# Area under curve

1. Jan 12, 2013

### bonfire09

1. The problem statement, all variables and given/known data
Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.

2. Relevant equations

The sketch of the curve is given too which I uploaded.

3. The attempt at a solution
Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)
0=2sin(6x)-1
sin(6x)=1/2
6x=sin^-1(1/2)
x=(sin^-1(1/2))/6

Then Im not sure what to do from here.

2. Jan 12, 2013

### HallsofIvy

Staff Emeritus
Good up to here.

either 2sin(6x)- 1= 0 or cos(x)= 0

Have you never learned that $sin(\pi/6)= 1/2$?

Draw an equilateral triangle with side length s. All three angles have measure $\pi/3$. Draw a perpendicular from one vertex to the opposite side forming a right triangle. That perpendicular bisects the angle and the opposite side. The angle at that vertex is $(\pi/3)/2= \pi/6$. The hypotenuse is a side of the original triangle and has length s. The "opposite side" has length s/2 so $sin(\pi/6)= (s/2)/s= 1/2$.

$6x= \pi/6$ so $x= \pi/36$

Of course, cos(x)= 0 for $x= \pi/2$.

Since sine and cosine are periodic, there are many solutions to these equations. Your graph should show which one you need to use.

3. Jan 12, 2013

### haruspex

Area or areas?
Where?
There's another solution, but I'll assume you already covered this in your sketch.
What angle has sine equal to 0.5?
Assuming the idea is to find the total area enclosed, treating each individual area as positive, you need to break the range of integration at each intersection. (If you don't do that some areas will come out negative and cancel against others.)
Within each subrange, figure out which function is the greater. If in the range [a,b] f(x)>g(x) then you integrate f-g over that range.