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Find the area bounded by the spiral $r=\ln \theta$ on the interval $\pi<=\theta<=2 \pi$. Which of the following is the right choice? Explain your answer.

(A) $2.405$ (B) $2.931$ (C) $3.743$ (D) $4.81$
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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 .
by Diamond (89,041 points)

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