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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) \(4.81\)
in Mathematics by Diamond (45,410 points) | 171 views

1 Answer

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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:
& u=(\ln \theta)^2 \\
& d v=d \theta
Now, differentiate \(u\) and integrate \(d v:\)
& d u=2 \ln \theta \cdot \frac{1}{\theta} d \theta=\frac{2 \ln \theta}{\theta} d \theta \\
& v=\theta

Apply integration by parts:
& \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}
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:
& (\ln 2)^2 \approx 0.480453 \\
& 2 \ln 2 \approx 1.386294
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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