Question

What is the surface area produced by rotating `f(x)=tanx-cos^2x, x in [0,pi/4]` around the x-axis?

Answer 1

About `1.483pi` `"u"^2`...

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For revolutions around the `x`-axis, 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)`