What is the area enclosed by #r=theta # for #theta in [0,pi]#?

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Answer 1 · 2016-05-07T11:54:13

The area = #pi^3/6=5.168#, areal units.
Also, the length of the spiral #=[{(1/2)[pi sqrt (pi^2+1)+ln(pi+sqrt(pi^2+1)}-{0}]=6.110#, nearly..

Area #=int(1/2)r^2 d theta=(1/2)int theta^2 d theta = (1/2)[theta^3]/3#, between the limits, 0 and #pi#
#=pi^3/6=5.168# areal units.

Added to the answer:
#r=theta#. So, r' = 1
Length of the spiral = #int sqrt(r^2+(r')^2) d theta#

#=int sqrt(theta^2+1 )d theta#, between the limits 0 and #pi#

#=.[(1/2)(theta sqrt( theta^2+1)+(1/2)ln(theta+sqrt(theta^2+1)]#, between the limits 0 and #pi#

#=[{(1/2)[pi sqrt (pi^2+1)+ln(pi+sqrt(pi^2+1)}-{0}]=6.110#, nearly..