What is the surface area produced by rotating #f(x)=tanx-cos^2x, x in [0,pi/4]# around the x-axis?
1 Answer
Jun 9, 2017
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For revolutions around the
#S = 2pi int_(alpha)^(beta) f(x) sqrt(1 + ((dy)/(dx))^2) \ dx#
Clearly, this is most suitable for very, very simple functions, and this is not one of those. Anyways, we should take the derivative and then square it.
#(dy)/(dx) = sec^2x + 2sinxcosx#
#=> ((dy)/(dx))^2 = sec^4x + 4sinxcosxsec^2x + 4sin^2xcos^2x#
So, the surface area integral becomes:
#S = 2pi int_0^(pi/4) (tanx - cos^2x)sqrt(1 + sec^4x + 4tanx + sin^2 2x) \ dx#
This is evidently a time sink to solve, so I will just plug it into Wolfram Alpha to evaluate like that. We then get:
#color(blue)(S ~~ 1.483pi)# #color(blue)("u"^2)#
