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

Jun 9, 2017

About $1.483 \pi$ ${\text{u}}^{2}$...

For revolutions around the $x$-axis, the surface area is given by:

$S = 2 \pi {\int}_{\alpha}^{\beta} f \left(x\right) \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx}$

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.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x + 2 \sin x \cos x$

$\implies {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\sec}^{4} x + 4 \sin x \cos x {\sec}^{2} x + 4 {\sin}^{2} x {\cos}^{2} x$

So, the surface area integral becomes:

$S = 2 \pi {\int}_{0}^{\frac{\pi}{4}} \left(\tan x - {\cos}^{2} x\right) \sqrt{1 + {\sec}^{4} x + 4 \tan x + {\sin}^{2} 2 x} \setminus \mathrm{dx}$

This is evidently a time sink to solve, so I will just plug it into Wolfram Alpha to evaluate like that. We then get:

$\textcolor{b l u e}{S \approx 1.483 \pi}$ $\textcolor{b l u e}{{\text{u}}^{2}}$