Asked by sk
Find the volume generated by rotating the area between
y = cos( 3 x )
and the x axis from x = 0 to x = π/ 12 around the x axis
y = cos( 3 x )
and the x axis from x = 0 to x = π/ 12 around the x axis
Answers
Answered by
Reiny
V = π∫cos(3x) dx from 0 to π/12
= π [ (1/3)sin(3x) ] from 0 to π/12
= π ( (1/3)sin(π/4) - (1/3)sin0 )
=(1/3) π (√2/2 - 0)
= √2 π/6
= π [ (1/3)sin(3x) ] from 0 to π/12
= π ( (1/3)sin(π/4) - (1/3)sin0 )
=(1/3) π (√2/2 - 0)
= √2 π/6
Answered by
Reiny
scrap that previous answer, I forgot to square it
V = π∫(cos(3x))^2 dx
We know
2cos^2 (3x) - 1= cos(6x)
cos^2 (3x) = 1/2 + (1/2)cos(6x)
then
V = π∫(1/2 + (1/2)cos(6x) ) form 0 to π/12
= π { x/2 + (1/12)sin(6x) ] from 0 to π/12
= π( π/24 + (1/12)sin(π/2) - 0 - 0 )
= π( π/24 + 0 )
= π^2 / 24
check my arithmetic
V = π∫(cos(3x))^2 dx
We know
2cos^2 (3x) - 1= cos(6x)
cos^2 (3x) = 1/2 + (1/2)cos(6x)
then
V = π∫(1/2 + (1/2)cos(6x) ) form 0 to π/12
= π { x/2 + (1/12)sin(6x) ] from 0 to π/12
= π( π/24 + (1/12)sin(π/2) - 0 - 0 )
= π( π/24 + 0 )
= π^2 / 24
check my arithmetic
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