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For example: The curve $$y= 0.001sec(x) $$ is rotated 2π rad about the x axis to form a soild.

Calculate the mean cross sectional area of this solid in the range $[0,\pi/4]$

I am having great trouble in figuring out a method to solve this questions. Any help will be great.

(Also I have though about using the mean value of $sec(x)$ with in the range but most likely that will not work as the cross sectional area will be $sec^2(x)dx$

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  • $\begingroup$ $\sec^2$ has an elementary antiderivative, so what's the issue? $\endgroup$ – Gae. S. Aug 13 '19 at 11:07
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A method:

The volume is given by $$\int_0^{π/4}{πy^2\mathrm d x},$$ so that the mean area $A_m$ satisfies the equation $$A_m\int_0^{π/4}{\mathrm d x}=\int_0^{π/4}{πy^2\mathrm d x}.$$

You should be able to do something now.

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