Question #f6b13

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Answer 1 · 2017-03-27T21:04:52

#= 8 pi#

The simplest thing to do here is to de-parameterise.

From #x(t) = 2t# we have #t = x/2#.

From #y(t) = 2 (1 - cos t)# we have #y(x) = 2 (1 - cos (x/2))#

For #t in [0 , 2 pi]#, we have #x in [0 , 4 pi]#

So we are looking at:

#2 int_(x=0)^(4 pi) 1 - cos (x/2 ) \ dx = 8 pi#

Have assumed that you know how to work with that integrand.