To find the area enclosed by the polar curve \(r=\ln \theta\) on the interval \(\pi \leq \theta \leq 2 \pi\), we can use the polar area formula:

\[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2(\theta) d\theta\]

In our case, \(r(\theta)=\ln \theta\), and we want to evaluate the integral on the interval \(\alpha=\pi\) to \(\beta=2 \pi\). Thus, we have:

\[A = \frac{1}{2} \int_{\pi}^{2\pi} (\ln{\theta})^2 d\theta\]

To solve this integral, we can use integration by parts. Let:

\[

\begin{aligned}

& u=(\ln \theta)^2 \\

& d v=d \theta

\end{aligned}

\]

Now, differentiate \(u\) and integrate \(d v:\)

\[

\begin{aligned}

& d u=2 \ln \theta \cdot \frac{1}{\theta} d \theta=\frac{2 \ln \theta}{\theta} d \theta \\

& v=\theta

\end{aligned}

\]

Apply integration by parts:

\[

\begin{aligned}

& \int_\pi^{2 \pi} \ln \theta d \theta=\left.u v\right|_\pi ^{2 \pi}-\int_\pi^{2 \pi} v d u \\

& \left.\ln \theta \theta\right|_\pi ^{2 \pi}-\int_\pi^{2 \pi} \theta \cdot \frac{1}{\theta} d \theta \\

& \left.\ln \theta \theta\right|_\pi ^{2 \pi}-\int_\pi^{2 \pi} d \theta \\

& \left.\ln \theta \theta\right|_\pi ^{2 \pi}-\left.\theta\right|_\pi ^{2 \pi}

\end{aligned}

\]

Plug this back into our original integral:

\[

\left.(\ln \theta)^2 \theta\right|_\pi ^{2 \pi}-2\left(\left.\ln \theta \theta\right|_\pi ^{2 \pi}-\left.\theta\right|_\pi ^{2 \pi}\right)

\]

Now, evaluate the limits of the integrals:

\[A = \frac{1}{2}[(\ln{2\pi})^2(2\pi) - (\ln{\pi})^2(\pi) - 2((\ln{2\pi})(2\pi) - (\ln\pi)(\pi)-(2 \pi-\pi))\]

Now simplify:

\[

A=\frac{1}{2}\left[(\ln 2)^2(2 \pi)-(\ln 1)^2(\pi)-2((\ln 2)(2 \pi)-\pi-\pi)\right]

\]

Since \(\ln 1=0\), the middle term disappears:

\[

A=\frac{1}{2}\left[(\ln 2)^2(2 \pi)-2((\ln 2)(2 \pi)-2 \pi)\right]

\]

Now distribute the constants:

\[

A=(\ln 2)^2 \pi-(\ln 2) 2 \pi+2 \pi

\]

Factor out the common term, \(\pi\) :

\[

A=\pi\left[\left((\ln 2)^2-2 \ln 2+2\right)\right]

\]

This is the area bounded by the spiral \(r=\ln \theta\) on the interval \(\pi \leq \theta \leq 2 \pi\).

To evaluate the expression, first compute the values within the brackets:

\[

(\ln 2)^2-2 \ln 2+2

\]

Using a calculator or software, we find:

\[

\begin{aligned}

& (\ln 2)^2 \approx 0.480453 \\

& 2 \ln 2 \approx 1.386294

\end{aligned}

\]

Now plug these values into the expression:

\[

0.480453-1.386294+2 \approx 1.094159

\]

Now multiply by \(\pi\) :

\[

A=\pi(1.094159) \approx 3.437447

\]

Thus, the area bounded by the spiral \(r=\ln \theta\) on the interval \(\pi \leq \theta \leq 2 \pi\) is approximately 3.437447 .