volume of a solid when the graph of $f(x)= \sec^2x$

Question

Find the volume of the solid of revolution obtained when the graph of $f(x)= \sec^2x$, from $x= -\frac{\pi}{4}$ to $x= \frac{\pi}{3}$, is rotated about the x-axis. Give your answer to four decimal places. I think I maybe getting confused as when to changed the limits when using u sub, I have rearranged $(\sec^2x)^2$ to $(\sec^2x)(1+\tan^2x)$ before expanding the brackets and integ.rating

Answer 1

So behind the integral sign you end up with $sec^4x$ and that indeed can be written as $sec^2x(1+tan^2x)$ Now distribute. The first anti derivative is of course just $tanx$ and the second one is a $tan^2x$ multiplied by its own derivative, so anti deriving that gives $\frac{tan^3x}{3}$ Now plug in upper and lower limits. Can you handle it from here?