# 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 **surface area** is given by:

#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)#