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 $sec^2(x)dx$

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.