y = 5 cos(3x), y = 5 sin(6x), x = 0, x = π/6

Question

y = 5 cos(3x),    y = 5 sin(6x),    x = 0,    x = π/6

Find the area:_______

Answer 1

Let me build on Patrick's work and explain why he does what he does. I am assuming that his math his correct. There is no reason for him to talk about limits, though. What he is doing is integrating in two parts because, for part of the interval, the first curve is above the second, and for part is below.

If we subtract curve 2 from curve 1,we get

5 cos(3x) - 5 sin(6x) =

5 cos(3x) - 10 sin(3x)cos(3x) =

5 cos(3x)(1 - 2 sin(3x))

Thus, points where cos(3x) = 0 and sin(3x) = 1/2 will determine where the first curve is above the second and where the second is above the first. As sin (x) = 1/2 at π/6 and 5π/6, and cos(x) = 0 at π/2 and 3π/2,

zeroes for this function will be at (remember to divide both values and period 2π by 3)

π/18 + 2πn/3, π/6 + 2πn/3, 5π/18 + 2πn/3, π/2 + 2πn/3, n an integer.

Then, the interval we are looking at, the only zero in the interior is at π/18. There will also be a 0 at the endpoint π/6/, but that does not affect things.

This is why we separate the integrals as we do.

Answer 2

integral of y1=cos(3x) is (1/3) sin(3x)

integral of y2=sin(6x) is (-1/6) cos(6x)

from 0 to pi/18, integral of y1-y2 is (1/3)sin(3x) + (1/6)cos(6x)

as x-->0 the limit is 1/6; as x--> pi/18 the limit is 1/6+ sqrt(3)/12

subtracting: (sqrt(3)+2)/12 - 2/12 = sqrt(3)/12

from pi/18 to pi/6 of y2-y1 is (-1/6) cos(6x) - (1/3)sin(3x)

as x-->pi/6, the limit is 1/6 - 1/3 = -1/6

as x--> pi/18, the limit is -sqrt(3)/2 - 1/2 = (-sqrt(3)-1)/2

subtracting: -sqrt(3)-1)/2 + 1/6 = (-3sqrt(3)-3 +1 )/6

Adding these limits: (-6sqrt(3) + 2 + sqrt(3))/12 = 2 -5sqrt(3)/12