Question #03feb

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Answer 1 · 2017-12-15T18:05:21

#=3.1323 u^2#

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#