Find the area under one arch of curve y=cos(x/4). This lesson is plane areas in rectangular coordinates. I don't know how to solve it. Thanks for your help.

Answer 1

First, we get the boundaries of the arch. The boundaries of the arch has y-coordinate of zero (when it crosses the x-axis), thus,
0 = cos(x/4)
cos^-1 (0) = x/4
π/2 = x/4
x = 2π

-π/2 = x/4
x = -2π

Thus, boundary is from -2π to 2π. Actually, there are several boundaries (like from 6π to 10π) if you plot y=cos(x/4), but here, lets just use -2π to 2π.

Then set-up the integral. Imagine vertical strips that fill the region inside the arch, that goes from -2π to 2π. Thus,
∫ y dx , from -2π to 2π
∫ cos(x/4) , from -2π to 2π
= 4 sin(x/4) , from -2π to 2π
= 4 sin(2π/4) - 4 sin(-2π/4)
= 4(1) - 4(-1)
= 4 + 4
= 8

You can also do the same for boundary from 6π to 10π, you should get the same answer.
hope this helps~ `u`