Question #03feb

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Question #03feb

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Answer 1 · 2017-12-15T18:05:21

$= 3.1323 u^{2}$ $=3.1323 u^2$

Explanation:

in polar coordinates the area is calculated

$A = \frac{1}{2} \int_{a}^{b} r^{2} d (θ)$ $A=1/2int_a^br^2d(theta)$

take the funtion

$r = \sqrt{ln (θ)}$ $r=sqrt(ln(theta)$

the area

$A = \frac{1}{2} \int_{1}^{2 π} (ln (θ)) d (θ)$ $A=1/2int_1^(2pi)(ln(theta))d(theta)$

integrate by parts

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

where

$u = ln (θ)$ $u=ln(theta)$ $d u = \frac{1}{θ} d (θ)$ $du=1/thetad(theta)$
$d v = d (θ)$ $dv=d(theta)$ $v = θ$ $v=theta$

$\frac{= ln (θ) {(θ)}_{1}^{2 π} - \int_{1}^{2 π} (θ) (\frac{1}{θ}) d (θ)}{2}$ $(=ln(theta)(theta)_1^(2pi)-int_1^(2pi)(theta)(1/theta)d(theta))/2$

$= \frac{ln (θ) {(θ)}_{1}^{2 π} - \int_{1}^{2 π} d (θ)}{2} = \frac{{(ln (θ) (θ) - (θ))}_{1}^{2 π}}{2}$ $=(ln(theta)(theta)_1^(2pi)-int_1^(2pi)d(theta))/2=((ln(theta)(theta)-(theta))_1^(2pi))/2$

$= \frac{((2 π) ln (2 π) - 1 ln (1)) - ((- 2 π) - (- 1))}{2}$ $=(((2pi)ln(2pi)-1ln(1))-((-2pi)-(-1)))/2$

$= \frac{((2 π) ln (2 π) - 1 ln (1)) - ((- 2 π) - (- 1)) = 1 - 2 π + 2 π ln (2 π)}{2} = \frac{6.2645 u^{2}}{2}$ $=(((2pi)ln(2pi)-1ln(1))-((-2pi)-(-1))=1-2pi+2piln(2pi))/2=(6.2645 u^2)/2$

$= 3.1323 u^{2}$ $=3.1323 u^2$

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Question #03feb

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