What is the surface area produced by rotating $f (x) = tan x - {cos}^{2} x, x \in [0, \frac{π}{4}]$ $f(x)=tanx-cos^2x, x in [0,pi/4]$ around the x-axis?

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What is the surface area produced by rotating $f (x) = tan x - {cos}^{2} x, x \in [0, \frac{π}{4}]$ $f(x)=tanx-cos^2x, x in [0,pi/4]$ around the x-axis?

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Answer 1 · 2017-06-09T01:28:58

About $1.483 π$ $1.483pi$ $u^{2}$ $"u"^2$ ...

For revolutions around the $x$ $x$ -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:

$color(blue)(S ~~ 1.483pi)$ $color(blue)("u"^2)$

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What is the surface area produced by rotating $f (x) = tan x - {cos}^{2} x, x \in [0, \frac{π}{4}]$ $f(x)=tanx-cos^2x, x in [0,pi/4]$ around the x-axis?

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What is the surface area produced by rotating f(x)=tanx-cos^2x, x in [0,pi/4]f(x)=tanx−cos2x,x∈[0,π4]f(x)=tanx-cos^2x, x in [0,pi/4] around the x-axis?

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What is the surface area produced by rotating $f (x) = tan x - {cos}^{2} x, x \in [0, \frac{π}{4}]$ $f(x)=tanx-cos^2x, x in [0,pi/4]$ around the x-axis?