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

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.

So, the surface area integral becomes:

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

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To find the surface area produced by rotating the function ( f(x) = \tan(x) - \cos^2(x) ) around the x-axis over the interval ( x \in [0, \frac{\pi}{4}] ), we use the formula for surface area of revolution:

[ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

Where:

- ( y = f(x) )
- ( \frac{dy}{dx} ) is the derivative of ( f(x) )

First, we find the derivative of ( f(x) ):

[ f'(x) = \sec^2(x) + 2\cos(x)\sin(x) ]

Next, we plug in the expressions for ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the given interval:

[ S = \int_{0}^{\frac{\pi}{4}} 2\pi (\tan(x) - \cos^2(x)) \sqrt{1 + (\sec^2(x) + 2\cos(x)\sin(x))^2} , dx ]

This integral represents the surface area of revolution for the function over the specified interval. Evaluating this integral will give the surface area.

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