int_0^1 y^2 \,dx$
Substituting $y = 2\ln(1+x^{3/2})$ into the formula:
$V = \pi \int_0^1 \left[ 2\ln(1+x^{3/2}) \right]^2 \,dx = 4\pi \int_0^1 \left(\ln(1+x^{3/2})\right)^2 \,dx$

4. Evaluate the definite integral using a graphing calculator.

Using a graphing calculator to evaluate the definite integral:
$\int_0^1 \left(\ln(1+x^{3/2})\right)^2 \,dx \approx 0.143592$
$V = 4\pi (0.143592) \approx 1.8044$
Rounding to 2 decimal places, $V \approx 1.80$.