# C4 integrationWatch

#1
I did the integral of (tan2x)^2, taking the limits to be pi/8 and 0. I got an answer of 0.5pi - pi^2/8 which I'm very sure in correct. I subtracted this answer from pi^2/8 (which is the volume of the cylinder) but still got the incorrect answer.

The mark scheme takes the integral of tan2x -1 instead and I'm not really sure why

0
1 year ago
#2
(Original post by Keira Larkin)
I did the integral of (tan2x)^2, taking the limits to be pi/8 and 0. I got an answer of 0.5pi - pi^2/8 which I'm very sure in correct. I subtracted this answer from pi^2/8 (which is the volume of the cylinder) but still got the incorrect answer.

The mark scheme takes the integral of tan2x -1 instead and I'm not really sure why

The volume of revolution is formed by rotating the area about the line y=1, NOT the x-axis (y=0).

To do this, them shift the entire graph down by 1, so you have tan2x -1, and THEN you can rotate it about the line y=0, i.e. the x-axis.
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#3
(Original post by ghostwalker)
The volume of revolution is formed by rotating the area about the line y=1, NOT the x-axis (y=0).

To do this, them shift the entire graph down by 1, so you have tan2x -1, and THEN you can rotate it about the line y=0, i.e. the x-axis.
Thank you!
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