Find the volume of the solid obtained by rotating the region in the first quadrant enclosed by the curves y= tanx and y=2cosx between the bounds 0 and pi/2.

Question

I need the setup.

Answer 1

The graph looks like this, and I see 2 different regions
in quadrant I bounded by the two curves.

www.wolframalpha.com/input/?i=plot+y+%3D+tanx+,+y+%3D+2cosx+from+0+to+%CF%80%2F2

- which one do you want
- you did not specify the line of rotation, e.g. the x-axis? the y-axis?

You will need their intersection
tanx = 2cosx
sinx/cosx = 2cosx
sinx = 2cos^2 x
sinx = 2(1 - sin^2 x)
sinx = 2 - 2sin^2 x
2sin^2 x + sinx - 2 = 0
sinx = (-1 ± √17)/4 = .78078 or -1.28... <---- not in quad I

This should get you going, once you decide what your rotation is

Answer 2

its around y=-1
then what do i do?

Answer 3

would it be

A= pi * ∫ 0 to 0.78 {(2cosx+1)^2 - (tanx+1)^2}