# The mean cross sectional area of a volume of revolution

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 $$sec^2(x)dx$$

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

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