Question #03feb

1 Answer
Dec 15, 2017

=3.1323 u^2

Explanation:

in polar coordinates the area is calculated

A=1/2int_a^br^2d(theta)

take the funtion

r=sqrt(ln(theta)

the area

A=1/2int_1^(2pi)(ln(theta))d(theta)

integrate by parts

intudv=uv-intvdu

where

u=ln(theta) du=1/thetad(theta)
dv=d(theta) v=theta

(=ln(theta)(theta)_1^(2pi)-int_1^(2pi)(theta)(1/theta)d(theta))/2

=(ln(theta)(theta)_1^(2pi)-int_1^(2pi)d(theta))/2=((ln(theta)(theta)-(theta))_1^(2pi))/2

=(((2pi)ln(2pi)-1ln(1))-((-2pi)-(-1)))/2

=(((2pi)ln(2pi)-1ln(1))-((-2pi)-(-1))=1-2pi+2piln(2pi))/2=(6.2645 u^2)/2

=3.1323 u^2