help! trig integration! (1 Viewer)

[r.haidar]

the arc of the curve y = cos2x between x=0 and x= pi/6 is rotated 360 degrees about the x-axis.

find the volume of the solid formed...

when i did it i got:

pi/2 x (integral between 0 and pi/6) cos4x + 1 dx

answers are diff... worked solutions anyone?

 

[damo676767]

the arc of the curve y = cos2x between x=0 and x= pi/6 is rotated 360 degrees about the x-axis.

find the volume of the solid formed...

when i did it i got:

pi/2 x (integral between 0 and pi/6) cos4x + 1 dx

answers are diff... worked solutions anyone?

A = pi I(pi/6 --> 0) cos²2x dx
= pi/2 I(pi/6 --> 0) (cos 4x + 1) dx
= pi/2 [(sin 4x)/4 + x] (pi/6 --> 0)
= pi/2 (root3 / 8 + pi/6)
= pi root3 / 16 + pi²/12

 

[r.haidar]

thanks mate, the answers have typos in them that stuffed me up lol

cheers!