Question

Question: Find the volume of revolution bounded by y-axis, y=cos(x), and y=sin(x) about the horizontal axis.

Since the rotation is happening at the horizontal axis, I thought the limits of integration would be [-1,1] and if the curves are rotated they would over lap so i used the shell method, thus I got v = ∫[-1,1] 2πy((arccos(y) - arcsin(y)) dy, I didn't get an answer in my calculator. Would doing Disk/Washer method work? I'm not sure if my limits of integration is wrong as well

Answer 1

It is true that both sinx and cosx are bounded by -1 and 1, but the curves intersect at x = π/4

So, the volume, using discs of thickness dx is

v = ∫[0,π/4] π(R^2-r^2) dx
where R=cosx and r=sinx
v = ∫[0,π/4] π(cos^2x-sin^2x) dx
= ∫[0,π/4] πcos(2x) dx = π/2

Using shells of thickness dy, the area has to be split where the curves intersect, so

v = ∫[0,1] 2πrh dy
where r=y and h changes from arcsin(y) to arccos(y) at y = 1/√2

v = ∫[0,1/√2] 2πy*arcsin(y) dy
+ ∫[1/√2,1] 2πy*arccos(y) dy
= π/4 + π/4 = π/2

discs is clearly much easier, since integrating arcsin needs some work.