# Find the Region enclosed

CellCoree
Find the area of the region enclosed between $$y=3sin(x)$$ and $$y=4cos(x)$$ from $$x=0$$ to $$x=0.9\pi$$ .

does that mean the integral would like this: $$\int 4cos(x) - 3sin(x)$$ (a=0, b = 0.9*pi) is that correct?

Homework Helper
The cosine is larger than the sine at x=0 and the sine is larger than the cosine at $0.9\pi$ so they cross each other somewhere in that interval.
Since area is always positive you should split up your integral, and find at which point they intersect.
A picture may help.

CellCoree
Galileo said:
The cosine is larger than the sine at x=0 and the sine is larger than the cosine at $0.9\pi$ so they cross each other somewhere in that interval.
Since area is always positive you should split up your integral, and find at which point they intersect.
A picture may help.

ah, 0.9pi is equal to 162 degrees right? i graphed it using my calculator. don't know why, but i thought of 0.9pi as 2.872433.

so would my equation look like this...

$$\int 4cos(x) -3sin(x) (a=0,b=.295\pi)$$ + $$\int 3sin(x) - 4cos(x) (a=.295\pi,b=9\pi)$$ is that correct?

$$\int_0^{0.295\pi}(4\cos x-3\sin x)dx+\int_{0.295\pi}^{0.9\pi}(3\sin x-4\cos x)dx$$